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Theorem 16.30. Let \( I \) be an ideal of \( R \) . The map \( \phi : R \rightarrow R/I \) defined by \( \phi \left( r\right) = r + I \) is a ring homomorphism of \( R \) onto \( R/I \) with kernel \( I \) . | Proof. Certainly \( \phi : R \rightarrow R/I \) is a surjective abelian group homomorphism. It remains to show that \( \phi \) works correctly under ring multiplication. Let \( r \) and \( s \) be in \( R \) . Then\n\n\[ \phi \left( r\right) \phi \left( s\right) = \left( {r + I}\right) \left( {s + I}\right) = {rs} + I ... | Yes |
Theorem 16.31 First Isomorphism Theorem. Let \( \psi : R \rightarrow S \) be a ring homomorphism. Then \( \ker \psi \) is an ideal of \( R \) . If \( \phi : R \rightarrow R/\ker \psi \) is the canonical homomorphism, then there exists a unique isomorphism \( \eta : R/\ker \psi \rightarrow \psi \left( R\right) \) such t... | Proof. Let \( K = \ker \psi \) . By the First Isomorphism Theorem for groups, there exists a well-defined group homomorphism \( \eta : R/K \rightarrow \psi \left( R\right) \) defined by \( \eta \left( {r + K}\right) = \psi \left( r\right) \) for the additive abelian groups \( R \) and \( R/K \) . To show that this is a... | Yes |
Theorem 16.35. Let \( R \) be a commutative ring with identity and \( M \) an ideal in \( R \) . Then \( M \) is a maximal ideal of \( R \) if and only if \( R/M \) is a field. | Proof. Let \( M \) be a maximal ideal in \( R \) . If \( R \) is a commutative ring, then \( R/M \) must also be a commutative ring. Clearly, \( 1 + M \) acts as an identity for \( R/M \) . We must also show that every nonzero element in \( R/M \) has an inverse. If \( a + M \) is a nonzero element in \( R/M \) , then ... | Yes |
Proposition 16.38. Let \( R \) be a commutative ring with identity 1, where \( 1 \neq 0 \) . Then \( P \) is a prime ideal in \( R \) if and only if \( R/P \) is an integral domain. | Proof. First let us assume that \( P \) is an ideal in \( R \) and \( R/P \) is an integral domain. Suppose that \( {ab} \in P \) . If \( a + P \) and \( b + P \) are two elements of \( R/P \) such that \( \left( {a + P}\right) \left( {b + P}\right) = 0 + P = P \) , then either \( a + P = P \) or \( b + P = P \) . This... | Yes |
Lemma 16.41. Let \( m \) and \( n \) be positive integers such that \( \gcd \left( {m, n}\right) = 1 \) . Then for \( a, b \in \mathbb{Z} \) the system\n\n\[ x \equiv a\;\left( {\;\operatorname{mod}\;m}\right) \]\n\n\[ x \equiv b\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nhas a solution. If \( {x}_{1} \) and \( {x}... | Proof. The equation \( x \equiv a\left( {\;\operatorname{mod}\;m}\right) \) has a solution since \( a + {km} \) satisfies the equation for all \( k \in \mathbb{Z} \) . We must show that there exists an integer \( {k}_{1} \) such that\n\n\[ a + {k}_{1}m \equiv b\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nThis is equ... | Yes |
Let us solve the system\n\n\[ \nx \equiv 3\;\left( {\;\operatorname{mod}\;4}\right) \]\n\n\[ \nx \equiv 4\;\left( {\;\operatorname{mod}\;5}\right) \]\n | Using the Euclidean algorithm, we can find integers \( s \) and \( t \) such that \( {4s} + {5t} = 1 \) . Two such integers are \( s = 4 \) and \( t = - 3 \) . Consequently,\n\n\[ \nx = a + {k}_{1}m = 3 + 4{k}_{1} = 3 + 4\left\lbrack {\left( {5 - 4}\right) 4}\right\rbrack = {19}. \]\n | No |
