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space isomorphism. If {v1,..., vn} is a basis of V, show that {T (v1),..., T (vn)} is a basis of W. Conclude that any vector space over a ﬁeld F of dimension n is isomorphic to F n. 17. Direct Sums. Let U and V be subspaces of a vector space W. The sum of U and V, denoted U + V, is deﬁned to be the set of all vectors of the form u + v, where u ∈ U and v ∈ V. (a) Prove that U + V and U ∩ V are subspaces of W. (b) If U + V = W and U ∩ V = 0, then W is said to be the direct sum of U and V and we write W = U ⊕ V. Show that every element w ∈ W can be written uniquely as w = u + v, where u ∈ U and v ∈ V. (c) Let U be a subspace of dimension k of a vector space W of dimension n. Prove that there exists a subspace V of dimension n − k such that W = U ⊕ V. Is the subspace V unique? (d) If U and V are arbitrary subspaces of a vector space W, show that dim(U + V ) = dim U + dim V − dim(U ∩ V ). 18. Dual Spaces. Let V and W be ﬁnite dimensional vector spaces over a ﬁeld F. (a) Show that the set of all linear transformations from V into W, denoted by Hom(V, W ), is a vector space over F, where we deﬁne vector addition as follows: (S + T )(v) = S(v) + T (v) (αS)(v) = αS(v), where S, T ∈ Hom(V, W ), α ∈ F, and v ∈ V. (b) Let V be an F -vector space. Deﬁne the dual space of V to be V ∗ = Hom(V, F ). Elements in the dual space of V are called linear functionals. Let v1,..., vn be an ordered basis for V. If v = α1
v1 + · · · + αnvn is any vector in V, deﬁne a linear functional φi : V → F by φi(v) = αi. Show that the φi’s form a basis for V ∗. This basis is called the dual basis of v1,..., vn (or simply the dual basis if the context makes the meaning clear). (c) Consider the basis {(3, 1), (2, −2)} for R2. What is the dual basis for (R2)∗? (d) Let V be a vector space of dimension n over a ﬁeld F and let V ∗∗ be the dual space V ∗. Show that each element v ∈ V gives rise to an element λv in V ∗∗ and that the map v → λv is an isomorphism of V with V ∗∗. EXERCISES 333 References and Suggested Readings [1] Beezer, R. A First Course in Linear Algebra. Available online at http://linear.ups.edu/. 2004. [2] Bretscher, O. Linear Algebra with Applications. 4th ed. Pearson, Upper Saddle River, NJ, 2009. [3] Curtis, C. W. Linear Algebra: An Introductory Approach. 4th ed. Springer, New York, 1984. [4] Hoﬀman, K. and Kunze, R. Linear Algebra. 2nd ed. Prentice-Hall, Englewood Cliﬀs, NJ, 1971. [5] Johnson, L. W., Riess, R. D., and Arnold, J. T. Introduction to Linear Algebra. 6th ed. Pearson, Upper Saddle River, NJ, 2011. [6] Leon, S. J. Linear Algebra with Applications. 8th ed. Pearson, Upper Saddle River, NJ, 2010. Sage Many of Sage’s computations, in a wide variety of algebraic settings, come from solving problems in linear algebra. So you will ﬁnd a wealth of linear algebra functionality. Further, you can use structures such as ﬁnite ﬁelds, to ﬁnd vector spaces in new settings. 21 Fields It is natural to ask whether or not some ﬁeld
F is contained in a larger ﬁeld. We think of the rational numbers, which reside inside the real numbers, while in turn, the real numbers live inside the complex numbers. We can also study the ﬁelds between Q and R and inquire as to the nature of these ﬁelds. More speciﬁcally if we are given a ﬁeld F and a polynomial p(x) ∈ F [x], we can ask whether or not we can ﬁnd a ﬁeld E containing F such that p(x) factors into linear factors over E[x]. For example, if we consider the polynomial p(x) = x4 − 5x2 + 6 in Q[x], then p(x) factors as (x2 − 2)(x2 − 3). However, both of these factors are irreducible in Q[x]. If we wish to ﬁnd a zero of p(x), we must go to a larger ﬁeld. Certainly the ﬁeld of real numbers will work, since p(x) = (x − √ 2)(x + √ 2)(x − √ 3)(x + √ 3). It is possible to ﬁnd a smaller ﬁeld in which p(x) has a zero, namely √ Q( √ 2) = {a + b 2 : a, b ∈ Q}. We wish to be able to compute and study such ﬁelds for arbitrary polynomials over a ﬁeld F. 21.1 Extension Fields A ﬁeld E is an extension ﬁeld of a ﬁeld F if F is a subﬁeld of E. The ﬁeld F is called the base ﬁeld. We write F ⊂ E. Example 1. For example, let √ F = Q( 2 ) = {a + b 2 : a, b ∈ Q} √ 334 21.1 EXTENSION FIELDS 335 √ √ 2+ 3 ) be the smallest ﬁeld containing both Q and and let E = Q( 3. Both E and F are extension ﬁelds of the rational numbers. We claim that E is an extension
ﬁeld of F. To see this, we need only show that 2 is in E. 2 must also be in E. Taking Since 3 must linear combinations of both be in E. 3 is in E, 1/( 2 + 2, we ﬁnd that 3 ) = √ 2 + √ 2 and 3 and 3 − √ 2 + 3 − 2+ √ √ √ √ √ √ √ √ √ √ √ √ Example 2. Let p(x) = x2 + x + 1 ∈ Z2[x]. Since neither 0 nor 1 is a root of this polynomial, we know that p(x) is irreducible over Z2. We will construct a ﬁeld extension of Z2 containing an element α such that p(α) = 0. By Theorem 17.13, the ideal p(x) generated by p(x) is maximal; hence, Z2[x]/p(x) is a ﬁeld. Let f (x) + p(x) be an arbitrary element of Z2[x]/p(x). By the division algorithm, f (x) = (x2 + x + 1)q(x) + r(x), where the degree of r(x) is less than the degree of x2 + x + 1. Therefore, f (x) + x2 + x + 1 = r(x) + x2 + x + 1. The only possibilities for r(x) are then 0, 1, x, and 1 + x. Consequently, E = Z2[x]/x2 + x + 1 is a ﬁeld with four elements and must be a ﬁeld extension of Z2, containing a zero α of p(x). The ﬁeld Z2(α) consists of elements 0 + 0α = 0 1 + 0α = 1 0 + 1α = α 1 + 1α = 1 + α. Notice that α2 + α + 1 = 0; hence, if we compute (1 + α)2, (1 + α)(1 + α) = 1 + α + α + (α)2 = α. Other calculations are accomplished in a similar manner. We summarize these computations in the following tables, which tell us how to add and multiply elements in E
336 CHAPTER 21 FIELDS · The following theorem, due to Kronecker, is so important and so basic to our understanding of ﬁelds that it is often known as the Fundamental Theorem of Field Theory. Theorem 21.1 Let F be a ﬁeld and let p(x) be a nonconstant polynomial in F [x]. Then there exists an extension ﬁeld E of F and an element α ∈ E such that p(α) = 0. Proof. To prove this theorem, we will employ the method that we used to construct Example 2. Clearly, we can assume that p(x) is an irreducible polynomial. We wish to ﬁnd an extension ﬁeld E of F containing an element α such that p(α) = 0. The ideal p(x) generated by p(x) is a maximal ideal in F [x] by Theorem 17.13; hence, F [x]/p(x) is a ﬁeld. We claim that E = F [x]/p(x) is the desired ﬁeld. We ﬁrst show that E is a ﬁeld extension of F. We can deﬁne a homomorphism of commutative rings by the map ψ : F → F [x]/p(x), where ψ(a) = a + p(x) for a ∈ F. It is easy to check that ψ is indeed a ring homomorphism. Observe that ψ(a) + ψ(b) = (a + p(x)) + (b + p(x)) = (a + b) + p(x) = ψ(a + b) and ψ(a)ψ(b) = (a + p(x))(b + p(x)) = ab + p(x) = ψ(ab). To prove that ψ is one-to-one, assume that a + p(x) = ψ(a) = ψ(b) = b + p(x). Then a−b is a multiple of p(x), since it lives in the ideal p(x). Since p(x) is a nonconstant polynomial, the only possibility is that a − b = 0
. Consequently, a = b and ψ is injective. Since ψ is one-to-one, we can identify F with the subﬁeld {a + p(x) : a ∈ F } of E and view E as an extension ﬁeld of F. 21.1 EXTENSION FIELDS 337 It remains for us to prove that p(x) has a zero α ∈ E. Set α = x + p(x). Then α is in E. If p(x) = a0 + a1x + · · · + anxn, then p(α) = a0 + a1(x + p(x)) + · · · + an(x + p(x))n = a0 + (a1x + p(x)) + · · · + (anxn + p(x)) = a0 + a1x + · · · + anxn + p(x) = 0 + p(x). Therefore, we have found an element α ∈ E = F [x]/p(x) such that α is a zero of p(x). Example 3. Let p(x) = x5 + x4 + 1 ∈ Z2[x]. Then p(x) has irreducible factors x2 + x + 1 and x3 + x + 1. For a ﬁeld extension E of Z2 such that p(x) has a root in E, we can let E be either Z2[x]/x2 +x+1 or Z2[x]/x3 +x+1. We will leave it as an exercise to show that Z2[x]/x3 + x + 1 is a ﬁeld with 23 = 8 elements. Algebraic Elements An element α in an extension ﬁeld E over F is algebraic over F if f (α) = 0 for some nonzero polynomial f (x) ∈ F [x]. An element in E that is not algebraic over F is transcendental over F. An extension ﬁeld E of a ﬁeld F is an algebraic extension of F if every element in E is algebraic over F. If E is a ﬁeld extension of F and α1,..., αn are contained in E, we denote
the smallest ﬁeld containing F and α1,..., αn by F (α1,..., αn). If E = F (α) for some α ∈ E, then E is a simple extension of F. √ 2 and i are algebraic over Q since they are zeros of the Example 4. Both polynomials x2 − 2 and x2 +1, respectively. Clearly π and e are algebraic over the real numbers; however, it is a nontrivial fact that they are transcendental over Q. Numbers in R that are algebraic over Q are in fact quite rare. Almost all real numbers are transcendental over Q.1 (In many cases we do not know whether or not a particular number is transcendental; for example, it is not known whether π + e is transcendental or algebraic.) A complex number that is algebraic over Q is an algebraic number. A transcendental number is an element of C that is transcendental over Q. √ Example 5. We will show that 3, then α2 = 2 + 2 + 3. Hence, α2 − 2 = 2 + √ √ 3 is algebraic over Q. If α = 3 and (α2 − 2)2 = 3. √ 1If we choose a number in R, then there is a probability of 1 that the number will be transcendental over Q. 338 CHAPTER 21 FIELDS Since α4 − 4α2 + 1 = 0, it must be true that α is a zero of the polynomial x4 − 4x2 + 1 ∈ Q[x]. It is very easy to give an example of an extension ﬁeld E over a ﬁeld F, where E contains an element transcendental over F. The following theorem characterizes transcendental extensions. Theorem 21.2 Let E be an extension ﬁeld of F and α ∈ E. Then α is transcendental over F if and only if F (α) is isomorphic to F (x), the ﬁeld of fractions of F [x]. Proof. Let φα : F [x] → E be the evaluation homomorphism for α. Then α is transcendental over F if and only if φα(p(x)) = p(α) = 0 for all nonconstant polynomials p(x
) ∈ F [x]. This is true if and only if ker φα = {0}; that is, it is true exactly when φα is one-to-one. Hence, E must contain a copy of F [x]. The smallest ﬁeld containing F [x] is the ﬁeld of fractions F (x). By Theorem 18.4, E must contain a copy of this ﬁeld. We have a more interesting situation in the case of algebraic extensions. Theorem 21.3 Let E be an extension ﬁeld of a ﬁeld F and α ∈ E with α algebraic over F. Then there is a unique irreducible monic polynomial p(x) ∈ F [x] of smallest degree such that p(α) = 0. If f (x) is another monic polynomial in F [x] such that f (α) = 0, then p(x) divides f (x). Proof. Let φα : F [x] → E be the evaluation homomorphism. The kernel of φα is a principal ideal generated by some p(x) ∈ F [x] with deg p(x) ≥ 1. We know that such a polynomial exists, since F [x] is a principal ideal domain and α is algebraic. The ideal p(x) consists exactly of those elements of F [x] having α as a zero. If f (α) = 0 and f (x) is not the zero polynomial, then f (x) ∈ p(x) and p(x) divides f (x). So p(x) is a polynomial of minimal degree having α as a zero. Any other polynomial of the same degree having α as a zero must have the form βp(x) for some β ∈ F. Suppose now that p(x) = r(x)s(x) is a factorization of p into polynomials of lower degree. Since p(α) = 0, r(α)s(α) = 0; consequently, either r(α) = 0 or s(α) = 0, which contradicts the fact that p is of minimal degree. Therefore, p(x) must be irreducible. Let E be an extension ﬁeld of F
and α ∈ E be algebraic over F. The unique monic polynomial p(x) of the last theorem is called the minimal polynomial for α over F. The degree of p(x) is the degree of α over F. Example 6. Let f (x) = x2 − 2 and g(x) = x4 − 4x2 + 1. These polynomials 2 and are the minimal polynomials of 3, respectively. 2 + √ √ 21.1 EXTENSION FIELDS 339 Proposition 21.4 Let E be a ﬁeld extension of F and α ∈ E be algebraic over F. Then F (α) ∼= F [x]/p(x), where p(x) is the minimal polynomial of α over F. Proof. Let φα : F [x] → E be the evaluation homomorphism. The kernel of this map is p(x), where p(x) is the minimal polynomial of α. By the First Isomorphism Theorem for rings, the image of φα in E is isomorphic to F (α) since it contains both F and α. Theorem 21.5 Let E = F (α) be a simple extension of F, where α ∈ E is algebraic over F. Suppose that the degree of α over F is n. Then every element β ∈ E can be expressed uniquely in the form β = b0 + b1α + · · · + bn−1αn−1 for bi ∈ F. Proof. Since φα(F [x]) ∼= F (α), every element in E = F (α) must be of the form φα(f (x)) = f (α), where f (α) is a polynomial in α with coeﬃcients in F. Let p(x) = xn + an−1xn−1 + · · · + a0 be the minimal polynomial of α. Then p(α) = 0; hence, αn = −an−1αn−1 − · · · − a0. Similarly, αn+1 = ααn = −an−1αn − an−2αn−1 − · · · − a0α = −an−1(−an−1αn−1 −
· · · − a0) − an−2αn−1 − · · · − a0α. Continuing in this manner, we can express every monomial αm, m ≥ n, as a linear combination of powers of α that are less than n. Hence, any β ∈ F (α) can be written as β = b0 + b1α + · · · + bn−1αn−1. To show uniqueness, suppose that β = b0 + b1α + · · · + bn−1αn−1 = c0 + c1α + · · · + cn−1αn−1 for bi and ci in F. Then g(x) = (b0 − c0) + (b1 − c1)x + · · · + (bn−1 − cn−1)xn−1 340 CHAPTER 21 FIELDS is in F [x] and g(α) = 0. Since the degree of g(x) is less than the degree of p(x), the irreducible polynomial of α, g(x) must be the zero polynomial. Consequently, b0 − c0 = b1 − c1 = · · · = bn−1 − cn−1 = 0, or bi = ci for i = 0, 1,..., n − 1. Therefore, we have shown uniqueness. Example 7. Since x2 + 1 is irreducible over R, x2 + 1 is a maximal ideal in R[x]. So E = R[x]/x2 + 1 is a ﬁeld extension of R that contains a root of x2 + 1. Let α = x + x2 + 1. We can identify E with the complex numbers. By Proposition 21.4, E is isomorphic to R(α) = {a + bα : a, b ∈ R}. We know that α2 = −1 in E, since α2 + 1 = (x + x2 + 1)2 + (1 + x2 + 1) = (x2 + 1) + x2 + 1 = 0. Hence, we have an isomorphism of R(α) with C deﬁned by the map that takes a + bα to a + bi. Let E be a ﬁeld extension of
a ﬁeld F. If we regard E as a vector space over F, then we can bring the machinery of linear algebra to bear on the problems that we will encounter in our study of ﬁelds. The elements in the ﬁeld E are vectors; the elements in the ﬁeld F are scalars. We can think of addition in E as adding vectors. When we multiply an element in E by an element of F, we are multiplying a vector by a scalar. This view of ﬁeld extensions is especially fruitful if a ﬁeld extension E of F is a ﬁnite dimensional vector space over F, and Theorem 21.5 states that E = F (α) is ﬁnite dimensional vector space over F with basis {1, α, α2,..., αn−1}. If an extension ﬁeld E of a ﬁeld F is a ﬁnite dimensional vector space over F of dimension n, then we say that E is a ﬁnite extension of degree n over F. We write to indicate the dimension of E over F. [E : F ] = n. Theorem 21.6 Every ﬁnite extension ﬁeld E of a ﬁeld F is an algebraic extension. Proof. Let α ∈ E. Since [E : F ] = n, the elements 1, α,..., αn 21.1 EXTENSION FIELDS 341 cannot be linearly independent. Hence, there exist ai ∈ F, not all zero, such that anαn + an−1αn−1 + · · · + a1α + a0 = 0. Therefore, p(x) = anxn + · · · + a0 ∈ F [x] is a nonzero polynomial with p(α) = 0. Remark. Theorem 21.6 says that every ﬁnite extension of a ﬁeld F is an algebraic extension. The converse is false, however. We will leave it as an exercise to show that the set of all elements in R that are algebraic over Q forms an inﬁnite ﬁeld extension of Q. The next theorem is a counting theorem, similar to Lagrange’s Theorem
in group theory. Theorem 21.6 will prove to be an extremely useful tool in our investigation of ﬁnite ﬁeld extensions. Theorem 21.7 If E is a ﬁnite extension of F and K is a ﬁnite extension of E, then K is a ﬁnite extension of F and [K : F ] = [K : E][E : F ]. Proof. Let {α1,..., αn} be a basis for E as a vector space over F and {β1,..., βm} be a basis for K as a vector space over E. We claim that {αiβj} is a basis for K over F. We will ﬁrst show that these vectors span K. Let u ∈ K. Then u = m i=1 aijαi, where bj ∈ E and aij ∈ F. Then j=1 bjβj and bj = n u = aijαi βj = m n j=1 i=1 i,j aij(αiβj). So the mn vectors αiβj must span K over F. We must show that {αiβj} are linearly independent. Recall that a set of vectors v1, v2,..., vn in a vector space V are linearly independent if implies that Let c1v1 + c2v2 + · · · + cnvn = 0 c1 = c2 = · · · = cn = 0. u = i,j cij(αiβj) = 0 342 CHAPTER 21 FIELDS for cij ∈ F. We need to prove that all of the cij’s are zero. We can rewrite u as m n j=1 i=1 cijαi βj = 0, i cijαi ∈ E. Since the βj’s are linearly independent over E, it must where be the case that n i=1 cijαi = 0 for all j. However, the αj are also linearly independent over F. Therefore, cij = 0 for all i and j, which completes the proof. The following corollary is easily proved using mathematical induction. Corollary 21.8 If Fi is a ﬁeld for i = 1,