Theorem 16.43 Chinese Remainder Theorem. Let \( {n}_{1},{n}_{2},\ldots ,{n}_{k} \) be positive integers such that \( \gcd \left( {{n}_{i},{n}_{j}}\right) = 1 \) for \( i \neq j \) . Then for any integers \( {a}_{1},\ldots ,{a}_{k} \), the system\n\n\[ x \equiv {a}_{1}\;\left( {\;\operatorname{mod}\;{n}_{1}}\right) \]\n... | Proof. We will use mathematical induction on the number of equations in the system. If there are \( k = 2 \) equations, then the theorem is true by Lemma 16.41. Now suppose that the result is true for a system of \( k \) equations or less and that we wish to find a solution of\n\n\[ x \equiv {a}_{1}\;\left( {\;\operato... | Yes |
Let us solve the system\n\n\[x \equiv 3\;\left( {\;\operatorname{mod}\;4}\right) \]\n\n\[x \equiv 4\;\left( {\;\operatorname{mod}\;5}\right) \]\n\n\[x \equiv 1\;\left( {\;\operatorname{mod}\;9}\right) \]\n\n\[x \equiv 5\;\left( {\;\operatorname{mod}\;7}\right) \] | From Example 16.42 we know that 19 is a solution of the first two congruences and any other solution of the system is congruent to 19 (mod 20). Hence, we can reduce the system to a system of three congruences:\n\n\[x \equiv {19}\;\left( {\;\operatorname{mod}\;{20}}\right) \]\n\n\[x \equiv 1\;\left( {\;\operatorname{mod... | Yes |
Suppose that we wish to multiply 2134 by 1531. We will use the integers \( {95},{97},{98} \), and 99 because they are relatively prime. We can break down each integer into four parts: | \[ {2134} \equiv {44}\;\left( {\;\operatorname{mod}\;{95}}\right) \] \[ {2134} \equiv 0\;\left( {\;\operatorname{mod}\;{97}}\right) \] \[ {2134} \equiv {76}\;\left( {\;\operatorname{mod}\;{98}}\right) \] \[ {2134} \equiv {55}\;\left( {\;\operatorname{mod}\;{99}}\right) \] and \[ {1531} \equiv {11}\;\left( {\;\operatorn... | Yes |
Suppose that\n\n\\[ p\\left( x\\right) = 3 + {0x} + 0{x}^{2} + 2{x}^{3} + 0{x}^{4} \\]\n\nand\n\n\\[ q\\left( x\\right) = 2 + {0x} - {x}^{2} + 0{x}^{3} + 4{x}^{4} \\]\n\nare polynomials in \\( \\mathbb{Z}\\left\\lbrack x\\right\\rbrack \\) . If the coefficient of some term in a polynomial is zero, then we usually just ... | \\[ p\\left( x\\right) q\\left( x\\right) = \\left( {3 + 2{x}^{3}}\\right) \\left( {2 - {x}^{2} + 4{x}^{4}}\\right) = 6 - 3{x}^{2} + 4{x}^{3} + {12}{x}^{4} - 2{x}^{5} + 8{x}^{7}, \\]\n\ncan be calculated either by determining the \\( {c}_{i}\\mathrm{\\;s} \\) in the definition or by simply multiplying polynomials in th... | Yes |
Theorem 17.3. Let \( R \) be a commutative ring with identity. Then \( R\left\lbrack x\right\rbrack \) is a commutative ring with identity. | Proof. Our first task is to show that \( R\left\lbrack x\right\rbrack \) is an abelian group under polynomial addition. The zero polynomial, \( f\left( x\right) = 0 \), is the additive identity. Given a polynomial \( p\left( x\right) = \) \( \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} \), the inverse of \( p\left... | No |
Proposition 17.4. Let \( p\left( x\right) \) and \( q\left( x\right) \) be polynomials in \( R\left\lbrack x\right\rbrack \), where \( R \) is an integral domain. Then \( \deg p\left( x\right) + \deg q\left( x\right) = \deg \left( {p\left( x\right) q\left( x\right) }\right) \) . Furthermore, \( R\left\lbrack x\right\rb... | Proof. Suppose that we have two nonzero polynomials\n\n\[ p\left( x\right) = {a}_{m}{x}^{m} + \cdots + {a}_{1}x + {a}_{0} \]\nand\n\n\[ q\left( x\right) = {b}_{n}{x}^{n} + \cdots + {b}_{1}x + {b}_{0} \]\n\nwith \( {a}_{m} \neq 0 \) and \( {b}_{n} \neq 0 \) . The degrees of \( p\left( x\right) \) and \( q\left( x\right)... | Yes |