..., k and Fi+1 is a ﬁnite extension of Fi, then Fk is a ﬁnite extension of F1 and [Fk : F1] = [Fk : Fk−1] · · · [F2 : F1]. Corollary 21.9 Let E be an extension ﬁeld of F. If α ∈ E is algebraic over F with minimal polynomial p(x) and β ∈ F (α) with minimal polynomial q(x), then deg q(x) divides deg p(x). Proof. We know that deg p(x) = [F (α) : F ] and deg q(x) = [F (β) : F ]. Since F ⊂ F (β) ⊂ F (α), [F (α) : F ] = [F (α) : F (β)][F (β) : F ]. Example 8. Let us determine an extension ﬁeld of Q containing is easy to determine that the minimal polynomial of It follows that 3 + √ √ √ √ 5. It 3+ 5 is x4 − 16x2 + 4. √ √ [Q( 3 + 5 ) : Q] = 4. √ √ √ √ 3 } is a basis for Q( 5 We know that {1, cannot be in Q( 3 ) either. There√ 3 ) and fore, {1, √ √ 5 ) over Q. {1, This example shows that it is possible that some extension F (α1,..., αn) is actually a simple extension of F even though n > 1. 3 ). It follows that 3, 15 } is a basis for Q( 5 cannot be in Q( √ √ 5 } is a basis for Q( √ √ 5 ) over Q( √ 3 + 5 ) = (Q( 3, 3 ))( 5 ) = Q( 3 ) over Q. Hence, 3, √ √ 21.1 EXTENSION FIELDS 343 5, 5, √ √ √ √ √ 5 )] = 2. 5 i), where √ 5 i /∈ Q( 3 √ Example 9. Let us compute a basis for Q( 3 √ positive square root of 5 and 3 �
� 5 ), so [Q( 3√ It is easy to determine that {1, √ √ We also know that {1, 3 5, ( 3 √ 5 ) over Q is basis for Q( 5, ( 3√ 5 i, 3√ 5 is the 5 is the real cube root of 5. We know that 5 i) : Q( 3√ √ 5i } is a basis for Q( 3 √ 5 )2} is a basis for Q( 3 √ 5 i) over Q( 3 5, 5 ). 5 ) over Q. Hence, a {1, √ 5 i is a zero of x6 + 5. We can show that this polynomial is Notice that 6 irreducible over Q using Eisenstein’s Criterion, where we let p = 5. Consequently, Q ⊂ Q( 6√ 5 i) ⊂ Q( 3√ √ But it must be the case that Q( 6 both of these extensions is 6. 5 i), since the degree of √ 5 i) = Q( 3 5 )7i = 5 6√ 5 )5i, ( 6√ 5 i or 6√ 5 )2, ( 6√ √ 5, 3 √ 5 i). √ 5 i}. √ 5, 5, Theorem 21.10 Let E be a ﬁeld extension of F. Then the following statements are equivalent. 1. E is a ﬁnite extension of F. 2. There exists a ﬁnite number of algebraic elements α1,..., αn ∈ E such that E = F (α1,..., αn). 3. There exists a sequence of ﬁelds E = F (α1,..., αn) ⊃ F (α1,..., αn−1) ⊃ · · · ⊃ F (α1) ⊃ F, where each ﬁeld F (α1,..., αi) is algebraic over F (α1,..., αi−1). Proof. (1) ⇒ (2). Let E be a ﬁnite algebraic extension of F. Then E is a ﬁnite dimensional vector space over F and
there exists a basis consisting of elements α1,..., αn in E such that E = F (α1,..., αn). Each αi is algebraic over F by Theorem 21.6. (2) ⇒ (3). Suppose that E = F (α1,..., αn), where every αi is algebraic over F. Then E = F (α1,..., αn) ⊃ F (α1,..., αn−1) ⊃ · · · ⊃ F (α1) ⊃ F, where each ﬁeld F (α1,..., αi) is algebraic over F (α1,..., αi−1). 344 CHAPTER 21 FIELDS (3) ⇒ (1). Let E = F (α1,..., αn) ⊃ F (α1,..., αn−1) ⊃ · · · ⊃ F (α1) ⊃ F, where each ﬁeld F (α1,..., αi) is algebraic over F (α1,..., αi−1). Since F (α1,..., αi) = F (α1,..., αi−1)(αi) is simple extension and αi is algebraic over F (α1,..., αi−1), it follows that [F (α1,..., αi) : F (α1,..., αi−1)] is ﬁnite for each i. Therefore, [E : F ] is ﬁnite. Algebraic Closure Given a ﬁeld F, the question arises as to whether or not we can ﬁnd a ﬁeld E such that every polynomial p(x) has a root in E. This leads us to the following theorem. Theorem 21.11 Let E be an extension ﬁeld of F. The set of elements in E that are algebraic over F form a ﬁeld. Proof. Let α, β ∈ E be algebraic over F. Then F (α, β) is a ﬁnite extension
of F. Since every element of F (α, β) is algebraic over F, α ± β, αβ, and α/β (β = 0) are all algebraic over F. Consequently, the set of elements in E that are algebraic over F forms a ﬁeld. Corollary 21.12 The set of all algebraic numbers forms a ﬁeld; that is, the set of all complex numbers that are algebraic over Q makes up a ﬁeld. Let E be a ﬁeld extension of a ﬁeld F. We deﬁne the algebraic closure of a ﬁeld F in E to be the ﬁeld consisting of all elements in E that are algebraic over F. A ﬁeld F is algebraically closed if every nonconstant polynomial in F [x] has a root in F. Theorem 21.13 A ﬁeld F is algebraically closed if and only if every nonconstant polynomial in F [x] factors into linear factors over F [x]. Proof. Let F be an algebraically closed ﬁeld. If p(x) ∈ F [x] is a nonconstant polynomial, then p(x) has a zero in F, say α. Therefore, x − α must be a 21.2 SPLITTING FIELDS 345 factor of p(x) and so p(x) = (x − α)q1(x), where deg q1(x) = deg p(x) − 1. Continue this process with q1(x) to ﬁnd a factorization p(x) = (x − α)(x − β)q2(x), where deg q2(x) = deg p(x) − 2. The process must eventually stop since the degree of p(x) is ﬁnite. Conversely, suppose that every nonconstant polynomial p(x) in F [x] factors into linear factors. Let ax − b be such a factor. Then p(b/a) = 0. Consequently, F is algebraically closed. Corollary 21.14 An algebraically closed ﬁeld F has no proper algebraic extension E. Proof. Let E be an algebraic extension of F ; then F ⊂
E. For α ∈ E, the minimal polynomial of α is x − α. Therefore, α ∈ F and F = E. Theorem 21.15 Every ﬁeld F has a unique algebraic closure. It is a nontrivial fact that every ﬁeld has a unique algebraic closure. The proof is not extremely diﬃcult, but requires some rather sophisticated set theory. We refer the reader to [3], [4], or [8] for a proof of this result. We now state the Fundamental Theorem of Algebra, ﬁrst proven by Gauss at the age of 22 in his doctoral thesis. This theorem states that every polynomial with coeﬃcients in the complex numbers has a root in the complex numbers. The proof of this theorem will be given in Chapter 23. Theorem 21.16 (Fundamental Theorem of Algebra) The ﬁeld of complex numbers is algebraically closed. 21.2 Splitting Fields Let F be a ﬁeld and p(x) be a nonconstant polynomial in F [x]. We already know that we can ﬁnd a ﬁeld extension of F that contains a root of p(x). However, we would like to know whether an extension E of F containing all of the roots of p(x) exists. In other words, can we ﬁnd a ﬁeld extension of F such that p(x) factors into a product of linear polynomials? What is the “smallest” extension containing all the roots of p(x)? Let F be a ﬁeld and p(x) = a0 + a1x + · · · + anxn be a nonconstant polynomial in F [x]. An extension ﬁeld E of F is a splitting ﬁeld of p(x) if there exist elements α1,..., αn in E such that E = F (α1,..., αn) and p(x) = (x − α1)(x − α2) · · · (x − αn). 346 CHAPTER 21 FIELDS A polynomial p(x) ∈ F [x] splits in E if it is the product of linear factors in E[x]. Example 10.
Let p(x) = x4 + 2x2 − 8 be in Q[x]. Then p(x) has irreducible factors x2 − 2 and x2 + 4. Therefore, the ﬁeld Q( 2, i) is a splitting ﬁeld for p(x). √ Example 11. Let p(x) = x3 − 3 be in Q[x]. Then p(x) has a root in the √ ﬁeld Q( 3 3 ). However, this ﬁeld is not a splitting ﬁeld for p(x) since the complex cube roots of 35i, √ are not in Q( 3 3 ). Theorem 21.17 Let p(x) ∈ F [x] be a nonconstant polynomial. Then there exists a splitting ﬁeld E for p(x). Proof. We will use mathematical induction on the degree of p(x). If deg p(x) = 1, then p(x) is a linear polynomial and E = F. Assume that the theorem is true for all polynomials of degree k with 1 ≤ k < n and let deg p(x) = n. We can assume that p(x) is irreducible; otherwise, by our induction hypothesis, we are done. By Theorem 21.1, there exists a ﬁeld K such that p(x) has a zero α1 in K. Hence, p(x) = (x − α1)q(x), where q(x) ∈ K[x]. Since deg q(x) = n − 1, there exists a splitting ﬁeld E ⊃ K of q(x) that contains the zeros α2,..., αn of p(x) by our induction hypothesis. Consequently, E = K(α2,..., αn) = F (α1,..., αn) is a splitting ﬁeld of p(x). The question of uniqueness now arises for splitting ﬁelds. This question is answered in the aﬃrmative. Given two splitting ﬁelds K and L of a polynomial p(x) ∈ F [x], there exists a ﬁeld isomorphism �
� : K → L that preserves F. In order to prove this result, we must ﬁrst prove a lemma. Lemma 21.18 Let φ : E → F be an isomorphism of ﬁelds. Let K be an extension ﬁeld of E and α ∈ K be algebraic over E with minimal polynomial p(x). Suppose that L is an extension ﬁeld of F such that β is root of the polynomial in F [x] obtained from p(x) under the image of φ. Then φ extends to a unique isomorphism ψ : E(α) → F (β) such that ψ(α) = β and ψ agrees with φ on E. 21.2 SPLITTING FIELDS 347 Proof. If p(x) has degree n, then by Theorem 21.5 we can write any element in E(α) as a linear combination of 1, α,..., αn−1. Therefore, the isomorphism that we are seeking must be ψ(a0 + a1α + · · · + an−1αn−1) = φ(a0) + φ(a1)β + · · · + φ(an−1)βn−1, where a0 + a1α + · · · + an−1αn−1 is an element in E(α). The fact that ψ is an isomorphism could be checked by direct computation; however, it is easier to observe that ψ is a composition of maps that we already know to be isomorphisms. We can extend φ to be an isomorphism from E[x] to F [x], which we will also denote by φ, by letting φ(a0 + a1x + · · · + anxn) = φ(a0) + φ(a1)x + · · · + φ(an)xn. This extension agrees with the original isomorphism φ : E → F, since constant polynomials get mapped to constant polynomials. By assumption, φ(p(x)) = q(x); hence, φ maps p(x) onto q(x). Consequently, we have an isomorphism φ : E[x]/ p
(x) → F [x]/ q(x). By Theorem 21.4, we have isomorphisms σ : E[x]/p(x) → F (α) and τ : F [x]/q(x) → F (β), deﬁned by evaluation at α and β, respectively. Therefore, ψ = τ −1φσ is the required isomorphism. E(α) σ E[x]/p(x) E ψ φ φ F (β) τ F [x]/q(x) F We leave the proof of uniqueness as a exercise. Theorem 21.19 Let φ : E → F be an isomorphism of ﬁelds and let p(x) be a nonconstant polynomial in E[x] and q(x) the corresponding polynomial in F [x] under the isomorphism. If K is a splitting ﬁeld of p(x) and L is a splitting ﬁeld of q(x), then φ extends to an isomorphism ψ : K → L. 348 CHAPTER 21 FIELDS Proof. We will use mathematical induction on the degree of p(x). We can assume that p(x) is irreducible over E. Therefore, q(x) is also irreducible over F. If deg p(x) = 1, then by the deﬁnition of a splitting ﬁeld, K = E and L = F and there is nothing to prove. Assume that the theorem holds for all polynomials of degree less than n. Since K is a splitting ﬁeld of E, all of the roots of p(x) are in K. Choose one of these roots, say α, such that E ⊂ E(α) ⊂ K. Similarly, we can ﬁnd a root β of q(x) in L such that F ⊂ F (β) ⊂ L. By Lemma 21.18, there exists an isomorphism φ : E(α) → F (β) such that φ(α) = β and φ agrees with φ on E. ψ φ φ K E(α) E L F (β) F Now write p(x)
= (x − α)f (x) and q(x) = (x − β)g(x), where the degrees of f (x) and g(x) are less than the degrees of p(x) and q(x), respectively. The ﬁeld extension K is a splitting ﬁeld for f (x) over E(α), and L is a splitting ﬁeld for g(x) over F (β). By our induction hypothesis there exists an isomorphism ψ : K → L such that ψ agrees with φ on E(α). Hence, there exists an isomorphism ψ : K → L such that ψ agrees with φ on E. Corollary 21.20 Let p(x) be a polynomial in F [x]. Then there exists a splitting ﬁeld K of p(x) that is unique up to isomorphism. 21.3 Geometric Constructions In ancient Greece, three classic problems were posed. These problems are geometric in nature and involve straightedge-and-compass constructions from what is now high school geometry; that is, we are allowed to use only a straightedge and compass to solve them. The problems can be stated as follows. 1. Given an arbitrary angle, can one trisect the angle into three equal subangles using only a straightedge and compass? 21.3 GEOMETRIC CONSTRUCTIONS 349 2. Given an arbitrary circle, can one construct a square with the same area using only a straightedge and compass? 3. Given a cube, can one construct the edge of another cube having twice the volume of the original? Again, we are only allowed to use a straightedge and compass to do the construction. After puzzling mathematicians for over two thousand years, each of these constructions was ﬁnally shown to be impossible. We will use the theory of ﬁelds to provide a proof that the solutions do not exist. It is quite remarkable that the long-sought solution to each of these three geometric problems came from abstract algebra. First, let us determine more speciﬁcally what we mean by a straightedge and compass, and also examine the nature of these problems in a bit more depth. To begin with, a straightedge is not a ruler. We cannot measure arbitrary lengths with a straightedge. It is merely a tool for drawing a line
through two points. The statement that the trisection of an arbitrary angle is impossible means that there is at least one angle that is impossible to trisect with a straightedge-and-compass construction. Certainly it is possible to trisect an angle in special cases. We can construct a 30◦ angle; hence, it is possible to trisect a 90◦ angle. However, we will show that it is impossible to construct a 20◦ angle. Therefore, we cannot trisect a 60◦ angle. Constructible Numbers A real number α is constructible if we can construct a line segment of length \|α\| in a ﬁnite number of steps from a segment of unit length by using a straightedge and compass. Theorem 21.21 The set of all constructible real numbers forms a subﬁeld F of the ﬁeld of real numbers. Proof. Let α and β be constructible numbers. We must show that α + β, α − β, αβ, and α/β (β = 0) are also constructible numbers. We can assume that both α and β are positive with α > β. It is quite obvious how to construct α + β and α − β. To ﬁnd a line segment with length αβ, we assume that β > 1 and construct the triangle in Figure 21.1 such that triangles ABC and ADE are similar. Since α/1 = x/β, the line segment x has length αβ. A similar construction can be made if β < 1. We will leave it as an exercise to show that the same triangle can be used to construct α/β for β = 0. 350 CHAPTER 21 FIELDS D C β α 1 B x E A Figure 21.1. Construction of products Lemma 21.22 If α is a constructible number, then number. √ α is a constructible Proof. similar; hence, 1/x = x/α, or x2 = α. In Figure 21.2 the triangles ABD, BCD, and ABC are B x 1 α A D C Figure 21.2. Construction of roots By Theorem 21.21, we can locate in the plane any point P = (p, q) that has rational coordinates p and q. We need to know what other points can be constructed with a compass and straightedge from points with rational coordinates. Lemma 21.23 Let F
be a subﬁeld of R. 1. If a line contains two points in F, then it has the equation ax+by+c = 0, where a, b, and c are in F. 21.3 GEOMETRIC CONSTRUCTIONS 351 2. If a circle has a center at a point with coordinates in F and a radius that is also in F, then it has the equation x2 + y2 + dx + ey + f = 0, where d, e, and f are in F. Proof. Let (x1, y1) and (x2, y2) be points on a line whose coordinates are in F. If x1 = x2, then the equation of the line through the two points is x − x1 = 0, which has the form ax + by + c = 0. If x1 = x2, then the equation of the line through the two points is given by y − y1 = y2 − y1 x2 − x1 (x − x1), which can also be put into the proper form. To prove the second part of the lemma, suppose that (x1, y1) is the center of a circle of radius r. Then the circle has the equation (x − x1)2 + (y − y1)2 − r2 = 0. This equation can easily be put into the appropriate form. Starting with a ﬁeld of constructible numbers F, we have three possible ways of constructing additional points in R with a compass and straightedge. 1. To ﬁnd possible new points in R, we can take the intersection of two lines, each of which passes through two known points with coordinates in F. 2. The intersection of a line that passes through two points that have coordinates in F and a circle whose center has coordinates in F with radius of a length in F will give new points in R. 3. We can obtain new points in R by intersecting two circles whose centers have coordinates in F and whose radii are of lengths in F. The ﬁrst case gives no new points in R, since the solution of two equations of the form ax + by + c = 0 having coeﬃcients in F will always be in F. The third case can be reduced to the second case. Let x2 + y2 + d1x + e1y + f1 = 0 x2 + y
2 + d2x + e2y + f2 = 0 be the equations of two circles, where di, ei, and fi are in F for i = 1, 2. These circles have the same intersection as the circle x2 + y2 + d1x + e1x + f1 = 0 352 and the line CHAPTER 21 FIELDS (d1 − d2)x + b(e2 − e1)y + (f2 − f1) = 0. The last equation is that of the chord passing through the intersection points of the two circles. Hence, the intersection of two circles can be reduced to the case of an intersection of a line with a circle. Considering the case of the intersection of a line and a circle, we must determine the nature of the solutions of the equations ax + by + c = 0 x2 + y2 + dx + ey + f = 0. If we eliminate y from these equations, we obtain an equation of the form Ax2 + Bx + C = 0, where A, B, and C are in F. The x coordinate of the intersection points is given by −B ± x = √ B2 − 4AC 2A α ), where α = B2 − 4AC > 0. We have proven the following √ and is in F ( lemma. Lemma 21.24 Let F be a ﬁeld of constructible numbers. Then the points determined by the intersections of lines and circles in F lie in the ﬁeld F ( α ) for some α in F. √ Theorem 21.25 A real number α is a constructible number if and only if there exists a sequence of ﬁelds Q = F0 ⊂ F1 ⊂ · · · ⊂ Fk such that Fi = Fi−1( exists an integer k > 0 such that [Q(α) : Q] = 2k. αi ) with αi ∈ Fi and α ∈ Fk. In particular, there √ Proof. The existence of the Fi’s and the αi’s is a direct consequence of Lemma 21.24 and of the fact that [Fk : Q] = [Fk : Fk−1][Fk−1 : Fk−2] · · · [F1 : Q] = 2k. Corollary 21.26 The ﬁ