Theorem 17.5. Let \( R \) be a commutative ring with identity and \( \alpha \in R \) . Then we have a ring homomorphism \( {\phi }_{\alpha } : R\left\lbrack x\right\rbrack \rightarrow R \) defined by\n\n\[ \n{\phi }_{\alpha }\left( {p\left( x\right) }\right) = p\left( \alpha \right) = {a}_{n}{\alpha }^{n} + \cdots + {a... | Proof. Let \( p\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} \) and \( q\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{m}{b}_{i}{x}^{i} \) . It is easy to show that \( {\phi }_{\alpha }(p\left( x\right) + \) \( q\left( x\right) ) = {\phi }_{\alpha }\left( {p\left( x\right) }\right) + {\phi }_{... | Yes |
Theorem 17.6 Division Algorithm. Let \( f\left( x\right) \) and \( g\left( x\right) \) be polynomials in \( F\left\lbrack x\right\rbrack \), where \( F \) is a field and \( g\left( x\right) \) is a nonzero polynomial. Then there exist unique polynomials \( q\left( x\right), r\left( x\right) \in \) \( F\left\lbrack x\ri... | Proof. We will first consider the existence of \( q\left( x\right) \) and \( r\left( x\right) \) . If \( f\left( x\right) \) is the zero polynomial, then\n\n\[ 0 = 0 \cdot g\left( x\right) + 0 \]\n\nhence, both \( q \) and \( r \) must also be the zero polynomial. Now suppose that \( f\left( x\right) \) is not the zero... | Yes |
The division algorithm merely formalizes long division of polynomials, a task we have been familiar with since high school. For example, suppose that we divide \( {x}^{3} - {x}^{2} + {2x} - 3 \) by \( x - 2 \) . | Hence, \( {x}^{3} - {x}^{2} + {2x} - 3 = \left( {x - 2}\right) \left( {{x}^{2} + x + 4}\right) + 5 \) . | Yes |
Corollary 17.8. Let \( F \) be a field. An element \( \alpha \in F \) is a zero of \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) if and only if \( x - \alpha \) is a factor of \( p\left( x\right) \) in \( F\left\lbrack x\right\rbrack \) . | Proof. Suppose that \( \alpha \in F \) and \( p\left( \alpha \right) = 0 \) . By the division algorithm, there exist polynomials \( q\left( x\right) \) and \( r\left( x\right) \) such that\n\n\[ p\left( x\right) = \left( {x - \alpha }\right) q\left( x\right) + r\left( x\right) \]\n\nand the degree of \( r\left( x\right... | Yes |
Let \( F \) be a field. A nonzero polynomial \( p\left( x\right) \) of degree \( n \) in \( F\left\lbrack x\right\rbrack \) can have at most \( n \) distinct zeros in \( F \) . | Proof. We will use induction on the degree of \( p\left( x\right) \) . If \( \deg p\left( x\right) = 0 \), then \( p\left( x\right) \) is a constant polynomial and has no zeros. Let \( \deg p\left( x\right) = 1 \) . Then \( p\left( x\right) = {ax} + b \) for some \( a \) and \( b \) in \( F \) . If \( {\alpha }_{1} \) ... | Yes |
The polynomial \( p\left( x\right) = {x}^{3} + {x}^{2} + 2 \) is irreducible over \( {\mathbb{Z}}_{3}\left\lbrack x\right\rbrack \) . | Suppose that this polynomial was reducible over \( {\mathbb{Z}}_{3}\left\lbrack x\right\rbrack \) . By the division algorithm there would have to be a factor of the form \( x - a \), where \( a \) is some element in \( {\mathbb{Z}}_{3}\left\lbrack x\right\rbrack \) . Hence, it would have to be true that \( p\left( a\ri... | Yes |
Lemma 17.13. Let \( p\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) . Then\n\n\[ p\left( x\right) = \frac{r}{s}\left( {{a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n}}\right) ,\]\n\nwhere \( r, s,{a}_{0},\ldots ,{a}_{n} \) are integers, the \( {a}_{i} \) ’s are relatively prime, and \( r \) and \( s \) are r... | Proof. Suppose that\n\n\[ p\left( x\right) = \frac{{b}_{0}}{{c}_{0}} + \frac{{b}_{1}}{{c}_{1}}x + \cdots + \frac{{b}_{n}}{{c}_{n}}{x}^{n},\]\n\nwhere the \( {b}_{i} \) ’s and the \( {c}_{i} \) ’s are integers. We can rewrite \( p\left( x\right) \) as\n\n\[ p\left( x\right) = \frac{1}{{c}_{0}\cdots {c}_{n}}\left( {{d}_{... | Yes |
Corollary 17.15. Let \( p\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} \) be a polynomial with coefficients in \( \mathbb{Z} \) and \( {a}_{0} \neq 0 \) . If \( p\left( x\right) \) has a zero in \( \mathbb{Q} \), then \( p\left( x\right) \) also has a zero \( \alpha \) in \( \mathbb{Z} \) . Furt... | Proof. Let \( p\left( x\right) \) have a zero \( a \in \mathbb{Q} \) . Then \( p\left( x\right) \) must have a linear factor \( x - a \) . By Gauss’s Lemma, \( p\left( x\right) \) has a factorization with a linear factor in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . Hence, for some \( \alpha \in \mathbb{Z} \n\n\[ \n... | Yes |
Let \( p\left( x\right) = {x}^{4} - 2{x}^{3} + x + 1 \) . We shall show that \( p\left( x\right) \) is irreducible over \( \mathbb{Q}\left\lbrack x\right\rbrack \) . | Assume that \( p\left( x\right) \) is reducible. Then either \( p\left( x\right) \) has a linear factor, say \( p\left( x\right) = \) \( \left( {x - \alpha }\right) q\left( x\right) \), where \( q\left( x\right) \) is a polynomial of degree three, or \( p\left( x\right) \) has two quadratic factors.\n\nIf \( p\left( x\... | Yes |
Theorem 17.17 Eisenstein's Criterion. Let \( p \) be a prime and suppose that\n\n\[ f\left( x\right) = {a}_{n}{x}^{n} + \cdots + {a}_{0} \in \mathbb{Z}\left\lbrack x\right\rbrack .\n\]\n\nIf \( p \mid {a}_{i} \) for \( i = 0,1,\ldots, n - 1 \), but \( p \nmid {a}_{n} \) and \( {p}^{2} \nmid {a}_{0} \), then \( f\left( ... | Proof. By Gauss’s Lemma, we need only show that \( f\left( x\right) \) does not factor into polynomials of lower degree in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . Let\n\n\[ f\left( x\right) = \left( {{b}_{r}{x}^{r} + \cdots + {b}_{0}}\right) \left( {{c}_{s}{x}^{s} + \cdots + {c}_{0}}\right)\n\]\n\nbe a factorizat... | Yes |
Theorem 17.20. If \( F \) is a field, then every ideal in \( F\left\lbrack x\right\rbrack \) is a principal ideal. | Proof. Let \( I \) be an ideal of \( F\left\lbrack x\right\rbrack \) . If \( I \) is the zero ideal, the theorem is easily true. Suppose that \( I \) is a nontrivial ideal in \( F\left\lbrack x\right\rbrack \), and let \( p\left( x\right) \in I \) be a nonzero element of minimal degree. If \( \deg p\left( x\right) = 0 ... | Yes |
It is not the case that every ideal in the ring \( F\left\lbrack {x, y}\right\rbrack \) is a principal ideal. | Consider the ideal of \( F\left\lbrack {x, y}\right\rbrack \) generated by the polynomials \( x \) and \( y \) . This is the ideal of \( F\left\lbrack {x, y}\right\rbrack \) consisting of all polynomials with no constant term. Since both \( x \) and \( y \) are in the ideal, no single polynomial can generate the entire... | Yes |