eld of all constructible numbers is an algebraic extension of Q. As we can see by the ﬁeld of constructible numbers, not every algebraic extension of a ﬁeld is a ﬁnite extension. 21.3 GEOMETRIC CONSTRUCTIONS 353 Doubling the Cube and Squaring the Circle We are now ready to investigate the classical problems of doubling the cube and squaring the circle. We can use the ﬁeld of constructible numbers to show exactly when a particular geometric construction can be accomplished. Doubling the cube is impossible. Given the edge of the cube, it is impossible to construct with a straightedge and compass the edge of the cube that has twice the volume of the original cube. Let the original cube have an edge of length 1 and, therefore, a volume of 1. If we could construct a cube √ having a volume of 2, then this new cube would have an edge of length 3 2. √ 2 is a zero of the irreducible polynomial x3 − 2 over Q; hence, However, 3 [Q( 3√ 2 ) : Q] = 3 This is impossible, since 3 is not a power of 2. Squaring the circle is impossible. Suppose that we have a circle of radius 1. The area of the circle is π; therefore, we must be able to construct a square with side π are both transcendental. Therefore, using a straightedge and compass, it is not possible to construct a square with the same area as the circle. π. This is impossible since π and consequently √ √ Trisecting an Angle Trisecting an arbitrary angle is impossible. We will show that it is impossible to construct a 20◦ angle. Consequently, a 60◦ angle cannot be trisected. We ﬁrst need to calculate the triple-angle formula for the cosine: cos 3θ = cos(2θ + θ) = cos 2θ cos θ − sin 2θ sin θ = (2 cos2 θ − 1) cos θ − 2 sin2 θ cos θ = (2 cos2 θ − 1) cos θ − 2(1 − cos2 θ) cos θ = 4 cos3 θ − 3 cos θ. The angle θ can be constructed if and only if α = cos θ is construct
ible. Let θ = 20◦. Then cos 3θ = cos 60◦ = 1/2. By the triple-angle formula for the cosine, 4α3 − 3α = 1 2. Therefore, α is a zero of 8x3 − 6x − 1. This polynomial has no factors in Z[x], and hence is irreducible over Q[x]. Thus, [Q(α) : Q] = 3. Consequently, α cannot be a constructible number. 354 CHAPTER 21 FIELDS Historical Note Algebraic number theory uses the tools of algebra to solve problems in number theory. Modern algebraic number theory began with Pierre de Fermat (1601–1665). Certainly we can ﬁnd many positive integers that satisfy the equation x2 + y2 = z2; Fermat conjectured that the equation xn + yn = zn has no positive integer solutions for n ≥ 3. He stated in the margin of his copy of the Latin translation of Diophantus’ Arithmetica that he had found a marvelous proof of this theorem, but that the margin of the book was too narrow to contain it. Building on work of other mathematicians, it was Andrew Wiles who ﬁnally succeeded in proving Fermat’s Last Theorem in the 1990s. Wiles’s achievement was reported on the front page of the New York Times. Attempts to prove Fermat’s Last Theorem have led to important contributions to algebraic number theory by such notable mathematicians as Leonhard Euler (1707– 1783). Signiﬁcant advances in the understanding of Fermat’s Last Theorem were made by Ernst Kummer (1810–1893). Kummer’s student, Leopold Kronecker (1823– 1891), became one of the leading algebraists of the nineteenth century. Kronecker’s theory of ideals and his study of algebraic number theory added much to the understanding of ﬁelds. David Hilbert (1862–1943) and Hermann Minkowski (1864–1909) were among the mathematicians who led the way in this subject at the beginning of the twentieth century. Hilbert and Minkowski were both mathematicians at G¨ottingen University in Germany. G¨ottingen was truly one the most important
centers of mathematical research during the last two centuries. The large number of exceptional mathematicians who studied there included Gauss, Dirichlet, Riemann, Dedekind, Noether, and Weyl. Andr´e Weil answered questions in number theory using algebraic geometry, a ﬁeld of mathematics that studies geometry by studying commutative rings. From about 1955 to 1970, A. Grothendieck dominated the ﬁeld of algebraic geometry. Pierre Deligne, a student of Grothendieck, solved several of Weil’s number-theoretic conjectures. One of the most recent contributions to algebra and number theory is Gerd Falting’s proof of the Mordell-Weil conjecture. This conjecture of Mordell and Weil essentially says that certain polynomials p(x, y) in Z[x, y] have only a ﬁnite number of integral solutions. Exercises 1. Show that each of the following numbers is algebraic over Q by ﬁnding the minimal polynomial of the number over Q. √ 7 √ √ 1/a) (b) (c) EXERCISES 355 (d) cos θ + i sin θ for θ = 2π/n with n ∈ N √ 3 (e) 2 − i 2. Find a basis for each of the following ﬁeld extensions. What is the degree of each extension? √ √ 3, 6 ) over Q 3 ) over Q √ 2, 3 2, i) over Q √ √ 3, 5, 2 ) over Q √ 2, 3 8 ) over Q( √ √ √ 2 ) 7 ) over Q √ √ (a) Q( √ (b) Q( 3 √ (c) Q( (d) Q( (e) Q( (f) Q( (g) Q(i, √ (h) Q( (i) Q( √ √ 2 + i, √ 3 + i) over Q √ 2 + √ 2, 5 ) over Q( 5 ) 10 ) over Q. Find the splitting ﬁeld for each of the following polynomials. (a) x4 − 10x2 + 21 over Q (b) x4 + 1 over Q (c
) x3 + 2x + 2 over Z3 (d) x3 − 3 over Q √ 4. Determine all of the subﬁelds of Q( 4 5. Show that Z2[x]/x3 + x + 1 is a ﬁeld with eight elements. Construct a 3, i). multiplication table for the multiplicative group of the ﬁeld. 6. Show that the regular 9-gon is not constructible with a straightedge and compass, but that the regular 20-gon is constructible. 7. Prove that the cosine of one degree (cos 1◦) is algebraic over Q but not constructible. 8. Can a cube be constructed with three times the volume of a given cube? 3,...) is an algebraic extension of Q but not a ﬁnite 9. Prove that Q( extension. √ √ 3, 4 √ 3, 8 10. Prove or disprove: π is algebraic over Q(π3). 11. Let p(x) be a nonconstant polynomial of degree n in F [x]. Prove that there 12. Prove or disprove: Q( exists a splitting ﬁeld E for p(x) such that [E : F ] ≤ n!. 2 ) ∼= Q( √ 13. Prove that the ﬁelds Q( 4 √ 3 ) and Q( 4 3 ). √ √ 3 i) are isomorphic but not equal. 356 CHAPTER 21 FIELDS 14. Let K be an algebraic extension of E, and E an algebraic extension of F. Prove that K is algebraic over F. [Caution: Do not assume that the extensions are ﬁnite.] 15. Prove or disprove: Z[x]/x3 − 2 is a ﬁeld. 16. Let F be a ﬁeld of characteristic p. Prove that p(x) = xp − a either is irreducible over F or splits in F. 17. Let E be the algebraic closure of a ﬁeld F. Prove that every polynomial p(x) in F [x] splits in E. 18. If every irreducible polynomial p(x) in F [x] is linear,
show that F is an algebraically closed ﬁeld. 19. Prove that if α and β are constructible numbers such that β = 0, then so is α/β. 20. Show that the set of all elements in R that are algebraic over Q form a ﬁeld extension of Q that is not ﬁnite. 21. Let E be an algebraic extension of a ﬁeld F, and let σ be an automorphism of E leaving F ﬁxed. Let α ∈ E. Show that σ induces a permutation of the set of all zeros of the minimal polynomial of α that are in E. √ √ √ √ 3 + 7 ). Extend your proof to show that 22. Show that Q( a, b ) = Q( Q( √ √ 3, √ 7 ) = Q( √ a + b ), where gcd(a, b) = 1. 23. Let E be a ﬁnite extension of a ﬁeld F. If [E : F ] = 2, show that E is a splitting ﬁeld of F. 24. Prove or disprove: Given a polynomial p(x) in Z6[x], it is possible to construct a ring R such that p(x) has a root in R. 25. Let E be a ﬁeld extension of F and α ∈ E. Determine [F (α) : F (α3)]. 26. Let α, β be transcendental over Q. Prove that either αβ or α + β is also transcendental. 27. Let E be an extension ﬁeld of F and α ∈ E be transcendental over F. Prove that every element in F (α) that is not in F is also transcendental over F. References and Suggested Readings [1] Dean, R. A. Elements of Abstract Algebra. Wiley, New York, 1966. [2] Dudley, U. A Budget of Trisections. Springer-Verlag, New York, 1987. An interesting and entertaining account of how not to trisect an angle. [3] Fraleigh, J. B. A First Course in Abstract Algebra. 7th ed. Pearson, Upper Saddle River, NJ, 2003. [4] Kaplansky
, I. Fields and Rings, 2nd ed. University of Chicago Press, Chicago, 1972. EXERCISES 357 [5] Klein, F. Famous Problems of Elementary Geometry. Chelsea, New York, 1955. [6] Martin, G. Geometric Constructions. Springer, New York, 1998. [7] H. Pollard and H. G. Diamond. Theory of Algebraic Numbers, Dover, Mineola, NY, 2010. [8] Walker, E. A. Introduction to Abstract Algebra. Random House, New York, 1987. This work contains a proof showing that every ﬁeld has an algebraic closure. Sage Extensions of the ﬁeld of rational numbers are a central object of study in number theory, so with Sage’s roots in this discipline, it is no surprise that there is extensive support for ﬁelds and for extensions of the rationals. Sage also contains an implementation of the entire ﬁeld of algebraic numbers, with exact representations. ﬁelds 22 Finite Fields Finite ﬁelds appear in many applications of algebra, including coding theory and cryptography. We already know one ﬁnite ﬁeld, Zp, where p is prime. In this chapter we will show that a unique ﬁnite ﬁeld of order pn exists for every prime p, where n is a positive integer. Finite ﬁelds are also called Galois ﬁelds in honor of ´Evariste Galois, who was one of the ﬁrst mathematicians to investigate them. 22.1 Structure of a Finite Field Recall that a ﬁeld F has characteristic p if p is the smallest positive integer such that for every nonzero element α in F, we have pα = 0. If no such integer exists, then F has characteristic 0. From Theorem 16.6 we know that p must be prime. Suppose that F is a ﬁnite ﬁeld with n elements. Then nα = 0 for all α in F. Consequently, the characteristic of F must be p, where p is a prime dividing n. This discussion is summarized in the following proposition. Proposition 22.1 If F is a ﬁnite ﬁeld, then the characteristic of F is p, where p is
prime. Throughout this chapter we will assume that p is a prime number unless otherwise stated. Proposition 22.2 If F is a ﬁnite ﬁeld of characteristic p, then the order of F is pn for some n ∈ N. Proof. Let φ : Z → F be the ring homomorphism deﬁned by φ(n) = n · 1. Since the characteristic of F is p, the kernel of φ must be pZ and the image of 358 22.1 STRUCTURE OF A FINITE FIELD 359 φ must be a subﬁeld of F isomorphic to Zp. We will denote this subﬁeld by K. Since F is a ﬁnite ﬁeld, it must be a ﬁnite extension of K and, therefore, an algebraic extension of K. Suppose that [F : K] = n is the dimension of F, where F is a K vector space. There must exist elements α1,..., αn ∈ F such that any element α in F can be written uniquely in the form α = a1α1 + · · · + anαn, where the ai’s are in K. Since there are p elements in K, there are pn possible linear combinations of the αi’s. Therefore, the order of F must be pn. Lemma 22.3 (Freshman’s Dream) Let p be prime and D be an integral domain of characteristic p. Then apn + bpn = (a + b)pn for all positive integers n. Proof. We will prove this lemma using mathematical induction on n. We can use the binomial formula (see Chapter 2, Example 3) to verify the case for n = 1; that is, (a + b)p = akbp−k. p k=0 p k If 0 < k < p, then p k = p! k!(p − k)! must be divisible by p, since p cannot divide k!(p − k)!. Note that D is an integral domain of characteristic p, so all but the ﬁrst and last terms in the sum must be zero. Therefore, (a + b)p = ap + bp. Now suppose that the result holds for all k, where 1 ≤ k ≤ n. By
the induction hypothesis, (a + b)pn+1 = ((a + b)p)pn = (ap + bp)pn = (ap)pn + (bp)pn = apn+1 + bpn+1. Therefore, the lemma is true for n + 1 and the proof is complete. Let F be a ﬁeld. A polynomial f (x) ∈ F [x] of degree n is separable if it has n distinct roots in the splitting ﬁeld of f (x); that is, f (x) is separable when it factors into distinct linear factors over the splitting ﬁeld of f. An 360 CHAPTER 22 FINITE FIELDS extension E of F is a separable extension of F if every element in E is the root of a separable polynomial in F [x]. √ √ Example 1. The polynomial x2 − 2 is separable over Q since it factors 2 ) is a separable extension of Q. Let as (x − α = a + b 2 ). If b = 0, then α is a root of x − a. If b = 0, then α is the root of the separable polynomial 2 ). In fact, Q( 2 be any element in Q( 2 )(x + √ √ √ x2 − 2ax + a2 − 2b2 = (x − (a + b √ 2 ))(x − (a − b √ 2 )). Fortunately, we have an easy test to determine the separability of any polynomial. Let f (x) = a0 + a1x + · · · + anxn be any polynomial in F [x]. Deﬁne the derivative of f (x) to be f (x) = a1 + 2a2x + · · · + nanxn−1. Lemma 22.4 Let F be a ﬁeld and f (x) ∈ F [x]. Then f (x) is separable if and only if f (x) and f (x) are relatively prime. Proof. Let f (x) be separable. Then f (x) factors over some extension ﬁeld of F as f (x) = (x − α1)(x − α2) · · · (x − αn), where
αi = αj for i = j. Taking the derivative of f (x), we see that f (x) = (x − α2) · · · (x − αn) + (x − α1)(x − α3) · · · (x − αn) + · · · + (x − α1) · · · (x − αn−1). Hence, f (x) and f (x) can have no common factors. To prove the converse, we will show that the contrapositive of the statement is true. Suppose that f (x) = (x−α)kg(x), where k > 1. Diﬀerentiating, we have f (x) = k(x − α)k−1g(x) + (x − α)kg(x). Therefore, f (x) and f (x) have a common factor. Theorem 22.5 For every prime p and every positive integer n, there exists a ﬁnite ﬁeld F with pn elements. Furthermore, any ﬁeld of order pn is isomorphic to the splitting ﬁeld of xpn − x over Zp. 22.1 STRUCTURE OF A FINITE FIELD 361 Proof. Let f (x) = xpn − x and let F be the splitting ﬁeld of f (x). Then by Lemma 22.4, f (x) has pn distinct zeros in F, since f (x) = pnxpn−1 − 1 = −1 is relatively prime to f (x). We claim that the roots of f (x) form a subﬁeld of F. Certainly 0 and 1 are zeros of f (x). If α and β are zeros of f (x), then α + β and αβ are also zeros of f (x), since αpn + βpn = (α + β)pn and αpnβpn = (αβ)pn. We also need to show that the additive inverse and the multiplicative inverse of each root of f (x) are roots of f (x). For any zero α of f (x), −α = (p − 1)α is also a zero of f (x). If α = 0, then (α−1)pn = (αpn)−1 = α
−1. Since the zeros of f (x) form a subﬁeld of F and f (x) splits in this subﬁeld, the subﬁeld must be all of F. Let E be any other ﬁeld of order pn. To show that E is isomorphic to F, we must show that every element in E is a root of f (x). Certainly 0 is a root of f (x). Let α be a nonzero element of E. The order of the multiplicative group of nonzero elements of E is pn − 1; hence, αpn−1 = 1 or αpn − α = 0. Since E contains pn elements, E must be a splitting ﬁeld of f (x); however, by Corollary 21.20, the splitting ﬁeld of any polynomial is unique up to isomorphism. The unique ﬁnite ﬁeld with pn elements is called the Galois ﬁeld of order pn. We will denote this ﬁeld by GF(pn). Theorem 22.6 Every subﬁeld of the Galois ﬁeld GF(pn) has pm elements, where m divides n. Conversely, if m \| n for m > 0, then there exists a unique subﬁeld of GF(pn) isomorphic to GF(pm). Proof. Let F be a subﬁeld of E = GF(pn). Then F must be a ﬁeld extension of K that contains pm elements, where K is isomorphic to Zp. Then m \| n, since [E : K] = [E : F ][F : K]. To prove the converse, suppose that m \| n for some m > 0. Then pm − 1 divides pn − 1. Consequently, xpm−1 − 1 divides xpn−1 − 1. Therefore, xpm − x must divide xpn − x, and every zero of xpm − x is also a zero of xpn − x. Thus, GF(pn) contains, as a subﬁeld, a splitting ﬁeld of xpm − x, which must be isomorphic to GF(pm). Example 2. The lattice of subﬁelds of GF(p24
) is given in Figure 22.1. With each ﬁeld F we have a multiplicative group of nonzero elements of F which we will denote by F ∗. The multiplicative group of any ﬁnite ﬁeld is cyclic. This result follows from the more general result that we will prove in the next theorem. 362 CHAPTER 22 FINITE FIELDS GF(p8) GF(p4) GF(p2) GF(p24) GF(p) GF(p12) GF(p6) GF(p3) Figure 22.1. Subﬁelds of GF(p24) Theorem 22.7 If G is a ﬁnite subgroup of F ∗, the multiplicative group of nonzero elements of a ﬁeld F, then G is cyclic. Proof. Let G be a ﬁnite subgroup of F ∗ with n = pe1 k elements, where pi’s are (not necessarily distinct) primes. By the Fundamental Theorem of Finite Abelian Groups (Theorem 13.3), 1 · · · pek G ∼= Z pe1 1 × · · · × Z p. ek k Let m be the least common multiple of pe1 k. Then G contains an element of order m. Since every α in G satisﬁes xr − 1 for some r dividing m, α must also be a root of xm − 1. Since xm − 1 has at most m roots in F, n ≤ m. On the other hand, we know that m ≤ \|G\|; therefore, m = n. Thus, G contains an element of order n and must be cyclic. 1,..., pek Corollary 22.8 The multiplicative group of all nonzero elements of a ﬁnite ﬁeld is cyclic. Corollary 22.9 Every ﬁnite extension E of a ﬁnite ﬁeld F is a simple extension of F. Proof. Let α be a generator for the cyclic group E∗ of nonzero elements of E. Then E = F (α). Example 3. The ﬁnite ﬁeld GF(24) is isomorphic to the ﬁeld Z2/1+x+x
4. Therefore, the elements of GF(24) can be taken to be {a0 + a1α + a2α2 + a3α3 : ai ∈ Z2 and 1 + α + α4 = 0}. 22.2 POLYNOMIAL CODES 363 Remembering that 1 + α + α4 = 0, we add and multiply elements of GF(24) exactly as we add and multiply polynomials. The multiplicative group of GF(24) is isomorphic to Z15 with generator α: α1 = α α2 = α2 α3 = α3 α4 = 1 + α α5 = α + α2 α6 = α2 + α3 α7 = 1 + α + α3 α8 = 1 + α2 α9 = α + α3 α10 = 1 + α + α2 α11 = α + α2 + α3 α12 = 1 + α + α2 + α3 α13 = 1 + α2 + α3 α14 = 1 + α3 α15 = 1. 22.2 Polynomial Codes With knowledge of polynomial rings and ﬁnite ﬁelds, it is now possible to derive more sophisticated codes than those of Chapter 7. First let us recall that an (n, k)-block code consists of a one-to-one encoding function E : Zk 2. The code is errorcorrecting if D is onto. A code is a linear code if it is the null space of a matrix H ∈ Mk×n(Z2). 2 and a decoding function D : Zn 2 → Zn 2 → Zk We are interested in a class of codes known as cyclic codes. Let φ : 2 → Zn Zk 2 be a binary (n, k)-block code. Then φ is a cyclic code if for every codeword (a1, a2,..., an), the cyclically shifted n-tuple (an, a1, a2,..., an−1) is also a codeword. Cyclic codes are particularly easy to implement on a computer using shift registers [2, 3]. Example 4. Consider the (6, 3)-linear codes generated by the two matrices G1 =          �
��       and G2 =                . Messages in the ﬁrst code are encoded as follows: (000) (001) (010) (011) → (000000) → (001001) → (010010) → (011011) (100) (101) (110) (111) → (100100) → (101101) → (110110) → (111111). 364 CHAPTER 22 FINITE FIELDS It is easy to see that the codewords form a cyclic code. In the second code, 3-tuples are encoded in the following manner: (000) (001) (010) (011) → (000000) → (001111) → (011110) → (010001) (100) (101) (110) (111) → (111100) → (110011) → (100010) → (101101). This code cannot be cyclic, since (101101) is a codeword but (011011) is not a codeword. Polynomial Codes We would like to ﬁnd an easy method of obtaining cyclic linear codes. To accomplish this, we can use our knowledge of ﬁnite ﬁelds and polynomial rings over Z2. Any binary n-tuple can be interpreted as a polynomial in Z2[x]. Stated another way, the n-tuple (a0, a1,..., an−1) corresponds to the polynomial f (x) = a0 + a1x + · · · + an−1xn−1, where the degree of f (x) is at most n − 1. For example, the polynomial corresponding to the 5-tuple (10011) is 1 + 0x + 0x2 + 1x3 + 1x4 = 1 + x3 + x4. Conversely, with any polynomial f (x) ∈ Z2[x] with deg f (x) < n we can associate a binary n-tuple. The
polynomial x + x2 + x4 corresponds to the 5-tuple (01101). Let us ﬁx a nonconstant polynomial g(x) in Z2[x] of degree n − k. We can deﬁne an (n, k)-code C in the following manner. If (a0,..., ak−1) is a k-tuple to be encoded, then f (x) = a0 + a1x + · · · + ak−1xk−1 is the corresponding polynomial in Z2[x]. To encode f (x), we multiply by g(x). The codewords in C are all those polynomials in Z2[x] of degree less than n that are divisible by g(x). Codes obtained in this manner are called polynomial codes. Example 5. If we let g(x) = 1 + x3, we can deﬁne a (6, 3)-code C as follows. To encode a 3-tuple (a0, a1, a2), we multiply the corresponding polynomial f (x) = a0 + a1x + a2x2 by 1 + x3. We are deﬁning a map φ : Z3 2 → Z6 2 It is easy to check that this map is a group by φ : f (x) → g(x)f (x). 2 as a vector space over Z2, φ is a homomorphism. In fact, if we regard Zn 22.2 POLYNOMIAL CODES 365 linear transformation of vector spaces (see Exercise 15, Chapter 20). Let us compute the kernel of φ. Observe that φ(a0, a1, a2) = (000000) exactly when 0 + 0x + 0x2 + 0x3 + 0x4 + 0x5 = (1 + x3)(a0 + a1x + a2x2) = a0 + a1x + a2x2 + a0x3 + a1x4 + a2x5. Since the polynomials over a ﬁeld form an integral domain, a0 + a1x + a2x2 must be the zero polynomial. Therefore, ker φ = {(000)} and φ is
one-to-one. To calculate a generator matrix for C, we merely need to examine the way the polynomials 1, x, and x2 are encoded: (1 + x3) · 1 = 1 + x3 (1 + x3)x = x + x4 (1 + x3)x2 = x2 + x5. We obtain the code corresponding to the generator matrix G1 in Example 4. The parity-check matrix for this code is  . Since the smallest weight of any nonzero codeword is 2, this code has the ability to detect all single errors. Rings of polynomials have a great deal of structure; therefore, our immediate goal is to establish a link between polynomial codes and ring theory. Recall that xn − 1 = (x − 1)(xn−1 + · · · + x + 1). The factor ring Rn = Z2[x]/xn − 1 can be considered to be the ring of polynomials of the form f (t) = a0 + a1t + · · · + an−1tn−1 that satisfy the condition tn = 1. It is an easy exercise to show that Zn 2 and Rn are isomorphic as vector spaces. We will often identify elements in Zn 2 with elements in Z[x]/xn − 1. In this manner we can interpret a linear code as a subset of Z[x]/xn − 1. The additional ring structure on polynomial codes is very powerful in describing cyclic codes. A cyclic shift of an n-tuple can be described by 366 CHAPTER 22 FINITE FIELDS polynomial multiplication. If f (t) = a0 + a1t + · · · + an−1tn−1 is a code polynomial in Rn, then tf (t) = an−1 + a0t + · · · + an−2tn−1 is the cyclically shifted word obtained from multiplying f (t) by t. The following theorem gives a beautiful classiﬁcation of cyclic codes in terms of the ideals of Rn. Theorem 22.10 A linear code C in Zn in Rn = Z[x]/xn − 1. 2 is cyclic if and only if it is an ideal Proof. Let C be a linear cyclic code and suppose
that f (t) is in C. Then tf (t) must also be in C. Consequently, tkf (t) is in C for all k ∈ N. Since C is a linear code, any linear combination of the codewords f (t), tf (t), t2f (t),..., tn−1f (t) is also a codeword; therefore, for every polynomial p(t), p(t)f (t) is in C. Hence, C is an ideal. Conversely, let C be an ideal in Z2[x]/xn + 1. Suppose that f (t) = a0 + a1t + · · · + an−1tn−1 is a codeword in C. Then tf (t) is a codeword in C; that is, (a1,..., an−1, a0) is in C. Theorem 22.10 tells us that knowing the ideals of Rn is equivalent to knowing the linear cyclic codes in Zn 2. Fortunately, the ideals in Rn are easy to describe. The natural ring homomorphism φ : Z2[x] → Rn deﬁned by φ[f (x)] = f (t) is a surjective homomorphism. The kernel of φ is the ideal generated by xn − 1. By Theorem 16.15, every ideal C in Rn is of the form φ(I), where I is an ideal in Z2[x] that contains xn − 1. By Theorem 17.12, we know that every ideal I in Z2[x] is a principal ideal, since Z2 is a ﬁeld. Therefore, I = g(x) for some unique monic polynomial in Z2[x]. Since xn − 1 is contained in I, it must be the case that g(x) divides xn − 1. Consequently, every ideal C in Rn is of the form C = g(t) = {f (t)g(t) : f (t) ∈ Rn and g(x) \| (xn − 1) in Z2[x]}. The unique monic polynomial of the smallest degree that generates C is called the minimal generator polynomial of C. Example 6. If we
factor x7 − 1 into irreducible components, we have x7 − 1 = (1 + x)(1 + x + x3)(1 + x2 + x3). We see that g(t) = (1 + t + t3) generates an ideal C in R7. This code is a (7, 4)-block code. As in Example 5, it is easy to calculate a generator matrix 22.2 POLYNOMIAL CODES 367 by examining what g(t) does to the polynomials 1, t, t2, and t3. A generator matrix for C is          . In general, we can determine a generator matrix for an (n, k)-code C by the manner in which the elements tk are encoded. Let xn − 1 = g(x)h(x) in Z2[x]. If g(x) = g0 + g1x + · · · + gn−kxn−k and h(x) = h0 + h1x + · · · + hkxk, then the n × k matrix G =             g0 g1... gn−k 0... 0 0 g0... gn−k−1 gn−k...... g0 g1... gn−k is a generator matrix for the code C with generator polynomial g(t). The parity-check matrix for C is the (n − k) × n matrix H =     0 0 · · · hk · · · · · · · · · · · · 0 0 · · · h0 0 hk · · · 0 hk · · · · · · 0 · · · h0 · · · · · ·     h0 0 · · · 0. We will leave the details of the proof of the following proposition as an exercise. Proposition 22.11 Let C = g(t) be a cyclic code in Rn and suppose that xn − 1 = g(x)h(x). Then G and H are
generator and parity-check matrices for C, respectively. Furthermore, HG = 0. Example 7. In Example 6, x7 − 1 = g(x)h(x) = (1 + x + x3)(1 + x + x2 + x4). 368 CHAPTER 22 FINITE FIELDS Therefore, a parity-check matrix for this code is  . To determine the error-detecting and error-correcting capabilities of a cyclic code, we need to know something about determinants. If α1,..., αn are elements in a ﬁeld F, then the n × n matrix        1 α1 α2 1... αn−1 1 1 α2 α2 2... αn− αn α2 n... · · · · · · · · ·... · · · αn−1 n is called the Vandermonde matrix. The determinant of this matrix is called the Vandermonde determinant. We will need the following lemma in our investigation of cyclic codes. Lemma 22.12 Let α1,..., αn be elements in a ﬁeld F with n ≥ 2. Then det        1 α1 α2 1... αn−1 1 1 α2 α2 2... αn−1 2 1 αn α2 n... · · · · · · · · ·... · · · αn−αi − αj). 1≤j<i≤n In particular, if the αi’s are distinct, then the determinant is nonzero. Proof. We will induct on n. If n = 2, then the determinant is α2 − α1. Let us assume the result for n − 1 and consider the polynomial p(x) deﬁned by p(x) = det        1 α1 α2 1... αn−1 1 1 α2 α2 2... αn−1 2 1 1 · · · x · · · αn−1 x2 · · · α2......... ·
· · αn−1 n−1 xn−1 n−1       . Expanding this determinant by cofactors on the last column, we see that p(x) is a polynomial of at most degree n − 1. Moreover, the roots of p(x) 22.2 POLYNOMIAL CODES 369 are α1,..., αn−1, since the substitution of any one of these elements in the last column will produce a column identical to the last column in the matrix. Remember that the determinant of a matrix is zero if it has two identical columns. Therefore, p(x) = (x − α1)(x − α2) · · · (x − αn−1)β, where β = (−1)n+n det        1 α1 α2 1... αn−2 1 1 α2 α2 2... αn− · · · · · · αn−1 · · · α2...... · · · αn−2 n−1 n−1 By our induction hypothesis, β = (−1)n+n (αi − αj). 1≤j<i≤n−1 If we let x = αn, the result now follows immediately. The following theorem gives us an estimate on the error detection and correction capabilities for a particular generator polynomial. Theorem 22.13 Let C = g(t) be a cyclic code in Rn and suppose that ω is a primitive nth root of unity over Z2. If s consecutive powers of ω are roots of g(x), then the minimum distance of C is at least s + 1. Proof. Suppose that g(ωr) = g(ωr+1) = · · · = g(ωr+s−1) = 0. Let f (x) be some polynomial in C with s or fewer nonzero coeﬃcients. We can assume that f (x) = ai0xi0 + ai1xi1 + · · · + ais−1xis−1 be some polynomial in C. It will suﬃce to show that all of the ai�
�s must be 0. Since g(ωr) = g(ωr+1) = · · · = g(ωr+s−1) = 0 and g(x) divides f (x), f (ωr) = f (ωr+1) = · · · = f (ωr+s−1) = 0. 370 CHAPTER 22 FINITE FIELDS Equivalently, we have the following system of equations: ai0(ωr)i0 + ai1(ωr)i1 + · · · + ais−1(ωr)is−1 = 0 ai0(ωr+1)i0 + ai1(ωr+1)i2 + · · · + ais−1(ωr+1)is−1 = 0... ai0(ωr+s−1)i0 + ai1(ωr+s−1)i1 + · · · + ais−1(ωr+s−1)is−1 = 0. Therefore, (ai0, ai1,..., ais−1) is a solution to the homogeneous system of linear equations (ωi0)rx0 + (ωi1)rx1 + · · · + (ωis−1)rxn−1 = 0 (ωi0)r+1x0 + (ωi1)r+1x1 + · · · + (ωis−1)r+1xn−1 = 0... (ωi0)r+s−1x0 + (ωi1)r+s−1x1 + · · · + (ωis−1)r+s−1xn−1 = 0. However, this system has a unique solution, since the determinant of the matrix      (ωi0)r (ωi0)r+1... (ωi0)r+s−1 (ωi1)r (ωi1)r+1... (ωi1)r+s−ωis−1)r (ωis−1)r+1... (ωis−1)r+s−1      can be shown to be nonzero using Lem
ma 22.12 and the basic properties of determinants (Exercise). Therefore, this solution must be ai0 = ai1 = · · · = ais−1 = 0. BCH Codes Some of the most important codes, discovered independently by A. Hocquenghem in 1959 and by R. C. Bose and D. V. Ray-Chaudhuri in 1960, are BCH codes. The European and transatlantic communication systems both use BCH codes. Information words to be encoded are of length 231, and a polynomial of degree 24 is used to generate the code. Since 231 + 24 = 255 = 28 − 1, we are dealing with a (255, 231)-block code. This BCH code will detect six errors and has a failure rate of 1 in 16 million. One advantage of BCH codes is that eﬃcient error correction algorithms exist for them. The idea behind BCH codes is to choose a generator polynomial of smallest degree that has the largest error detection and error correction 22.2 POLYNOMIAL CODES 371 capabilities. Let d = 2r + 1 for some r ≥ 0. Suppose that ω is a primitive nth root of unity over Z2, and let mi(x) be the minimal polynomial over Z2 of ωi. If g(x) = lcm[m1(x), m2(x),..., m2r(x)], then the cyclic code g(t) in Rn is called the BCH code of length n and distance d. By Theorem 22.13, the minimum distance of C is at least d. Theorem 22.14 Let C = g(t) be a cyclic code in Rn. The following statements are equivalent. 1. The code C is a BCH code whose minimum distance is at least d. 2. A code polynomial f (t) is in C if and only if f (ωi) = 0 for 1 ≤ i < d. 3. The matrix H =  ω2 1 ω ω4 1 ω2   ω6 1 ω3  .........   1 ω2r ω4r ωn−1 ω(n−1)(2) ω(n−1)(3)... ·
· · · · · · · ·... · · · ω(n−1)(2r)        is a parity-check matrix for C. Proof. (1) ⇒ (2). If f (t) is in C, then g(x) \| f (x) in Z2[x]. Hence, for i = 1,..., 2r, f (ωi) = 0 since g(ωi) = 0. Conversely, suppose that f (ωi) = 0 for 1 ≤ i ≤ d. Then f (x) is divisible by each mi(x), since mi(x) is the minimal polynomial of ωi. Therefore, g(x) \| f (x) by the deﬁnition of g(x). Consequently, f (x) is a codeword. sponding n-tuple in Zn (2) ⇒ (3). Let f (t) = a0 + a1t + · · · + an−1vtn−1 be in Rn. The corre2 is x = (a0a1 · · · an−1)t. By (2),     Hx =     a0 + a1ω + · · · + an−1ωn−1 a0 + a1ω2 + · · · + an−1(ω2)n−1... a0 + a1ω2r + · · · + an−1(ω2r)n−ω) f (ω2)... f (ω2r)     = 0 exactly when f (t) is in C. Thus, H is a parity-check matrix for C. (3) ⇒ (1). By (3), a code polynomial f (t) = a0 + a1t + · · · + an−1tn−1 is in C exactly when f (ωi) = 0 for i = 1,..., 2r. The smallest such polynomial is g(t) = lcm[m1(t),..., m2r(t)]. Therefore, C
= g(t). Example 8. It is easy to verify that x15 − 1 ∈ Z2[x] has a factorization x15 − 1 = (x + 1)(x2 + x + 1)(x4 + x + 1)(x4 + x3 + 1)(x4 + x3 + x2 + x + 1), 372 CHAPTER 22 FINITE FIELDS where each of the factors is an irreducible polynomial. Let ω be a root of 1 + x + x4. The Galois ﬁeld GF(24) is {a0 + a1ω + a2ω2 + a3ω3 : ai ∈ Z2 and 1 + ω + ω4 = 0}. By Example 3, ω is a primitive 15th root of unity. The minimal polynomial of ω is m1(x) = 1 + x + x4. It is easy to see that ω2 and ω4 are also roots of m1(x). The minimal polynomial of ω3 is m2(x) = 1 + x + x2 + x3 + x4. Therefore, g(x) = m1(x)m2(x) = 1 + x4 + x6 + x7 + x8 has roots ω, ω2, ω3, ω4. Since both m1(x) and m2(x) divide x15 −1, the BCH code is a (15, 7)-code. If x15 − 1 = g(x)h(x), then h(x) = 1 + x4 + x6 + x7; therefore, a parity-check matrix for this code is                        . Exercises 1. Calculate each of the following. (a) [GF(36) : GF(33)] (b) [GF(128) : GF(16)] (c) [GF(625) : GF(25)] (d) [GF(p12) : GF(p2)] 2
. Calculate [GF(pm) : GF(pn)], where n \| m. 3. What is the lattice of subﬁelds for GF(p30)? 4. Let α be a zero of x3 + x2 + 1 over Z2. Construct a ﬁnite ﬁeld of order 8. Show that x3 + x2 + 1 splits in Z2(α). 5. Construct a ﬁnite ﬁeld of order 27. 6. Prove or disprove: Q∗ is cyclic. 7. Factor each of the following polynomials in Z2[x]. EXERCISES 373 (a) x5 − 1 (b) x6 + x5 + x4 + x3 + x2 + x + 1 (c) x9 − 1 (d) x4 + x3 + x2 + x + 1 8. Prove or disprove: Z2[x]/x3 + x + 1 ∼= Z2[x]/x3 + x2 + 1. 9. Determine the number of cyclic codes of length n for n = 6, 7, 8, 10. 10. Prove that the ideal t + 1 in Rn is the code in Zn 2 consisting of all words of even parity. 11. Construct all BCH codes of (a) length 7. (b) length 15. 12. Prove or disprove: There exists a ﬁnite ﬁeld that is algebraically closed. 13. Let p be prime. Prove that the ﬁeld of rational functions Zp(x) is an inﬁnite ﬁeld of characteristic p. 14. Let D be an integral domain of characteristic p. Prove that (a − b)pn = apn − bpn for all a, b ∈ D. 15. Show that every element in a ﬁnite ﬁeld can be written as the sum of two squares. 16. Let E and F be subﬁelds of a ﬁnite ﬁeld K. If E is isomorphic to F, show that E = F. 17. Let F ⊂ E ⊂ K be ﬁelds. If K is separable over F, show that K
is also separable over E. 18. Let E be an extension of a ﬁnite ﬁeld F, where F has q elements. Let α ∈ E be algebraic over F of degree n. Prove that F (α) has qn elements. 19. Show that every ﬁnite extension of a ﬁnite ﬁeld F is simple; that is, if E is a ﬁnite extension of a ﬁnite ﬁeld F, prove that there exists an α ∈ E such that E = F (α). 20. Show that for every n there exists an irreducible polynomial of degree n in Zp[x]. 21. Prove that the Frobenius map φ : GF(pn) → GF(pn) given by φ : α → αp is an automorphism of order n. 22. Show that every element in GF(pn) can be written in the form ap for some unique a ∈ GF(pn). 23. Let E and F be subﬁelds of GF(pn). If \|E\| = pr and \|F \| = ps, what is the order of E ∩ F? 24. Wilson’s Theorem. Let p be prime. Prove that (p − 1)! ≡ −1 (mod p). 374 CHAPTER 22 FINITE FIELDS 25. If g(t) is the minimal generator polynomial for a cyclic code C in Rn, prove that the constant term of g(x) is 1. 26. Often it is conceivable that a burst of errors might occur during transmission, as in the case of a power surge. Such a momentary burst of interference might alter several consecutive bits in a codeword. Cyclic codes permit the detection of such error bursts. Let C be an (n, k)-cyclic code. Prove that any error burst up to n − k digits can be detected. 27. Prove that the rings Rn and Zn 2 are isomorphic as vector spaces. 28. Let C be a code in Rn that is generated by g(t). If f (t) is another code in Rn, show that g(t) ⊂ f (t) if and only if f (x) divides g(x) in Z2[x]. 29. Let C