Theorem 17.22. Let \( F \) be a field and suppose that \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) . Then the ideal generated by \( p\left( x\right) \) is maximal if and only if \( p\left( x\right) \) is irreducible. | Proof. Suppose that \( p\left( x\right) \) generates a maximal ideal of \( F\left\lbrack x\right\rbrack \) . Then \( \langle p\left( x\right) \rangle \) is also a prime ideal of \( F\left\lbrack x\right\rbrack \) . Since a maximal ideal must be properly contained inside \( F\left\lbrack x\right\rbrack, p\left( x\right)... | Yes |
Lemma 18.1. The relation \( \sim \) between elements of \( S \) is an equivalence relation. | Proof. Since \( D \) is commutative, \( {ab} = {ba} \) ; hence, \( \sim \) is reflexive on \( D \) . Now suppose that \( \left( {a, b}\right) \sim \left( {c, d}\right) \) . Then \( {ad} = {bc} \) or \( {cb} = {da} \) . Therefore, \( \left( {c, d}\right) \sim \left( {a, b}\right) \) and the relation is symmetric. Finall... | Yes |
Lemma 18.2. The operations of addition and multiplication on \( {F}_{D} \) are well-defined. | Proof. We will prove that the operation of addition is well-defined. The proof that multiplication is well-defined is left as an exercise. Let \( \left\lbrack {{a}_{1},{b}_{1}}\right\rbrack = \left\lbrack {{a}_{2},{b}_{2}}\right\rbrack \) and \( \left\lbrack {{c}_{1},{d}_{1}}\right\rbrack = \left\lbrack {{c}_{2},{d}_{2... | No |
Lemma 18.3. The set of equivalence classes of \( S,{F}_{D} \), under the equivalence relation \( \sim \) , together with the operations of addition and multiplication defined by\n\n\[ \left\lbrack {a, b}\right\rbrack + \left\lbrack {c, d}\right\rbrack = \left\lbrack {{ad} + {bc},{bd}}\right\rbrack \]\n\n\[ \left\lbrack... | Proof. The additive and multiplicative identities are \( \left\lbrack {0,1}\right\rbrack \) and \( \left\lbrack {1,1}\right\rbrack \), respectively. To show that \( \left\lbrack {0,1}\right\rbrack \) is the additive identity, observe that\n\n\[ \left\lbrack {a, b}\right\rbrack + \left\lbrack {0,1}\right\rbrack = \left\... | No |
Not every integral domain is a unique factorization domain. | The subring \( \mathbb{Z}\left\lbrack {\sqrt{3}i}\right\rbrack = \{ a + b\sqrt{3}i\} \) of the complex numbers is an integral domain (Exercise 16.6.12, Chapter 16). Let \( z = a + b\sqrt{3}i \) and define \( \nu : \mathbb{Z}\left\lbrack {\sqrt{3}i}\right\rbrack \rightarrow \mathbb{N} \cup \{ 0\} \) by \( \nu \left( z\r... | Yes |
Lemma 18.11. Let \( D \) be an integral domain and let \( a, b \in D \) . Then\n\n1. \( a \mid b \) if and only if \( \langle b\rangle \subset \langle a\rangle \) . | Proof. (1) Suppose that \( a \mid b \) . Then \( b = {ax} \) for some \( x \in D \) . Hence, for every \( r \) in \( D,{br} = \left( {ax}\right) r = a\left( {xr}\right) \) and \( \langle b\rangle \subset \langle a\rangle \) . Conversely, suppose that \( \langle b\rangle \subset \langle a\rangle \) . Then \( b \in \lang... | Yes |
Theorem 18.12. Let \( D \) be a PID and \( \langle p\rangle \) be a nonzero ideal in \( D \) . Then \( \langle p\rangle \) is a maximal ideal if and only if \( p \) is irreducible. | Proof. Suppose that \( \langle p\rangle \) is a maximal ideal. If some element \( a \) in \( D \) divides \( p \), then \( \langle p\rangle \subset \langle a\rangle \) . Since \( \langle p\rangle \) is maximal, either \( D = \langle a\rangle \) or \( \langle p\rangle = \langle a\rangle \) . Consequently, either \( a \)... | Yes |
Corollary 18.13. Let \( D \) be a PID. If \( p \) is irreducible, then \( p \) is prime. | Proof. Let \( p \) be irreducible and suppose that \( p \mid {ab} \) . Then \( \langle {ab}\rangle \subset \langle p\rangle \) . By Corollary 16.40, since \( \langle p\rangle \) is a maximal ideal, \( \langle p\rangle \) must also be a prime ideal. Thus, either \( a \in \langle p\rangle \) or \( b \in \langle p\rangle ... | Yes |