= g(t) be a cyclic code in Rn and suppose that xn − 1 = g(x)h(x), where g(x) = g0 + g1x + · · · + gn−kxn−k and h(x) = h0 + h1x + · · · + hkxk. Deﬁne G to be the n × k matrix G =             g0 g1... gn−k 0... 0 0 g0... gn−k−1 gn−k...... g0 g1... gn−k and H to be the (n − k) × n matrix H =     0 0 · · · hk · · · · · · · · · · · · 0 0 · · · h0 0 hk · · · 0 hk · · · · · · 0 · · · h0 · · · · · · (a) Prove that G is a generator matrix for C. (b) Prove that H is a parity-check matrix for C. (c) Show that HG = 0.     h0 0 · · · 0. Additional Exercises: Error Correction for BCH Codes BCH codes have very attractive error correction algorithms. Let C be a BCH code in Rn, and suppose that a code polynomial c(t) = c0 + c1t + · · · + cn−1tn−1 is transmitted. Let w(t) = w0 + w1t + · · · wn−1tn−1 be the polynomial in Rn that is received. If errors have occurred in bits a1,..., ak, then w(t) = c(t) + e(t), where e(t) = ta1 + ta2 + · · · + tak is the error polynomial. The decoder must determine the integers ai and then recover c(t) from w(t) by ﬂipping the aith bit. From w(t) we can compute w(ωi) = si for i = 1,..., 2r
, where ω is a primitive nth root of unity over Z2. We say the syndrome of w(t) is s1,..., s2r. EXERCISES 375 1. Show that w(t) is a code polynomial if and only if si = 0 for all i. 2. Show that si = w(ωi) = e(ωi) = ωia1 + ωia2 + · · · + ωiak for i = 1,..., 2r. The error-locator polynomial is deﬁned to be s(x) = (x + ωa1)(x + ωa2) · · · (x + ωak ). 3. Recall the (15, 7)-block BCH code in Example 7. By Theorem 8.3, this code is capable of correcting two errors. Suppose that these errors occur in bits a1 and a2. The error-locator polynomial is s(x) = (x + ωa1 )(x + ωa2). Show that s(x) = x2 + s1x + s2 1 +. s3 s1 4. Let w(t) = 1 + t2 + t4 + t5 + t7 + t12 + t13. Determine what the originally transmitted code polynomial was. References and Suggested Readings [1] Childs, L. A Concrete Introduction to Higher Algebra. 2nd ed. Springer-Verlag, New York, 1995.. [2] G˚ading, L. and Tambour, T. Algebra for Computer Science. Springer-Verlag, New York, 1988. [3] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York, 1998. An excellent presentation of ﬁnite ﬁelds and their applications. [4] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985. [5] Roman, S. Coding and Information Theory. Springer-Verlag, New York, 1992. [6] van Lint, J. H. Introduction to Coding Theory. Springer, New York, 1999. Sage Finite ﬁelds are important in a variety of applied disciplines, such as cryptography and coding
theory (see introductions to these topics in other chapters). Sage has excellent support for ﬁnite ﬁelds allowing for a wide variety of computations. 23 Galois Theory A classic problem of algebra has been to ﬁnd the solutions of a polynomial equation. The solution to the quadratic equation was known in antiquity. Italian mathematicians found general solutions to the general cubic and quartic equations in the sixteenth century; however, attempts to solve the general ﬁfth-degree, or quintic, polynomial were repulsed for the next three hundred years. Certainly, equations such as x5 − 1 = 0 or x6 − x3 − 6 = 0 could be solved, but no solution like the quadratic formula was found for the general quintic, ax5 + bx4 + cx3 + dx2 + ex + f = 0. Finally, at the beginning of the nineteenth century, Ruﬃni and Abel both found quintics that could not be solved with any formula. It was Galois, however, who provided the full explanation by showing which polynomials could and could not be solved by formulas. He discovered the connection between groups and ﬁeld extensions. Galois theory demonstrates the strong interdependence of group and ﬁeld theory, and has had far-reaching implications beyond its original purpose. In this chapter we will prove the Fundamental Theorem of Galois Theory. This result will be used to establish the insolvability of the quintic and to prove the Fundamental Theorem of Algebra. 23.1 Field Automorphisms Our ﬁrst task is to establish a link between group theory and ﬁeld theory by examining automorphisms of ﬁelds. Proposition 23.1 The set of all automorphisms of a ﬁeld F is a group under composition of functions. 376 23.1 FIELD AUTOMORPHISMS 377 Proof. If σ and τ are automorphisms of E, then so are στ and σ−1. The identity is certainly an automorphism; hence, the set of all automorphisms of a ﬁeld F is indeed a group. Proposition 23.2 Let E be a ﬁeld extension of F. Then the set of all automorphisms of E that ﬁx F elementwise is a group;
that is, the set of all automorphisms σ : E → E such that σ(α) = α for all α ∈ F is a group. Proof. We need only show that the set of automorphisms of E that ﬁx F elementwise is a subgroup of the group of all automorphisms of E. Let σ and τ be two automorphisms of E such that σ(α) = α and τ (α) = α for all α ∈ F. Then στ (α) = σ(α) = α and σ−1(α) = α. Since the identity ﬁxes every element of E, the set of automorphisms of E that leave elements of F ﬁxed is a subgroup of the entire group of automorphisms of E. Let E be a ﬁeld extension of F. We will denote the full group of automorphisms of E by Aut(E). We deﬁne the Galois group of E over F to be the group of automorphisms of E that ﬁx F elementwise; that is, G(E/F ) = {σ ∈ Aut(E) : σ(α) = α for all α ∈ F }. If f (x) is a polynomial in F [x] and E is the splitting ﬁeld of f (x) over F, then we deﬁne the Galois group of f (x) to be G(E/F ). Example 1. Complex conjugation, deﬁned by σ : a + bi → a − bi, is an automorphism of the complex numbers. Since σ(a) = σ(a + 0i) = a − 0i = a, the automorphism deﬁned by complex conjugation must be in G(C/R). Example 2. Consider the ﬁelds Q ⊂ Q( a, b ∈ Q( 5 ), √ √ √ 5 ) ⊂ Q( √ 3, 5 ). Then for is an automorphism of Q( 3, 5 ) leaving Q( 5 ) ﬁxed. Similarly, √ √ σ(a + b √ √ τ ( √ is an automorphism of Q
( µ = στ moves both the Galois group of Q( group is isomorphic to Z2 × Z2. 3, 3 and √ √ 3, √ 5 ) leaving Q( 3 ) ﬁxed. The automorphism 5. It will soon be clear that {id, σ, τ, µ} is 5 ) over Q. The following table shows that this 378 CHAPTER 23 GALOIS THEORY τ τ µ id σ µ µ τ σ id id id σ τ µ id σ τ µ σ σ id µ τ √ 3, √ We may also regard the ﬁeld Q( √ 5, 3, has basis {1, 5 ) : Q)] = 4. [Q( 3, √ √ √ √ 15 }. It is no coincidence that \|G(Q( 5 ) as a vector space over Q that 5 )/Q)\| = √ √ 3, Proposition 23.3 Let E be a ﬁeld extension of F and f (x) be a polynomial in F [x]. Then any automorphism in G(E/F ) deﬁnes a permutation of the roots of f (x) that lie in E. Proof. Let f (x) = a0 + a1x + a2x2 + · · · + anxn and suppose that α ∈ E is a zero of f (x). Then for σ ∈ G(E/F ), 0 = σ(0) = σ(f (α)) = σ(a0 + a1α + a2α2 + · · · + anαn) = a0 + a1σ(α) + a2[σ(α)]2 + · · · + an[σ(α)]n; therefore, σ(α) is also a zero of f (x). Let E be an algebraic extension of a ﬁeld F. Two elements α, β ∈ E are conjugate over F if they have the same minimal polynomial. For example, in the ﬁeld Q( 2 are conjugate over Q since they are both roots of the irreducible polynomial x2 − 2. 2 ) the elements 2 and − √ √ √ A converse of the last proposition
exists. The proof follows directly from Lemma 21.18. Proposition 23.4 If α and β are conjugate over F, there exists an isomorphism σ : F (α) → F (β) such that σ is the identity when restricted to F. Theorem 23.5 Let f (x) be a polynomial in F [x] and suppose that E is the splitting ﬁeld for f (x) over F. If f (x) has no repeated roots, then \|G(E/F )\| = [E : F ]. 23.1 FIELD AUTOMORPHISMS 379 Proof. The proof is similar to the proof of Theorem 21.19. We will use mathematical induction on the degree of f (x). If the degree of f (x) is 0 or 1, then E = F and there is nothing to show. Assume that the result holds for all polynomials of degree k with 0 ≤ k < n. Let p(x) be an irreducible factor of f (x) of degree r. Since all of the roots of p(x) are in E, we can choose one of these roots, say α, so that F ⊂ F (α) ⊂ E. If β is any other root of p(x), then F ⊂ F (β) ⊂ E. By Lemma 21.18, there exists a unique isomorphism σ : F (α) → F (β) for each such β that ﬁxes F elementwise. Since E is a splitting ﬁeld of F (β), there are exactly r such isomorphisms. We can factor p(x) in F (α) as p(x) = (x − α)p1(x). The degrees of p1(x) and q1(x) are both less than r. Since we know that E is the splitting ﬁeld of p1(x) over F (α), we can apply the induction hypothesis to conclude that \|G(E/F (α))\| = [E : F (α)]. Consequently, there are [E : F ] = [E : F (α)][F (α) : F ] possible automorphisms of E that ﬁx F, or \|G(E/F )\| = [E : F ].
Corollary 23.6 Let F be a ﬁnite ﬁeld with a ﬁnite extension E such that [E : F ] = k. Then G(E/F ) is cyclic or order k. Proof. Let p be the characteristic of E and F and assume that the orders of E and F are pm and pn, respectively. Then nk = m. We can also assume that E is the splitting ﬁeld of xpm − x over a subﬁeld of order p. Therefore, E must also be the splitting ﬁeld of xpm − x over F. Applying Theorem 23.5, we ﬁnd that \|G(E/F )\| = k. To prove that G(E/F ) is cyclic, we must ﬁnd a generator for G(E/F ). Let σ : E → E be deﬁned by σ(α) = αpn. We claim that σ is the element in G(E/F ) that we are seeking. We ﬁrst need to show that σ is in Aut(E). If α and β are in E, σ(α + β) = (α + β)pn = αpn + βpn = σ(α) + σ(β) by Lemma 22.3. Also, it is easy to show that σ(αβ) = σ(α)σ(β). Since σ is a nonzero homomorphism of ﬁelds, it must be injective. It must also be onto, since E is a ﬁnite ﬁeld. We know that σ must be in G(E/F ), since F is the splitting ﬁeld of xpn − x over the base ﬁeld of order p. This means that σ leaves every element in F ﬁxed. Finally, we must show that the order of σ is k. By Theorem 23.5, we know that σk(α) = αpk = α is the identity 380 CHAPTER 23 GALOIS THEORY of G(E/F ). However, σr cannot be the identity for 1 ≤ r < k; otherwise, xprk − x would have pm roots, which
is impossible. Example 3. We can now conﬁrm that the Galois group of Q( 5 ) over Q in Example 2 is indeed isomorphic to Z2 × Z2. Certainly the group H = {id, σ, τ, µ} is a subgroup of G(Q( 5 )/Q); however, H must be all 5 )/Q), since of G(Q( √ √ √ √ 3, 3, 3, √ √ √ \|H\| = [Q( √ 3, 5 ) : Q] = \|G(Q( √ √ 3, 5 )/Q)\| = 4. Example 4. Let us compute the Galois group of f (x) = x4 + x3 + x2 + x + 1 over Q. We know that f (x) is irreducible by Exercise 19 in Chapter 17. Furthermore, since (x − 1)f (x) = x5 − 1, we can use DeMoivre’s Theorem to determine that the roots of f (x) are ωi, where i = 1,..., 4 and ω = cos(2π/5) + i sin(2π/5). Hence, the splitting ﬁeld of f (x) must be Q(ω). We can deﬁne automorphisms σi of Q(ω) by σi(ω) = ωi for i = 1,..., 4. It is easy to check that these are indeed distinct automorphisms in G(Q(ω)/Q). Since [Q(ω) : Q] = \|G(Q(ω)/Q)\| = 4, the σi’s must be all of G(Q(ω)/Q). Therefore, G(Q(ω)/Q) ∼= Z4 since ω is a generator for the Galois group. Separable Extensions Many of the results that we have just proven depend on the fact that a polynomial f (x) in F [x] has no repeated roots in its splitting ﬁeld. It is evident that we need to know exactly when a polynomial factors into distinct linear factors in its splitting ﬁeld. Let E be the splitting ﬁeld of a polynomial f (x) in F [x
]. Suppose that f (x) factors over E as f (x) = (x − α1)n1(x − α2)n2 · · · (x − αr)nr = r i=1 (x − αi)ni. 23.1 FIELD AUTOMORPHISMS 381 We deﬁne the multiplicity of a root αi of f (x) to be ni. A root with multiplicity 1 is called a simple root. Recall that a polynomial f (x) ∈ F [x] of degree n is separable if it has n distinct roots in its splitting ﬁeld E. Equivalently, f (x) is separable if it factors into distinct linear factors over E[x]. An extension E of F is a separable extension of F if every element in E is the root of a separable polynomial in F [x]. Also recall that f (x) is separable if and only if gcd(f (x), f (x)) = 1 (Lemma 22.4). Proposition 23.7 Let f (x) be an irreducible polynomial over F. If the characteristic of F is 0, then f (x) is separable. If the characteristic of F is p and f (x) = g(xp) for some g(x) in F [x], then f (x) is also separable. Proof. First assume that charF = 0. Since deg f (x) < deg f (x) and f (x) is irreducible, the only way gcd(f (x), f (x)) = 1 is if f (x) is the zero polynomial; however, this is impossible in a ﬁeld of characteristic zero. If charF = p, then f (x) can be the zero polynomial if every coeﬃcient of f (x) is a multiple of p. This can happen only if we have a polynomial of the form f (x) = a0 + a1xp + a2x2p + · · · + anxnp. Certainly extensions of a ﬁeld F of the form F (α) are some of the easiest to study and understand. Given a ﬁeld extension E of F, the obvious question to ask is when it is possible to
ﬁnd an element α ∈ E such that E = F (α). In this case, α is called a primitive element. We already know that primitive elements exist for certain extensions. For example, √ √ √ √ Q( 3, 5 ) = Q( 3 + 5 ) and Q( 3√ 5, √ 5 i) = Q( 6√ 5 i). Corollary 22.9 tells us that there exists a primitive element for any ﬁnite extension of a ﬁnite ﬁeld. The next theorem tells us that we can often ﬁnd a primitive element. Theorem 23.8 (Primitive Element Theorem) Let E be a ﬁnite separable extension of a ﬁeld F. Then there exists an α ∈ E such that E = F (α). Proof. We already know that there is no problem if F is a ﬁnite ﬁeld. Suppose that E is a ﬁnite extension of an inﬁnite ﬁeld. We will prove the result for F (α, β). The general case easily follows when we use mathematical induction. Let f (x) and g(x) be the minimal polynomials of α and β, respectively. Let K be the ﬁeld in which both f (x) and g(x) split. Suppose 382 CHAPTER 23 GALOIS THEORY that f (x) has zeros α = α1,..., αn in K and g(x) has zeros β = β1,..., βm in K. All of these zeros have multiplicity 1, since E is separable over F. Since F is inﬁnite, we can ﬁnd an a in F such that a = αi − α β − βj for all i and j with j = 1. Therefore, a(β − βj) = αi − α. Let γ = α + aβ. Then γ = α + aβ = αi + aβj; hence, γ − aβj = αi for all i, j with j = 1. Deﬁne h(x) ∈ F (γ)[x] by h(x) = f (γ − ax
). Then h(β) = f (α) = 0. However, h(βj) = 0 for j = 1. Hence, h(x) and g(x) have a single common factor in F (γ)[x]; that is, the irreducible polynomial of β over F (γ) must be linear, since β is the only zero common to both g(x) and h(x). So β ∈ F (γ) and α = γ − aβ is in F (γ). Hence, F (α, β) = F (γ). 23.2 The Fundamental Theorem The goal of this section is to prove the Fundamental Theorem of Galois Theory. This theorem explains the connection between the subgroups of G(E/F ) and the intermediate ﬁelds between E and F. Proposition 23.9 Let {σi : i ∈ I} be a collection of automorphisms of a ﬁeld F. Then F{σi} = {a ∈ F : σi(a) = a for all σi} is a subﬁeld of F. Proof. Let σi(a) = a and σi(b) = b. Then σi(a ± b) = σi(a) ± σi(b) = a ± b and σi(ab) = σi(a)σi(b) = ab. If a = 0, then σi(a−1) = [σi(a)]−1 = a−1. Finally, σi(0) = 0 and σi(1) = 1 since σi is an automorphism. Corollary 23.10 Let F be a ﬁeld and let G be a subgroup of Aut(F ). Then FG = {α ∈ F : σ(α) = α for all σ ∈ G} is a subﬁeld of F. 23.2 THE FUNDAMENTAL THEOREM 383 The subﬁeld F{σi} of F is called the ﬁxed ﬁeld of {σi}. The ﬁeld ﬁxed for a subgroup G of Aut(F ) will be denoted by FG. �
� √ √ Example 5. Let σ : Q( 3 to − maps 3, 3. Then Q( √ √ 5 ) → Q( 3, 5 ) is the subﬁeld of Q( √ √ √ 5 ) be the automorphism that 5 ) left ﬁxed by σ. √ 3, Proposition 23.11 Let E be a splitting ﬁeld over F of a separable polynomial. Then EG(E/F ) = F. Proof. Let G = G(E/F ). Clearly, F ⊂ EG ⊂ E. Also, E must be a splitting ﬁeld of EG and G(E/F ) = G(E/EG). By Theorem 23.5, \|G\| = [E : EG] = [E : F ]. Therefore, [EG : F ] = 1. Consequently, EG = F. A large number of mathematicians ﬁrst learned Galois theory from Emil Artin’s monograph on the subject [1]. The very clever proof of the following lemma is due to Artin. Lemma 23.12 Let G be a ﬁnite group of automorphisms of E and let F = EG. Then [E : F ] ≤ \|G\|. Proof. Let \|G\| = n. We must show that any set of n + 1 elements α1,..., αn+1 in E is linearly dependent over F ; that is, we need to ﬁnd elements ai ∈ F, not all zero, such that a1α1 + a2α2 + · · · + an+1αn+1 = 0. Suppose that σ1 = id, σ2,..., σn are the automorphisms in G. The homogeneous system of linear equations σ1(α1)x1 + σ1(α2)x2 + · · · + σ1(αn+1)xn+1 = 0 σ2(α1)x1 + σ2(α2)x2 + · · · + σ2(αn+1)xn+1 = 0... σn(α1)x1 + σn(α2)x2 + · · · + σ
n(αn+1)xn+1 = 0 has more unknowns than equations. From linear algebra we know that this system has a nontrivial solution, say xi = ai for i = 1, 2,..., n + 1. Since σ1 is the identity, the ﬁrst equation translates to a1α1 + a2α2 + · · · + an+1αn+1 = 0. 384 CHAPTER 23 GALOIS THEORY The problem is that some of the ai’s may be in E but not in F. We must show that this is impossible. Suppose that at least one of the ai’s is in E but not in F. By rearranging the αi’s we may assume that a1 is nonzero. Since any nonzero multiple of a solution is also a solution, we can also assume that a1 = 1. Of all possible solutions ﬁtting this description, we choose the one with the smallest number of nonzero terms. Again, by rearranging α2,..., αn+1 if necessary, we can assume that a2 is in E but not in F. Since F is the subﬁeld of E that is ﬁxed elementwise by G, there exists a σi in G such that σi(a2) = a2. Applying σi to each equation in the system, we end up with the same homogeneous system, since G is a group. Therefore, x1 = σi(a1) = 1, x2 = σi(a2),..., xn+1 = σi(an+1) is also a solution of the original system. We know that a linear combination of two solutions of a homogeneous system is also a solution; consequently, x1 = 1 − 1 = 0 x2 = a2 − σi(a2)... xn+1 = an+1 − σi(an+1) must be another solution of the system. This is a nontrivial solution because σi(a2) = a2, and has fewer nonzero entries than our original solution. This is a contradiction, since the number of nonzero solutions to our original solution was assumed to be minimal. We can therefore conclude that a1 = · · ·
= an+1 = 0. Let E be an algebraic extension of F. If every irreducible polynomial in F [x] with a root in E has all of its roots in E, then E is called a normal extension of F ; that is, every irreducible polynomial in F [x] containing a root in E is the product of linear factors in E[x]. Theorem 23.13 Let E be a ﬁeld extension of F. Then the following statements are equivalent. 1. E is a ﬁnite, normal, separable extension of F. 2. E is a splitting ﬁeld over F of a separable polynomial. 3. F = EG for some ﬁnite group of automorphisms of E. Proof. (1) ⇒ (2). Let E be a ﬁnite, normal, separable extension of F. By the Primitive Element Theorem, we can ﬁnd an α in E such that E = F (α). 23.2 THE FUNDAMENTAL THEOREM 385 Let f (x) be the minimal polynomial of α over F. The ﬁeld E must contain all of the roots of f (x) since it is a normal extension F ; hence, E is a splitting ﬁeld for f (x). (2) ⇒ (3). Let E be the splitting ﬁeld over F of a separable polynomial. By Proposition 23.11, EG(E/F ) = F. Since \|G(E/F )\| = [E : F ], this is a ﬁnite group. (3) ⇒ (1). Let F = EG for some ﬁnite group of automorphisms G of E. Since [E : F ] ≤ \|G\|, E is a ﬁnite extension of F. To show that E is a ﬁnite, normal extension of F, let f (x) ∈ F [x] be an irreducible monic polynomial that has a root α in E. We must show that f (x) is the product of distinct linear factors in E[x]. By Proposition 23.3, automorphisms in G permute the roots of f (x) lying in E. Hence, if we
let G act on α, we can obtain distinct roots α1 = α, α2,..., αn in E. Let g(x) = n i=1(x − αi). Then g(x) is separable over F and g(α) = 0. Any automorphism σ in G permutes the factors of g(x) since it permutes these roots; hence, when σ acts on g(x), it must ﬁx the coeﬃcients of g(x). Therefore, the coeﬃcients of g(x) must be in F. Since deg g(x) ≤ deg f (x) and f (x) is the minimal polynomial of α, f (x) = g(x). Corollary 23.14 Let K be a ﬁeld extension of F such that F = KG for some ﬁnite group of automorphisms G of K. Then G = G(K/F ). Proof. Since F = KG, G is a subgroup of G(K/F ). Hence, [K : F ] ≤ \|G\| ≤ \|G(K/F )\| = [K : F ]. It follows that G = G(K/F ), since they must have the same order. Before we determine the exact correspondence between ﬁeld extensions and automorphisms of ﬁelds, let us return to a familiar example. Example 6. In Example 2 we examined the automorphisms of Q( 5 ) ﬁxing Q. Figure 23.1 compares the lattice of ﬁeld extensions of Q with the lattice of subgroups of G(Q( 5 )/Q). The Fundamental Theorem of Galois Theory tells us what the relationship is between the two lattices. √ √ 3, 3, √ √ We are now ready to state and prove the Fundamental Theorem of Galois Theory. Theorem 23.15 (Fundamental Theorem of Galois Theory) Let F be a ﬁnite ﬁeld or a ﬁeld of characteristic zero. If E is a ﬁnite normal extension of F with Galois group G(E/F ), then the following statements are true. 386 CHAPTER 23 GALOIS THEORY {id, σ, τ,
µ} √ Q( 3, √ 5 ) {id, σ} {id, τ } {id, µ} √ Q( √ 3 ) Q( 5 ) Q( √ 15 ) {id} Q Figure 23.1. G(Q( √ √ 3, 5 )/Q) 1. The map K → G(E/K) is a bijection of subﬁelds K of E containing F with the subgroups of G(E/F ). 2. If F ⊂ K ⊂ E, then [E : K] = \|G(E/K)\| and [K : F ] = [G(E/F ) : G(E/K)]. 3. F ⊂ K ⊂ L ⊂ E if and only if {id} ⊂ G(E/L) ⊂ G(E/K) ⊂ G(E/F ). 4. K is a normal extension of F if and only if G(E/K) is a normal subgroup of G(E/F ). In this case G(K/F ) ∼= G(E/F )/G(E/K). Proof. (1) Suppose that G(E/K) = G(E/L) = G. Both K and L are ﬁxed ﬁelds of G; hence, K = L and the map deﬁned by K → G(E/K) is one-to-one. To show that the map is onto, let G be a subgroup of G(E/F ) and K be the ﬁeld ﬁxed by G. Then F ⊂ K ⊂ E; consequently, E is a normal extension of K. Thus, G(E/K) = G and the map K → G(E/K) is a bijection. (2) By Theorem 23.5, \|G(E/K)\| = [E : K]; therefore, \|G(E/F )\| = [G(E/F ) : G(E/K)] · \|G(E/K)\| = [E : F ] = [E : K][K : F ]. Thus, [K : F ] = [G(E/F ) : G(E/K
)]. (3) Statement (3) is illustrated in Figure 23.2. We leave the proof of this property as an exercise. (4) This part takes a little more work. Let K be a normal extension of F. If σ is in G(E/F ) and τ is in G(E/K), we need to show that σ−1τ σ 23.2 THE FUNDAMENTAL THEOREM 387 E L K F {id} G(E/L) G(E/K) G(E/F ) Figure 23.2. Subgroups of G(E/F ) and subﬁelds of E is in G(E/K); that is, we need to show that σ−1τ σ(α) = α for all α ∈ K. Suppose that f (x) is the minimal polynomial of α over F. Then σ(α) is also a root of f (x) lying in K, since K is a normal extension of F. Hence, τ (σ(α)) = σ(α) or σ−1τ σ(α) = α. Conversely, let G(E/K) be a normal subgroup of G(E/F ). We need to show that F = KG(K/F ). Let τ ∈ G(E/K). For all σ ∈ G(E/F ) there exists a τ ∈ G(E/K) such that τ σ = στ. Consequently, for all α ∈ K τ (σ(α)) = σ(τ (α)) = σ(α); hence, σ(α) must be in the ﬁxed ﬁeld of G(E/K). Let σ be the restriction of σ to K. Then σ is an automorphism of K ﬁxing F, since σ(α) ∈ K for all α ∈ K; hence, σ ∈ G(K/F ). Next, we will show that the ﬁxed ﬁeld of G(K/F ) is F. Let β be an element in K that is ﬁxed by all automorphisms In particular, σ(β) = β for all σ ∈ G(E/F ). Therefore
, β in G(K/F ). belongs to the ﬁxed ﬁeld F of G(E/F ). Finally, we must show that when K is a normal extension of F, G(K/F ) ∼= G(E/F )/G(E/K). For σ ∈ G(E/F ), let σK be the automorphism of K obtained by restricting σ to K. Since K is a normal extension, the argument in the preceding paragraph shows that σK ∈ G(K/F ). Consequently, we have a map φ : G(E/F ) → G(K/F ) deﬁned by σ → σK. This map is a group homomorphism since φ(στ ) = (στ )K = σKτK = φ(σ)φ(τ ). 388 CHAPTER 23 GALOIS THEORY The kernel of φ is G(E/K). By (2), \|G(E/F )\|/\|G(E/K)\| = [K : F ] = \|G(K/F )\|. Hence, the image of φ is G(K/F ) and φ is onto. Applying the First Isomorphism Theorem, we have G(K/F ) ∼= G(E/F )/G(E/K). 2, i). To see this, notice that f (x) factors as (x2 + √ 2 and ± 4 Example 7. In this example we will illustrate the Fundamental Theorem of Galois Theory by determining the lattice of subgroups of the Galois group of f (x) = x4 − 2. We will compare this lattice to the lattice of ﬁeld extensions of Q that are contained in the splitting ﬁeld of x4 − 2. The splitting ﬁeld of √ f (x) is Q( 4 2 ); √ hence, the roots of f (x) are ± 4 2 to √ Q and then adjoin the root i of x2 + 1 to Q( 4 2 ). The splitting ﬁeld of f (x) √ √ 2 )(i) = Q( 4 is then Q( 4 √ √ Since [Q( 4 2 ) : Q] =
4 and i is not in Q( 4 √ √ √ 2 )] = 2. Hence, [Q( 4 2, i) : Q( 4 [Q( 4 {1, 4√ √ is a basis of Q( 4 √ in Q( 4 2 )(x2 − √ 2 i. We ﬁrst adjoin the root 4 2, i) over Q. The lattice of ﬁeld extensions of Q contained 2, i) is illustrated in Figure 23.3(a). 2 ), it must be the case that 2, i) : Q] = 8. The set 2 )3, i, i 4√ 2 )2, i( 4√ 2 )2, ( 4√ 2, i( 4√ 2, ( 4√ 2 )3} 2, i). √ √ √ 2 ) = i 4 The Galois group G of f (x) must be of order 8. Let σ be the automorphism √ deﬁned by σ( 4 2 and σ(i) = i, and τ be the automorphism deﬁned by complex conjugation; that is, τ (i) = −i. Then G has an element of order 4 and an element of order 2. It is easy to verify by direct computation that the elements of G are {id, σ, σ2, σ3, τ, στ, σ2τ, σ3τ } and that the relations τ 2 = id, σ4 = id, and τ στ = σ−1 are satisﬁed; hence, G must be isomorphic to D4. The lattice of subgroups of G is illustrated in Figure 23.3(b). Historical Note Solutions for the cubic and quartic equations were discovered in the 1500s. Attempts to ﬁnd solutions for the quintic equations puzzled some of history’s best mathematicians. In 1798, P. Ruﬃni submitted a paper that claimed no such solution could be found; however, the paper was not well received. In 1826, Niels Henrik Abel (1802–1829) ﬁnally oﬀered the ﬁrst correct proof that quintics are not always solvable by radicals. 23.2
THE FUNDAMENTAL THEOREM 389 √ Q( 4 2, i) √ Q( 4 2 ) √ Q( 4 2 i) √ √ 2, i) Q((1 + i) 4 √ 2 ) Q((1 − i) 4 Q( 2 ) √ Q( 2 ) Q(i) √ Q( 2 i) Q D4 (a) {id, σ2, τ, σ2τ } {id, σ, σ2, σ3}{id, σ2, στ, σ3τ } {id, τ } {id, σ2τ } {id, σ2} {id, στ } {id, σ3τ } {id} (b) Figure 23.3. Galois group of x4 − 2 Abel inspired the work of ´Evariste Galois. Born in 1811, Galois began to display extraordinary mathematical talent at the age of 14. He applied for entrance to the ´Ecole Polytechnique several times; however, he had great diﬃculty meeting the formal entrance requirements, and the examiners failed to recognize his mathematical genius. He was ﬁnally accepted at the ´Ecole Normale in 1829. Galois worked to develop a theory of solvability for polynomials. In 1829, at the age of 17, Galois presented two papers on the solution of algebraic equations to the Acad´emie des Sciences de Paris. These papers were sent to Cauchy, who subsequently lost them. A third paper was submitted to Fourier, who died before he could read the paper. Another paper was presented, but was not published until 1846. 390 CHAPTER 23 GALOIS THEORY Galois’ democratic sympathies led him into the Revolution of 1830. He was expelled from school and sent to prison for his part in the turmoil. After his release in 1832, he was drawn into a duel over a love aﬀair. Certain that he would be killed, he spent the evening before his death outlining his work and his basic ideas for research in a long letter to his friend Chevalier. He was indeed dead the next day, at the age of 20. 23.3 Applications Solvability by Radicals Throughout this section we shall assume that all ﬁeld
s have characteristic zero to ensure that irreducible polynomials do not have multiple roots. The immediate goal of this section is to determine when the roots of a polynomial f (x) can be computed in a ﬁnite number of operations on the coeﬃcients of f (x). The allowable operations are addition, subtraction, multiplication, division, and the extraction of nth roots. Certainly the solution to the quadratic equation, ax2 + bx + c = 0, illustrates this process: −b ± x = √ b2 − 4ac 2a. The only one of these operations that might demand a larger ﬁeld is the taking of nth roots. We are led to the following deﬁnition. An extension ﬁeld E of a ﬁeld F is an extension by radicals if there are elements α1,..., αr ∈ K and positive integers n1,..., nr such that E = F (α1,..., αr), where αn1 1 ∈ F and αni i ∈ F (α1,..., αi−1) for i = 2,..., r. A polynomial f (x) is solvable by radicals over F if the splitting ﬁeld K of f (x) over F is contained in an extension of F by radicals. Our goal is to arrive at criteria that will tell us whether or not a polynomial f (x) is solvable by radicals by examining the Galois group f (x). The easiest polynomial to solve by radicals is one of the form xn − a. As we discussed in Chapter 4, the roots of xn − 1 are called the nth roots of unity. These roots are a ﬁnite subgroup of the splitting ﬁeld of xn − 1. By 23.3 APPLICATIONS 391 Theorem 22.7, the nth roots of unity form a cyclic group. Any generator of this group is called a primitive nth root of unity. Example 8. The polynomial xn − 1 is solvable by radicals over Q. The roots of this polynomial are 1, ω, ω2,..., ωn−1, where ω = cos 2π n +
i sin 2π n. The splitting ﬁeld of xn − 1 over Q is Q(ω). Recall that a subnormal series of a group G is a ﬁnite sequence of subgroups G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}, where Hi is normal in Hi+1. A subnormal series is a composition series if all the factor groups are simple; that is, if none of the factor groups of the series contains a normal subgroup. A group G is solvable if it has a composition series {Hi} such that all of the factor groups Hi+1/Hi are abelian. For example, if we examine the series {id} ⊂ A3 ⊂ S3, we see that A3 is solvable. On the other hand, S5 is not solvable, by Theorem 10.6. Lemma 23.16 Let F be a ﬁeld of characteristic zero and E be the splitting ﬁeld of xn − a over F with a ∈ F. Then G(E/F ) is a solvable group. √ a, ω n √ a,..., ωn−1 n Proof. First suppose that F contains all of its nth roots of unity. The roots √ of xn − a are n a, where ω is a primitive nth root of unity. If ζ is one of these roots, then distinct roots of xn − 1 are ζ, ωζ,..., ωn−1ζ, and E = F (ζ). Since G(E/F ) permutes the roots xn − 1, the elements in G(E/F ) must be determined by their action on these roots. Let σ and τ be in G(E/F ) and suppose that σ(ζ) = ωiζ and τ (ζ) = ωjζ. If F contains the roots of unity, then στ (ζ) = σ(ωjζ) = ωjσ(ζ) = ωijζ = ωiτ (ζ) = τ (ωiζ) = τ σ(ζ). Therefore, στ = τ
σ and G(E/F ) is abelian, and G(E/F ) is solvable. Suppose that F does not contain a primitive nth root of unity. Let ω be a generator of the cyclic group of the nth roots of unity. Let α be a zero of xn − a. Since α and ωα are both in the splitting ﬁeld of xn − a, ω = (ωα)/α is also in E. Let K = F (ω). Then F ⊂ K ⊂ E. Since K is the splitting ﬁeld of xn − 1, K is a normal extension of F. Any automorphism σ in G(F (ω)/F ) is determined by σ(ω). It must be the case that σ(ω) = ωi for some integer i 392 CHAPTER 23 GALOIS THEORY since all of the zeros of xn − 1 are powers of ω. If τ (ω) = ωj is in G(F (ω)/F ), then στ (ω) = σ(ωj) = [σ(ω)]j = ωij = [τ (ω)]i = τ (ωi) = τ σ(ω). Therefore, G(F (ω)/F ) is abelian. By the Fundamental Theorem of Galois Theory the series {id} ⊂ G(E/F (ω)) ⊂ G(E/F ) is a normal series. Since G(E/F (ω)) and G(E/F )/G(E/F (ω)) ∼= G(F (ω)/F ) are both abelian, G(E/F ) is solvable. Lemma 23.17 Let F be a ﬁeld of characteristic zero and let E be a radical extension of F. Then there exists a normal radical extension K of F that contains E. Proof. Since E is a radical extension of F, there exist elements α1,..., αr ∈ K and positive integers n1,..., nr such that E = F (α1,..., αr), where αn1 1 ∈ F and αni i ∈ F (α1,..., αi−1) for i = 2
,..., r. Let f (x) = f1(x) · · · fr(x), where fi is the minimal polynomial of αi over F, and let K be the splitting ﬁeld of K over F. Every root of f (x) in K is of the form σ(αi), where σ ∈ G(K/F ). Therefore, for any σ ∈ G(K/F ), we have [σ(α1)]n1 ∈ F and [σ(αi)]ni ∈ F (α1,..., αi−1) for i = 2,..., r. Hence, if G(K/F ) = {σ1 = id, σ2,..., σk}, then K = F (σ1(αj)) is a radical extension of F. We will now prove the main theorem about solvability by radicals. Theorem 23.18 Let f (x) be in F [x], where charF = 0. If f (x) is solvable by radicals, then the Galois group of f (x) over F is solvable. Proof. Let K be a splitting ﬁeld of f (x) over F. Since f (x) is solvable, there exists an extension E of radicals F = F0 ⊂ F1 ⊂ · · · Fn = E. Since Fi is normal over Fi−1, we know by Lemma 23.17 that E is a normal extension of each Fi. By the Fundamental Theorem of Galois Theory, G(E/Fi) is a 23.3 APPLICATIONS 393 normal subgroup of G(E/Fi−1). Therefore, we have a subnormal series of subgroups of G(E/F ): {id} ⊂ G(E/Fn−1) ⊂ · · · ⊂ G(E/F1) ⊂ G(E/F ). Again by the Fundamental Theorem of Galois Theory, we know that G(E/Fi−1)/G(E/Fi) ∼= G(Fi/Fi−1). By Lemma 23.16, G(Fi/Fi−1) is solvable; hence, G(E/F ) is also solvable. The converse of Theorem 23.
18 is also true. For a proof, see any of the references at the end of this chapter. Insolvability of the Quintic We are now in a position to ﬁnd a ﬁfth-degree polynomial that is not solvable by radicals. We merely need to ﬁnd a polynomial whose Galois group is S5. We begin by proving a lemma. Lemma 23.19 If p is prime, then any subgroup of Sp that contains a transposition and a cycle of length p must be all of Sp. Proof. Let G be a subgroup of Sp that contains a transposition σ and τ a cycle of length p. We may assume that σ = (12). The order of τ is p and τ n must be a cycle of length p for 1 ≤ n < p. Therefore, we may assume that µ = τ n = (12i3... ip) for some n, where 1 ≤ n < p (see Exercise 13 in Chapter 5). Noting that (12)(12i3... ip) = (2i3... ip) and (2i3... ip)k(12)(2i3... ip)−k = (1ik), we can obtain all the transpositions of the form (1n) for 1 ≤ n < p. However, these transpositions generate all transpositions in Sp, since (1j)(1i)(1j) = (ij). The transpositions generate Sp. Example 9. We will show that f (x) = x5 − 6x3 − 27x − 3 ∈ Q[x] is not solvable. We claim that the Galois group of f (x) over Q is S5. By Eisenstein’s Criterion, f (x) is irreducible and, therefore, must be separable. The derivative of f (x) is f (x) = 5x4 − 18x2 − 27; hence, setting f (x) = 0 and solving, we ﬁnd that the only real roots of f (x) are x = ± √ 6 6 + 9 5. 394 CHAPTER 23 GALOIS THEORY y f (x) = x5 − 6x3 − 27x − 3 40 -40 2 4 x -4 -2 Figure 23.4.