Lemma 18.14. Let \( D \) be a PID. Let \( {I}_{1},{I}_{2},\ldots \) be a set of ideals such that \( {I}_{1} \subset {I}_{2} \subset \cdots \) . Then there exists an integer \( N \) such that \( {I}_{n} = {I}_{N} \) for all \( n \geq N \) . | Proof. We claim that \( I = \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{I}_{i} \) is an ideal of \( D \) . Certainly \( I \) is not empty, since \( {I}_{1} \subset I \) and \( 0 \in I \) . If \( a, b \in I \), then \( a \in {I}_{i} \) and \( b \in {I}_{j} \) for some \( i \) and \( j \) in \( \mathbb{N} \) . Without l... | Yes |
Every PID is a UFD, but it is not the case that every UFD is a PID. | In Corollary 18.31, we will prove that \( \mathbb{Z}\left\lbrack x\right\rbrack \) is a UFD. However, \( \mathbb{Z}\left\lbrack x\right\rbrack \) is not a PID. Let \( I = \{ {5f}\left( x\right) + {xg}\left( x\right) : f\left( x\right), g\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \} \) . We can easily sho... | Yes |
We claim that \( \nu \left( {a + {bi}}\right) = {a}^{2} + {b}^{2} \) is a Euclidean valuation on \( \mathbb{Z}\left\lbrack i\right\rbrack \) . Let \( z, w \in \mathbb{Z}\left\lbrack i\right\rbrack \) . Then \( \nu \left( {zw}\right) = {\left| zw\right| }^{2} = {\left| z\right| }^{2}{\left| w\right| }^{2} = \nu \left( z... | Next, we must show that for any \( z = a + {bi} \) and \( w = c + {di} \) in \( \mathbb{Z}\left\lbrack i\right\rbrack \) with \( w \neq 0 \), there exist elements \( q \) and \( r \) in \( \mathbb{Z}\left\lbrack i\right\rbrack \) such that \( z = {qw} + r \) with either \( r = 0 \) or \( \nu \left( r\right) < \nu \left... | No |
Theorem 18.21. Every Euclidean domain is a principal ideal domain. | Proof. Let \( D \) be a Euclidean domain and let \( \nu \) be a Euclidean valuation on \( D \) . Suppose \( I \) is a nontrivial ideal in \( D \) and choose a nonzero element \( b \in I \) such that \( \nu \left( b\right) \) is minimal for all \( a \in I \) . Since \( D \) is a Euclidean domain, there exist elements \(... | Yes |
Theorem 18.24 Gauss’s Lemma. Let \( D \) be a UFD and let \( f\left( x\right) \) and \( g\left( x\right) \) be primitive polynomials in \( D\left\lbrack x\right\rbrack \) . Then \( f\left( x\right) g\left( x\right) \) is primitive. | Proof. Let \( f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i} \) and \( g\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{b}_{i}{x}^{i} \) . Suppose that \( p \) is a prime dividing the coefficients of \( f\left( x\right) g\left( x\right) \) . Let \( r \) be the smallest integer such that \( p... | Yes |
Lemma 18.25. Let \( D \) be a UFD, and let \( p\left( x\right) \) and \( q\left( x\right) \) be in \( D\left\lbrack x\right\rbrack \) . Then the content of \( p\left( x\right) q\left( x\right) \) is equal to the product of the contents of \( p\left( x\right) \) and \( q\left( x\right) \) . | Proof. Let \( p\left( x\right) = c{p}_{1}\left( x\right) \) and \( q\left( x\right) = d{q}_{1}\left( x\right) \), where \( c \) and \( d \) are the contents of \( p\left( x\right) \) and \( q\left( x\right) \), respectively. Then \( {p}_{1}\left( x\right) \) and \( {q}_{1}\left( x\right) \) are primitive. We can now wr... | Yes |