The graph of f (x) = x5 − 6x3 − 27x − 3 Therefore, f (x) can have at most one maximum and one minimum. It is easy to show that f (x) changes sign between −3 and −2, between −2 and 0, and once again between 0 and 4 (Figure 23.4). Therefore, f (x) has exactly three distinct real roots. The remaining two roots of f (x) must be complex conjugates. Let K be the splitting ﬁeld of f (x). Since f (x) has ﬁve distinct roots in K and every automorphism of K ﬁxing Q is determined by the way it permutes the roots of f (x), we know that G(K/Q) is a subgroup of S5. Since f is irreducible, there is an element in σ ∈ G(K/Q) such that σ(a) = b for two roots a and b of f (x). The automorphism of C that takes a + bi → a − bi leaves the real roots ﬁxed and interchanges the complex roots; consequently, G(K/Q) ⊂ S5. By Lemma 23.19, S5 is generated by a transposition and an element of order 5; therefore, G(K/Q) must be all of S5. By Theorem 10.6, S5 is not solvable. Consequently, f (x) cannot be solved by radicals. 23.3 APPLICATIONS 395 The Fundamental Theorem of Algebra It seems ﬁtting that the last theorem that we will state and prove is the Fundamental Theorem of Algebra. This theorem was ﬁrst proven by Gauss in his doctoral thesis. Prior to Gauss’s proof, mathematicians suspected that there might exist polynomials over the real and complex numbers having no solutions. The Fundamental Theorem of Algebra states that every polynomial over the complex numbers factors into distinct linear factors. Theorem 23.20 (Fundamental Theorem of Algebra) The ﬁeld of complex numbers is algebraically closed; that is, every polynomial in C[x] has a root in C. For our proof we shall assume two facts from calculus. We need the results that every polynomial of odd degree over R has a real root
and that every positive real number has a square root. Proof. Suppose that E is a proper ﬁnite ﬁeld extension of the complex numbers. Since any ﬁnite extension of a ﬁeld of characteristic zero is a simple extension, there exists an α ∈ E such that E = C(α) with α the root of an irreducible polynomial f (x) in C[x]. The splitting ﬁeld L of f (x) is a ﬁnite normal separable extension of C that contains E. We must show that it is impossible for L to be a proper extension of C. Suppose that L is a proper extension of C. Since L is the splitting ﬁeld of f (x)(x2 + 1) over R, L is a ﬁnite normal separable extension of R. Let K be the ﬁxed ﬁeld of a Sylow 2-subgroup G of G(L/R). Then L ⊃ K ⊃ R and \|G(L/K)\| = [L : K]. Since [L : R] = [L : K][K : R], we know that [K : R] must be odd. Consequently, K = R(β) with β having a minimal polynomial f (x) of odd degree. Therefore, K = R. We now know that G(L/R) must be a 2-group. It follows that G(L/C) is a 2-group. We have assumed that L = C; therefore, \|G(L/C)\| ≥ 2. By the ﬁrst Sylow Theorem and the Fundamental Theorem of Galois Theory, there exists a subgroup G of G(L/C) of index 2 and a ﬁeld E ﬁxed elementwise by G. Then [E : C] = 2 and there exists an element γ ∈ E with minimal polynomial x2 + bx + c in C[x]. This polynomial has roots (−b ± b2 − 4c )/2 that are in C, since b2 − 4c is in C. This is impossible; hence, L = C. √ Although our proof was strictly algebraic, we were forced to rely on results from calculus. It is necessary to assume
the completeness axiom from analysis to show that every polynomial of odd degree has a real root and that every positive real number has a square root. It seems that there is no possible way to avoid this diﬃculty and formulate a purely algebraic 396 CHAPTER 23 GALOIS THEORY argument. It is somewhat amazing that there are several elegant proofs of the Fundamental Theorem of Algebra that use complex analysis. It is also interesting to note that we can obtain a proof of such an important theorem from two very diﬀerent ﬁelds of mathematics. Exercises 1. Compute each of the following Galois groups. Which of these ﬁeld extensions are normal ﬁeld extensions? If the extension is not normal, ﬁnd a normal extension of Q in which the extension ﬁeld is contained. √ 2, 3 (d) G(Q( 2, i)/Q) √ √ (a) G(Q( √ (b) G(Q( 4 √ (c) G(Q( 30 )/Q) 5 )/Q) √ 3, 2, √ 5 )/Q) √ (e) G(Q( 6, i)/Q) 2. Determine the separability of each of the following polynomials. (a) x3 + 2x2 − x − 2 over Q (b) x4 + 2x2 + 1 over Q (c) x4 + x2 + 1 over Z3 (d) x3 + x2 + 1 over Z2 3. Give the order and describe a generator of the Galois group of GF(729) over GF(9). 4. Determine the Galois groups of each of the following polynomials in Q[x]; hence, determine the solvability by radicals of each of the polynomials. (a) x5 − 12x2 + 2 (b) x5 − 4x4 + 2x + 2 (c) x3 − 5 (d) x4 − x2 − 6 (e) x5 + 1 (f) (x2 − 2)(x2 + 2) (g) x8 − 1 (h) x8 + 1 (i) x4 − 3x2 − 10 5. Find a primitive element in the splitting ﬁeld
of each of the following polyno- mials in Q[x]. (a) x4 − 1 (b) x4 − 8x2 + 15 (c) x4 − 2x2 − 15 (d) x3 − 2 6. Prove that the Galois group of an irreducible quadratic polynomial is isomor- phic to Z2. 7. Prove that the Galois group of an irreducible cubic polynomial is isomorphic to S3 or Z3. EXERCISES 397 8. Let F ⊂ K ⊂ E be ﬁelds. If E is a normal extension of F, show that E must also be a normal extension of K. 9. Let G be the Galois group of a polynomial of degree n. Prove that \|G\| divides n!. 10. Let F ⊂ E. If f (x) is solvable over F, show that f (x) is also solvable over E. 11. Construct a polynomial f (x) in Q[x] of degree 7 that is not solvable by radicals. 12. Let p be prime. Prove that there exists a polynomial f (x) ∈ Q[x] of degree p with Galois group isomorphic to Sp. Conclude that for each prime p with p ≥ 5 there exists a polynomial of degree p that is not solvable by radicals. 13. Let p be a prime and Zp(t) be the ﬁeld of rational functions over Zp. Prove that f (x) = xp − t is an irreducible polynomial in Zp(t)[x]. Show that f (x) is not separable. 14. Let E be an extension ﬁeld of F. Suppose that K and L are two intermediate ﬁelds. If there exists an element σ ∈ G(E/F ) such that σ(K) = L, then K and L are said to be conjugate ﬁelds. Prove that K and L are conjugate if and only if G(E/K) and G(E/L) are conjugate subgroups of G(E/F ). 15. Let σ ∈ Aut(R). If a is a positive real number, show
that σ(a) > 0. 16. Let K be the splitting ﬁeld of x3 + x2 + 1 ∈ Z2[x]. Prove or disprove that K is an extension by radicals. 17. Let F be a ﬁeld such that char F = 2. Prove that the splitting ﬁeld of f (x) = ax2 + bx + c is F ( α ), where α = b2 − 4ac. √ 18. Prove or disprove: Two diﬀerent subgroups of a Galois group will have diﬀerent ﬁxed ﬁelds. 19. Let K be the splitting ﬁeld of a polynomial over F. If E is a ﬁeld extension of F contained in K and [E : F ] = 2, then E is the splitting ﬁeld of some polynomial in F [x]. 20. We know that the cyclotomic polynomial Φp(x) = xp − 1 x − 1 = xp−1 + xp−2 + · · · + x + 1 is irreducible over Q for every prime p. Let ω be a zero of Φp(x), and consider the ﬁeld Q(ω). (a) Show that ω, ω2,..., ωp−1 are distinct zeros of Φp(x), and conclude that they are all the zeros of Φp(x). (b) Show that G(Q(ω)/Q) is abelian of order p − 1. (c) Show that the ﬁxed ﬁeld of G(Q(ω)/Q) is Q. 398 CHAPTER 23 GALOIS THEORY 21. Let F be a ﬁnite ﬁeld or a ﬁeld of characteristic zero. Let E be a ﬁnite normal extension of F with Galois group G(E/F ). Prove that F ⊂ K ⊂ L ⊂ E if and only if {id} ⊂ G(E/L) ⊂ G(E/K) ⊂ G(E/F ). 22. Let F be a ﬁeld of characteristic zero
and let f (x) ∈ F [x] be a separable polynomial of degree n. If E is the splitting ﬁeld of f (x), let α1,..., αn be the roots of f (x) in E. Let ∆ = i=j(αi − αj). We deﬁne the discriminant of f (x) to be ∆2. (a) If f (x) = ax2 + bx + c, show that ∆2 = b2 − 4ac. (b) If f (x) = x3 + px + q, show that ∆2 = −4p3 − 27q2. (c) Prove that ∆2 is in F. (d) If σ ∈ G(E/F ) is a transposition of two roots of f (x), show that σ(∆) = − ∆. (e) If σ ∈ G(E/F ) is an even permutation of the roots of f (x), show that σ(∆) = ∆. (f) Prove that G(E/F ) is isomorphic to a subgroup of An if and only if ∆ ∈ F. (g) Determine the Galois groups of x3 + 2x − 4 and x3 + x − 3. References and Suggested Readings [1] Artin, E. Theory: Lectures Delivered at the University of Notre Dame (Notre Dame Mathematical Lectures, Number 2). Dover, Mineola, NY, 1997. [2] Edwards, H. M. Galois Theory. Springer-Verlag, New York, 1984. [3] Fraleigh, J. B. A First Course in Abstract Algebra. 7th ed. Pearson, Upper Saddle River, NJ, 2003. [4] Gaal, L. Classical Galois Theory with Examples. American Mathematical Society, Providence, 1979. [5] Garling, D. J. H. A Course in Galois Theory. Cambridge University Press, Cambridge, 1986. [6] Kaplansky, I. Fields and Rings. 2nd ed. University of Chicago Press, Chicago, 1972. [7] Rothman, T. “The Short Life of ´Evariste Galois,” Scientiﬁc American
, April 1982, 136–49. Sage Fields, ﬁeld extensions, roots of polynomials, and group theory — Sage has it all, and so it is possible to carefully study very complicated examples from Galois theory with Sage. Hints and Solutions Chapter 1. Preliminaries 1. (a) {2}. (b) {5}. 2. (a) {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3), (c, 1), (c, 2), (c, 3)}. (d) ∅. 6. If x ∈ A ∪ (B ∩ C), then either x ∈ A or and A ∪ C ⇒ x ∈ (A ∪ B) ∩ (A ∪ C) ⇒ A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C). Conversely, x ∈ (A ∪ B) ∩ (A ∪ C) ⇒ x ∈ A ∪ B and A ∪ C ⇒ x ∈ A or x is in both B and C⇒ x ∈ A∪(B∩C) ⇒ (A∪B)∩(A∪C) ⊂ A∪(B∩C). Hence, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). 10. (A ∩ B) ∪ (A \ B) ∪ (B \ A) = (A ∩ B) ∪ (A ∩ B) ∪ (B ∩ A) = [A ∩ (B ∪ B)] ∪ (B ∩ A) = A ∪ (B ∩ A) = (A ∪ B) ∩ (A ∪ A) = A ∪ B. 14. A \ (B ∪ C) = A ∩ (B ∪ C) = (A ∩ A) ∩ (B ∩ C ) = (A ∩ B) ∩ (A ∩ C ) = (A \ B) ∩ (A \ C). 17. (a) Not a map. f (2/3) is undeﬁned. (c) Not a map
. f (1/2) = 3/4 and f (2/4) = 3/8. 18. (a) One-to-one but not onto. f (R) = {x ∈ R : x > 0}. (c) Neither one-to-one nor onto. 20. (a) f (n) = n + 1. 22. (a) Let x, y ∈ A. Then g(f (x)) = (g ◦ f )(x) = (g ◦ f )(y) = g(f (y)) ⇒ f (x) = f (y) ⇒ x = y, so g ◦ f is one-to-one. (b) Let c ∈ C, then c = (g ◦ f )(x) = g(f (x)) for some x ∈ A. Since f (x) ∈ B, g is onto. 23. f −1(x) = (x + 1)/(x − 1). 24. (a) Let y ∈ f (A1 ∪ A2) ⇒ there exists an x ∈ A1 ∪ A2 such that f (x) = y ⇒ y ∈ f (A1) or f (A2) ⇒ y ∈ f (A1) ∪ f (A2) ⇒ f (A1 ∪ A2) ⊂ f (A1) ∪ f (A2). 399 400 HINTS AND SOLUTIONS Conversely, let y ∈ f (A1) ∪ f (A2) ⇒ y ∈ f (A1) or f (A2) ⇒ there exists an x ∈ A1 or there exists an x ∈ A2 such that f (x) = y ⇒ there exists an x ∈ A1 ∪ A2 such that f (x) = y ⇒ f (A1) ∪ f (A2) ⊂ f (A1 ∪ A2). Hence, f (A1 ∪ A2) = f (A1) ∪ f (A2). 25. (a) Not an equivalence relation. Fails to be symmetric. (c) Not an equivalence relation. Fails to be transitive. 28. Let X = N ∪ { 2 } and de�
�ne x ∼ y if x + y ∈ N. √ Chapter 2. The Integers 1. S(1) : [1(1 + 1)(2(1) + 1)]/6 = 1 = 12 is true. Assume S(k) : 12 + 22 + · · · + k2 = [k(k + 1)(2k + 1)]/6 is true. Then 12 + 22 + · · · + k2 + (k + 1)2 = [k(k + 1)(2k + 1)]/6 + (k + 1)2 = [(k + 1)((k + 1) + 1)(2(k + 1) + 1)]/6, so S(k + 1) is true. Thus S(n) is true for all positive integers n. 3. S(4) : 4! = 24 > 16 = 24 is true. Assume S(k) : k! > 2k is true. Then (k + 1)! = k!(k + 1) > 2k · 2 = 2k+1, so S(k + 1) is true. Thus S(n) is true for all positive integers n. 8. Look at the proof in Example 3. 11. S(0) : (1 + x) is true. Assume S(k) : (1 + x)k − 1 ≥ kx is true. Then (1 + x)k+1 − 1 = (1 + x)(1 + x)k − 1 = (1 + x)k + x(1 + x)k − 1 ≥ kx + x(1 + x)k ≥ kx + x = (k + 1)x, so S(k + 1) is true. Thus S(n) is true for all positive integers n. 15. (a) (14)14 + (−5)39 = 1. (c) (3709)1739 + (−650)9923 = 1. (e) (881)23771 + (−1050)19945 = 1. 17. (b) Use mathematical induction. (c) Show that f1 = 1, f2 = 1, and fn+2 = fn+1 + fn. (d) Use part (c). (e) Use part (b) and Problem 16. 19. Use the Fundamental Theorem of Arithmetic. 23
. Let S = {s ∈ N : a \| s, b \| s}. S = ∅, since \|ab\| ∈ S. By the Principle of Well-Ordering, S contains a least element m. To show uniqueness, suppose that a \| n and b \| n for some n ∈ N. By the division algorithm, there exist unique integers q and r such that n = mq + r, where 0 ≤ r < m. a \| m, b \| m, a \| n, b \| n ⇒ a \| r, b \| r ⇒ r = 0 by the minimality of m. Therefore, m \| n. 27. Since gcd(a, b) = 1, there exist integers r and s such that ar + bs = 1 ⇒ acr + bcs = c. Since a \| a and a \| bc, a \| c. 29. Let p = p1p2 · · · pk + 1, where p1 = 2, p2 = 3,..., pk are the ﬁrst k primes. Show that p is prime. HINTS AND SOLUTIONS 401 Chapter 3. Groups 1. (a) {..., −4, 3, 10,...}. (c) {..., −8, 18, 44,...}. (e) {..., −1, 5, 11,...}. 2. (a) Not a group. (c) A group. 6. · 1 5 7 11 1 1 5 7 11 5 5 1 11 7 7 7 11 1 5 11 11 7 5 1 8. Pick two matrices. Almost any pair will work. 15. There is a group of order 6 that is nonabelian. 16. Look at the symmetry group of an equilateral triangle or a square. 17. There are actually ﬁve diﬀerent groups of order 8. 18. Let σ = 1 2 a1 a2 · · · · · · n an be in Sn. All of the ai’s must be distinct. There are n ways to choose a1, n − 1 ways to choose a2,..., 2 ways to choose an−1, and only one way to choose an. Therefore, we can form σ in n(n − 1) · · · 2 · 1 = n!
ways. 24. (aba−1)n = (aba−1)(aba−1) · · · (aba−1) = ab(aa−1)b(aa−1)b · · · (aa−1)ba−1 = abna−1. 30. abab = (ab)2 = e = a2b2 = aabb ⇒ ba = ab. 34. H1 = {id}, H2 = {id, ρ1, ρ2}, H3 = {id, µ1}, H4 = {id, µ2}, H5 = {id, µ3}, S3. 40. id = 1 = 1 + 0 √ √ √ 2, (a + b 2 )(c + d √ √ 2 ) = (ac + 2bd) + (ad + bc) √ 2, and (a + b 2 )−1 = a/(a2 − 2b2) − b 2/(a2 − 2b2). 45. Not a subgroup. Look at S3. 48. a4b = ba ⇒ b = a6b = a2ba ⇒ ab = a3ba = ba. Chapter 4. Cyclic Groups 1. (a) False. (c) False. (e) True. 2. (a) 12. (c) Inﬁnite. (e) 10. 3. (a) 7Z = {..., −7, 0, 7, 14,...}. (b) {0, 3, 6, 9, 12, 15, 18, 21}. (c) {0}, {0, 6}, {0, 4, 8}, {0, 3, 6, 9}, {0, 2, 4, 6, 8, 10}. (g) {1, 3, 7, 9}. (j) {1, −1, i, −i}. 402 4. (a) (c) 1 0 0 1, −1 0 0 −1 HINTS AND SOLUTIONS 0 −1 0 1, 0 −1, 1 0. 1 0 0 −1 0 1 1 1,, 1 −1 0 1 0 −1 1 −1 −1 −1 1 0, −1 0 0 −1.,, 10. (a) 0, 1, −1. (b) 1, −1.
11. 1, 2, 3, 4, 6, 8, 12, 24. 15. (a) 3i − 3. (c) 43 − 18i. (e) i. √ √ 16. (a) 17. (a) 3 + i. (c) −3. 2 cis(7π/4). (c) 2 √ 18. (a) (1 − i)/2. (c) 16(i − 22. (a) 292. (c) 1523. 27. \|g ∩ h\| = 1. 2 cis(π/4). (e) 3 cis(3π/2). √ 3 ). (e) −1/4. 31. The identity element in any group has ﬁnite order. Let g, h ∈ G have orders m and n, respectively. Since (g−1)m = e and (gh)mn = e, the elements of ﬁnite order in G form a subgroup of G. 37. If g is an element distinct from the identity in G, g must generate G; otherwise, g is a nontrivial proper subgroup of G. Chapter 5. Permutation Groups 1. (a) (12453). (c) (13)(25). 2. (a) (135)(24). (c) (14)(23). (e) (1324). (g) (134)(25). (n) (17352). 3. (a) (16)(15)(13)(14). (c) (16)(14)(12). 4. (a1, an, an−1,..., a2). 5. (a) {(13), (13)(24), (132), (134), (1324), (1342)}. Not a subgroup. 8. (12345)(678). 11. Permutations of the form (1), (a1, a2)(a3, a4), (a1, a2, a3), (a1, a2, a3, a4, a5) are possible for A5. 17. (123)(12) = (13) = (23) = (12)(123). 25. Use the fact that (ab)(bc) = (abc) and (ab)(cd) = (abc)(bcd). 30. (a) Show
that στ σ−1(i) = (σ(a1), σ(a2),..., σ(ak))(i) for 1 ≤ i ≤ n. HINTS AND SOLUTIONS 403 Chapter 6. Cosets and Lagrange’s Theorem 1. The order of g and the order h must both divide the order of G. The smallest number that 5 and 7 both divide is lcm(5, 7) = 35. 2. 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. 3. False. 4. False. 5. (a) (c) H = {0, 8, 16} 1 + H = {1, 9, 17} 2 + H = {2, 10, 18} 3 + H = {3, 11, 19} 4 + H = {4, 12, 20} 5 + H = {5, 13, 21} 6 + H = {6, 14, 22} 7 + H = {7, 15, 23}. 3Z = {..., −3, 0, 3, 6,...} 1 + 3Z = {..., −2, 1, 4, 7,...} 2 + 3Z = {..., −1, 2, 5, 8,...}. 7. 4φ(15) ≡ 48 ≡ 1 (mod 15). 12. Let g1 ∈ gH. Then there exists an h ∈ H such that g1 = gh = ghg−1g ⇒ g1 ∈ Hg ⇒ gH ⊂ Hg. Similarly, Hg ⊂ gH. Therefore, gH = Hg. 17. If a /∈ H, then a−1 /∈ H ⇒ a−1 ∈ aH = a−1H = bH ⇒ there exist h1, h2 ∈ H such that a−1h1 = bh2 ⇒ ab = h1h−1 2 ∈ H. Chapter 7. Introduction to Cryptography 1. LAORYHAPDWK. 3. Hint: Q = E, F = X, A = R. 4. 26! − 1. 7. (a) 2791. (c) 112135 25032 442. 9. (
a) 31. (c) 14. 10. (a) n = 11 · 41. (c) n = 8779 · 4327. Chapter 8. Algebraic Coding Theory 2. (0000) /∈ C. 3. (a) 2. (c) 2. 4. (a) 3. (c) 4. 6. (a) dmin = 2. (c) dmin = 1. 404 HINTS AND SOLUTIONS 7. (a) (00000), (00101), (10011), (10110b) (00000), (010111), (101101), (111010. Multiple errors occur in one of the received words. 11. (a) A canonical parity-check matrix with standard generator matrix c) A canonical parity-check matrix with standard generator matrix. 12. (a) All possible syndromes occur. 15. (a) The cosets of C are C Cosets (00000) (00101) (10011) (10110) (10000) + C (10000) (10101) (00011) (00110) (01000) + C (01000) (01101) (11011) (11110) (00100) + C (00100) (00001) (10111) (10010) (00010) + C (00010) (00111) (10001) (10100) (11000) + C (11000) (11101) (01011) (01110) (01100) + C (01100) (01001) (11111) (11010) (01010) + C (01010) (01111) (11001) (11100) HINTS AND SOLUTIONS 405 A decoding table does not exist for C since it is only single error-detecting. 19. Let x ∈ C have odd weight and deﬁne a map from the set of odd codewords to the set of even codewords by y → x + y. Show that this map is a bijection. 23. For 20 information positions, at least six check bits are needed to ensure an error-correcting code. Chapter 9. Isomorphisms 1. The group nZ is an inﬁnite cyclic group generated by n. Every inﬁnite cyclic group is isomorphic to Z
. 2. Deﬁne φ : C∗ → GL2(R) by φ(a + bi) = a b −b a. 3. False. 6. Deﬁne a map from Zn into the nth roots of unity by k → cis(2kπ/n). 8. Assume that Q is cyclic and try to ﬁnd a generator. 11. D4, Q8, Z8, Z2 × Z4, Z2 × Z2 × Z2. 16. (a) 12. (c) 5. 20. True. 25. Z2 × Z2 × Z13 is not cyclic. 27. Let a be a generator for G. If φ : G → H is an isomorphism, show that φ(a) is a generator for H. 38. Any automorphism of Z6 must send 1 to another generator of Z6. 45. To show that φ is one-to-one, let g1 = h1k1 and g2 = h2k2. Then φ(g1) = φ(g2) ⇒ φ(h1k1) = φ(h2k2) ⇒ (h1, k1) = (h2, k2) ⇒ h1 = h2, k1 = k2 ⇒ g1 = g2. Chapter 10. Normal Subgroups and Factor Groups 1. (a) A4 (12) A4 A4 A4 (12)A4 (12)A4 (12)A4 A4 (c) D4 is not normal in S4. 8. If a ∈ G is a generator for G, then aH is a generator for G/H. 406 HINTS AND SOLUTIONS 13. Since eg = ge for all g ∈ G, the identity is in C(g). If x, y ∈ C(g), then xyg = xgy = gxy ⇒ xy ∈ C(g). If xg = gx, then x−1g = gx−1 ⇒ x−1 ∈ C(g) ⇒ C(g) is a subgroup of G. If g is normal in G, then g1xg−1 for all g1 ∈ G. 1 g = gg1xg−1
1 15. (a) Let g ∈ G and h ∈ G. If h = aba−1b−1, then ghg−1 = gaba−1b−1g−1 = (gag−1)(gbg−1)(ga−1g−1)(gb−1g−1) = (gag−1)(gbg−1)(gag−1)−1(gbg−1)−1. We also need to show that if h = h1 · · · hn with hi = aibia−1, then ghg−1 is a product of elements of the same type. However, ghg−1 = gh1 · · · hng−1 = (gh1g−1)(gh2g−1) · · · (ghng−1). i b−1 i Chapter 11. Homomorphisms 2. (a) A homomorphism. (c) Not a homomorphism. 4. φ(m + n) = 7(m + n) = 7m + 7n = φ(m) + φ(n). The kernel of φ is {0} and the image of φ is 7Z. 5. For any homomorphism φ : Z24 → Z18, the kernel of φ must be a subgroup of Z24 and the image of φ must be a subgroup of Z18. 9. Let a, b ∈ G. Then φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a). Chapter 12. Matrix Groups and Symmetry 1. 1 2 x + y2 + x2 − y2 = x + y, x + y − x2 − y2 x2 + 2x, y + y2 − x2 − y2 1 2 1 2 = x, y. = 3. (a) An element of SO(2). (c) Not in O(3). 5. (a) x, y = x1y1 + · · · + xnyn = y1x1 + · · · + ynxn = y, x. 7. Use the unimodular matrix 5 2 2 1. 10. Show that the kernel of the map det : O(n) → R∗ is SO(n). 13. True
. 17. p6m. HINTS AND SOLUTIONS 407 Chapter 13. The Structure of Groups 1. Since 40 = 23 · 5, the possible abelian groups of order 40 are Z40 Z5 × Z4 × Z2, and Z5 × Z2 × Z2 × Z2. ∼= Z8 × Z5, 4. (a) {0} ⊂ 6 ⊂ 3 ⊂ Z12. (e) {((1), 0)} ⊂ {(1), (123), (132)} × {0} ⊂ S3 × {0} ⊂ S3 × 2 ⊂ S3 × Z4. 7. Use the Fundamental Theorem of Finitely Generated Abelian Groups. 12. If N and G/N are solvable, then they have solvable series N = Nn ⊃ Nn−1 ⊃ · · · ⊃ N1 ⊃ N0 = {e} G/N = Gn/N ⊃ Gn−1/N ⊃ · · · G1/N ⊃ G0/N = {N }. The series G = Gn ⊃ Gn−1 ⊃ · · · ⊃ G0 = N = Nn ⊃ Nn−1 ⊃ · · · ⊃ N1 ⊃ N0 = {e} is a subnormal series. The factors of this series are abelian since Gi+1/Gi (Gi+1/N )/(Gi/N ). ∼= 16. Use the fact that Dn has a cyclic subgroup of index 2. 21. G/G is abelian. Chapter 14. Group Actions 1. Example 1. 0, R2 \ {0}. Example 2. X = {1, 2, 3, 4}. 2. (a) X(1) = {1, 2, 3}, X(12) = {3}, X(13) = {2}, X(23) = {1}, X(123) = X(132) = ∅. G1 = {(1), (23)}, G2 = {(1), (13)}, G3 = {(1), (12)}. 3. (a) O1 = O2 = O3 = {1, 2, 3}. 6. (a) O
(1) = {(1)}, O(12) = {(12), (13), (14), (23), (24), (34)}, O(12)(34) = {(12)(34), (13)(24), (14)(23)}, O(123) = {(123), (132), (124), (142), (134), (143), (234), (243)}, O(1234) = {(1234), (1243), (1324), (1342), (1423), (1432)}. The class equation is 1 + 3 + 6 + 6 + 8 = 24. 8. (34 + 31 + 32 + 31 + 32 + 32 + 33 + 33)/8 = 21. 11. (1 · 34 + 6 · 33 + 11 · 32 + 6 · 31)/24 = 15. 15. (1 · 26 + 3 · 24 + 4 · 23 + 2 · 22 + 2 · 21)/12 = 13. 17. (1 · 28 + 3 · 26 + 2 · 24)/6 = 80. 22. x ∈ gC(a)g−1 ⇐⇒ g−1xg ∈ C(a) ⇐⇒ ag−1xg = g−1xga ⇐⇒ gag−1x = xgag−1 ⇐⇒ x ∈ C(gag−1). 408 HINTS AND SOLUTIONS Chapter 15. The Sylow Theorems 1. If \|G\| = 18 = 2 · 32, then the order of a Sylow 2-subgroup is 2, and the order of a Sylow 3-subgroup is 9. If \|G\| = 54 = 2 · 33, then the order of a Sylow 2-subgroup is 2, and the order of a Sylow 3-subgroup is 27. 2. The four Sylow 3-subgroups of S4 are P1 = {(1), (123), (132)}, P2 = {(1), (124), (142)}, P3 = {(1), (134), (143)}, P4 = {(1), (234), (243)}. 5. Since \|G\| = 96 = 25 · 3, G has either one or three Sylow 2-subgroups by the Third Sylow Theorem. If there is only one subgroup
, we are done. If there are three Sylow 2-subgroups, let H and K be two of them. \|H ∩ K\| ≥ 16; otherwise, HK would have (32 · 32)/8 = 128 elements, which is impossible. H ∩ K is normal in both H and K since it has index 2 in both groups. Hence, N (H ∩ K) contains both H and K. Therefore, \|N (H ∩ K)\| must be a multiple of 32 greater than 1 and still divide 96, so N (H ∩ K) = G. 8. G has a Sylow q-subgroup of order q2. Since the number of such subgroups is congruent to 1 modulo q and divides p2q2, there must be either 1, p, or p2 Sylow q-subgroups. Since q \|p2 − 1 = (p − 1)(p + 1), there can be only one Sylow q-subgroup, say Q. Similarly, we can show that there is a single Sylow p-subgroup P. Every element in Q other than the identity has order q or q2, so P ∩ Q = {e}. Now show that hk = kh for h ∈ P and k ∈ Q. Deduce that G = P × Q is abelian. 10. False. 17. If G is abelian, then G is cyclic, since \|G\| = 3 · 5 · 17. Now look at Example 5. 23. Deﬁne a mapping between the right cosets of N (H) in G and the conjugates of H in G by N (H)g → g−1Hg. Prove that this map is a bijection. 26. Let aG, bG ∈ G/G. Then (aG)(bG) = abG = ab(b−1a−1ba)G = (abb−1a−1)baG = baG. Chapter 16. Rings 1. (a) 7Z is a ring but not a ﬁeld. (c) Q( √ 2 ) is a ﬁeld. (f) R is not a ring. 3. (a) {1, 3, 7, 9}. (c) {1, 2, 3, 4, 5, 6}. (e. (a)
{0}, {0, 9}, {0, 6, 12}, {0, 3, 6, 9, 12, 15}, {0, 2, 4, 6, 8, 10, 12, 14, 16}. (c) There are no nontrivial ideals. HINTS AND SOLUTIONS 409 7. Assume there is an isomorphism φ : C → R with φ(i) = a. 8. False. Assume there is an isomorphism φ : Q( 2 ) → Q( √ √ 3 ) such that √ φ( 2 ) = a. 13. (a) x ≡ 17 (mod 55). (c) x ≡ 214 (mod 2772). 16. If I = {0}, show that 1 ∈ I. 19. (a) φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a). 27. Let a ∈ R with a = 0. The principal ideal generated by a is R ⇒ there exists a b ∈ R such that ab = 1. 29. Compute (a + b)2 and (−ab)2. 35. Let a/b, c/d ∈ Z(p). Then a/b + c/d = (ad + bc)/bd and (a/b) · (c/d) = (ac)/(bd) are both in Z(p), since gcd(bd, p) = 1. 39. Suppose that x2 = x and x = 0. Since R is an integral domain, x = 1. To ﬁnd a nontrivial idempotent, look in M2(R). Chapter 17. Polynomials 2. (a) 9x2 + 2x + 5. (b) 8x4 + 7x3 + 2x2 + 7x. 3. (a) 5x3 + 6x2 − 3x + 4 = (5x22x + 1)(x − 2) + 6. (c) 4x5 − x3 + x2 + 4 = (4x2 + 4)(x3 + 3) + 4x2 + 2. 5. (a) No zeros in Z12. (c) 3, 4. 7. (2x + 1)2 = 1. 8. (a
) Reducible. (c) Irreducible. 10. x2 + x + 8 = (x + 2)(x + 9) = (x + 7)(x + 4). 13. Z is not a ﬁeld. 14. False. x2 + 1 = (x + 1)(x + 1). 16. Let φ : R → S be an isomorphism. Deﬁne φ : R[x] → S[x] by φ(a0 + a1x + · · · + anxn) = φ(a0) + φ(a1)x + · · · + φ(an)xn. 19. Deﬁne g(x) by g(x) = Φp(x + 1) and show that g(x) is irreducible over Q. 25. Find a nontrivial proper ideal in F [x]. 410 HINTS AND SOLUTIONS Chapter 18. Integral Domains 1. z−1 = 1/(a + b √ 3 i) = (a − b √ 3 i)/(a2 + 3b2) is in Z[ √ 3 i] if and only if a2 + 3b2 = 1. The only integer solutions to the equation are a = ±1, b = 0. 2. (a) 5 = 1 + 2i)(1 − 2i). (c) 6 + 8i = (−1 + 7i)(1 − i). 4. True. 9. Let z = a + bi and w = c + di = 0 be in Z[i]. Prove that z/w ∈ Q(i). 15. Let a = ub with u a unit. Then ν(b) ≤ ν(ub) ≤ ν(a). Similarly, ν(a) ≤ ν(b). 16. Show that 21 can be factored in two diﬀerent ways. Chapter 19. Lattices and Boolean Algebras 2. 30 10 15. False. 6. (a) (a ∨ b ∨ a) ∧ a. (c) a ∨ (a ∧ b). 8. Not equivalent. HINTS AND SOLUTIONS 411 10. a ∧ [(a ∧ b) ∨ b] = a ∧ (a ∨ b
). 15. Let I, J be ideals in R. We need to show that I +J = {r+s : r ∈ I and s ∈ J} is the smallest ideal in R containing both I and J. If r1, r2 ∈ I and s1, s2 ∈ J, then (r1 + s1) + (r2 + s2) = (r1 + r2) + (s1 + s2) is in I + J. For a ∈ R, a(r1 + s1) = ar1 + as1 ∈ I + J; hence, I + J is an ideal in R. 19. (a) No. 21. (⇒). a = b ⇒ (a ∧ b) ∨ (a ∧ b) = (a ∧ a) ∨ (a ∧ a) = O ∨ O = O. (⇐). (a∧b)∨(a ∧b) = O ⇒ a∨b = (a∨a)∨b = a∨(a∨b) = a∨[I ∧(a∨b)] = a∨[(a∨a)∧(a∨b)] = [a∨(a∧b)]∨[a∨(a∧b)] = a∨[(a∧b)∨(a∧b)] = a∨0 = a. A symmetric argument shows that a ∨ b = b. Chapter 20. Vector Spaces √ √ √ √ 2, 3 ) has basis {1, 2, 3. Q( √ 3, 6 } over Q. 5. Pn has basis {1, x, x2,..., xn−1}. 7. (a) Subspace of dimension 2 with basis {(1, 0, −3), (0, 1, 2)}. (d) Not a subspace. 10. 0 = α0 = α(−v + v) = α(−v) + αv ⇒ −αv = α(−v). 12. Let v0 = 0, v1,..., vn ∈ V and α0 = 0, α1,..., αn ∈ F.
Then α0v0 + · · · + αnvn = 0. 15. (a) Let u, v ∈ ker(T ) and α ∈ F. Then T (u + v) = T (u) + T (v) = 0 T (αv) = αT (v) = α0 = 0. Hence, u + v, αv ∈ ker(T ) ⇒ ker(T ) is a subspace of V. (c) T (u) = T (v) ⇔ T (u − v) = T (u) − T (v. 17. (a) Let u, u ∈ U and v, v ∈ V. Then (u + v) + (u + v) = (u + u) + (v + v) ∈ U + V α(u + v) = αu + αv ∈ U + V. Chapter 21. Fields 3 x2 − 62 9. (c) x4 − 2x2 + 25. √ √ √ 6 }. (c) {1, i, 3, 2, 1. (a) x4 − 2 √ 2. (a) {1, √ 2, √ √ 2 i}. (e) {1, 21/6, 21/3, 21/2, 22/3, 25/6}. 3, 3. (a) Q( 5. Use the fact that the elements of Z2[x]/x3 + x + 1 are 0, 1, α, 1 + α, α2, 7 ). 1 + α2, α + α2, 1 + α + α2 and the fact that α3 + α + 1 = 0. 412 8. False. HINTS AND SOLUTIONS 14. Suppose that E is algebraic over F and K is algebraic over E. Let α ∈ K. It suﬃces to show that α is algebraic over some ﬁnite extension of F. Since α is algebraic over E, it must be the zero of some polynomial p(x) = β0 + β1x + · · · + βnxn in E[x]. Hence α is algebraic over F (β0,..., βn). √ √ √ √ √ √ √ 3, 7 )
⊃ Q( over Q. Since [Q( degree of the minimal polynomial of 7 ) since {1, 3, 7 ) : Q] = 4, [Q( 3 + 3 + √ 3, √ 7, 3 + 7 is 4, Q( 21 } is a basis for Q( 7 ) √ 7 ) : Q] = 2 or 4. Since the √ √ 7 ) = Q( 7 ). 3 + √ √ 3, 3, √ √ √ 22. Q( √ √ 27. Let β ∈ F (α) not in F. Then β = p(α)/q(α), where p and q are polynomials in α with q(α) = 0 and coeﬃcients in F. If β is algebraic over F, then there exists a polynomial f (x) ∈ F [x] such that f (β) = 0. Let f (x) = a0 +a1x+· · ·+anxn. Then 0 = f (β) = f p(α) q(α) = a0 + a1 p(α) q(α) + · · · + an p(α) q(α) n. Now multiply both sides by q(α)n to show that there is a polynomial in F [x] that has α as a zero. Chapter 22. Finite Fields 1. (a) 2. (c) 2. 4. There are eight elements in Z2(α). Exhibit two more zeros of x3 + x2 + 1 other than α in these eight elements. 5. Find an irreducible polynomial p(x) in Z3[x] of degree 3 and show that Z3[x]/p(x) has 27 elements. 7. (a) x5 − 1 = (x + 1)(x4 + x3 + x2 + x + 1). (c) x9 − 1 = (x + 1)(x2 + x + 1)(x6 + x3 + 1). 8. True. 11. (a) Use the fact that x7 − 1 = (x + 1)(x3 + x + 1)(x3 + x2 + 1). 12. False. 17. If p(x) ∈ F [x], then p(x
) ∈ E[x]. 18. Since α is algebraic over F of degree n, we can write any element β ∈ F (α) uniquely as β = a0 + a1α + · · · + an−1αn−1 with ai ∈ F. There are qn possible n-tuples (a0, a1,..., an−1). 24. Factor xp−1 − 1 over Zp. HINTS AND SOLUTIONS 413 Chapter 23. Galois Theory 1. (a) Z2. (c) Z2 × Z2 × Z2. 2. (a) Separable. (c) Not separable. 3. [GF(729) : GF(9)] = [GF(729) : GF(3)]/[GF(9) : GF(3)] = 6/2 = 3 ⇒ G(GF(729)/GF(9)) ∼= Z3. A generator for G(GF(729)/GF(9)) is σ, where σ36 (α) = α36 4. (a) S5. (c) S3. 5. (a) Q(i). = α729 for α ∈ GF(729). 7. Let E be the splitting ﬁeld of a cubic polynomial in F [x]. Show that [E : F ] is less than or equal to 6 and is divisible by 3. Since G(E/F ) is a subgroup of S3 whose order is divisible by 3, conclude that this group must be isomorphic to Z3 or S3. 9. G is a subgroup of Sn. 16. True. 20. (a) Clearly ω, ω2,..., ωp−1 are distinct since ω = 1 or 0. To show that ωi is a zero of Φp, calculate Φp(ωi). (b) The conjugates of ω are ω, ω2,..., ωp−1. Deﬁne a map φi : Q(ω) → Q(ωi) by φi(a0 + a1ω + · · · + ap−2ωp−2) = a0 + a1ωi + · · · + cp−2(ωi)p−2,
where ai ∈ Q. Prove that φi is an isomorphism of ﬁelds. Show that φ2 generates G(Q(ω)/Q). (c) Show that {ω, ω2,..., ωp−1} is a basis for Q(ω) over Q, and consider which linear combinations of ω, ω2,..., ωp−1 are left ﬁxed by all elements of G(Q(ω)/Q). GNU Free Documentation License Version 1.2, November 2002 Copyright 2000,2001,2002 Free Software Foundation, Inc. 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. Preamble The purpose of this License is to make a manual, textbook, or other functional and useful document “free” in the sense of freedom: to assure everyone the eﬀective freedom to copy and redistribute it, with or without modifying it, either commercially or noncommercially. Secondarily, this License preserves for the author and publisher a way to get credit for their work, while not being considered responsible for modiﬁcations made by others. This License is a kind of “copyleft”, which means that derivative works of the document must themselves be free in the same sense. It complements the GNU General Public License, which is a copyleft license designed for free software. We have designed this License in order to use it for manuals for free software, because free software needs free documentation: a free program should come with manuals providing the same freedoms that the software does. But this License is not limited to software manuals; it can be used for any textual work, regardless of subject matter or whether it is published as a printed book. We recommend this License principally for works whose purpose is instruction or reference. 1. Applicability And Deﬁnitions This License applies to any manual or other work, in any medium, that contains a notice placed by the copyright holder saying it can be distributed under the terms of this License. Such a notice grants a world-wide, royalty-free license, unlimited in duration, to use that work under the conditions stated herein. The “Document”, 414 GFDL LICENSE 415 below, refers to any such manual or

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