Lemma 18.26. Let \( D \) be a UFD and \( F \) its field of fractions. Suppose that \( p\left( x\right) \in D\left\lbrack x\right\rbrack \) and \( p\left( x\right) = f\left( x\right) g\left( x\right) \), where \( f\left( x\right) \) and \( g\left( x\right) \) are in \( F\left\lbrack x\right\rbrack \) . Then \( p\left( x... | Proof. Let \( a \) and \( b \) be nonzero elements of \( D \) such that \( {af}\left( x\right) ,{bg}\left( x\right) \) are in \( D\left\lbrack x\right\rbrack \) . We can find \( {a}_{1},{b}_{2} \in D \) such that \( {af}\left( x\right) = {a}_{1}{f}_{1}\left( x\right) \) and \( {bg}\left( x\right) = {b}_{1}{g}_{1}\left(... | Yes |
Let \( X \) be any set. We will define the power set of \( X \) to be the set of all subsets of \( X \) . We denote the power set of \( X \) by \( \mathcal{P}\left( X\right) \) . For example, let \( X = \{ a, b, c\} \) . Then \( \mathcal{P}\left( X\right) \) is the set of all subsets of the set \( \{ a, b, c\} \) : | \[ \varnothing \;\{ a\} \;\{ c\} \;\{ c\} \] \[ \{ a, b\} \;\{ a, c\} \;\{ b, c\} \;\{ a, b, c\} . \] | No |
There can be more than one partial order on a particular set. We can form a partial order on \( \mathbb{N} \) by \( a \preccurlyeq b \) if \( a \mid b \) . | The relation is certainly reflexive since \( a \mid a \) for all \( a \in \mathbb{N} \) . If \( m \mid n \) and \( n \mid m \), then \( m = n \) ; hence, the relation is also antisymmetric. The relation is transitive, because if \( m\left| {n\text{and}n}\right| p \), then \( m \mid p \) . | Yes |
Theorem 19.9. Let \( Y \) be a nonempty subset of a poset \( X \). If \( Y \) has a least upper bound, then \( Y \) has a unique least upper bound. If \( Y \) has a greatest lower bound, then \( Y \) has a unique greatest lower bound. | Proof. Let \( {u}_{1} \) and \( {u}_{2} \) be least upper bounds for \( Y \). By the definition of the least upper bound, \( {u}_{1} \preccurlyeq u \) for all upper bounds \( u \) of \( Y \). In particular, \( {u}_{1} \preccurlyeq {u}_{2} \). Similarly, \( {u}_{2} \preccurlyeq {u}_{1} \). Therefore, \( {u}_{1} = {u}_{2... | Yes |
Let \( X \) be a set. Then the power set of \( X,\mathcal{P}\left( X\right) \), is a lattice. For two sets \( A \) and \( B \) in \( \mathcal{P}\left( X\right) \), the least upper bound of \( A \) and \( B \) is \( A \cup B \). | Certainly \( A \cup B \) is an upper bound of \( A \) and \( B \), since \( A \subset A \cup B \) and \( B \subset A \cup B \) . If \( C \) is some other set containing both \( A \) and \( B \), then \( C \) must contain \( A \cup B \) ; hence, \( A \cup B \) is the least upper bound of \( A \) and \( B \) . Similarly,... | Yes |
Theorem 19.13. If \( L \) is a lattice, then the binary operations \( \vee \) and \( \land \) satisfy the following properties for \( a, b, c \in L \). 1. Commutative laws: \( a \vee b = b \vee a \) and \( a \land b = b \land a \). 2. Associative laws: \( a \vee \left( {b \vee c}\right) = \left( {a \vee b}\right) \vee ... | Proof. By the Principle of Duality, we need only prove the first statement in each part. (1) By definition \( a \vee b \) is the least upper bound of \( \{ a, b\} \), and \( b \vee a \) is the least upper bound of \( \{ b, a\} \) ; however, \( \{ a, b\} = \{ b, a\} \). (2) We will show that \( a \vee \left( {b \vee c}\... | Yes |
Theorem 19.14. Let \( L \) be a nonempty set with two binary operations \( \vee \) and \( \land \) satisfying the commutative, associative, idempotent, and absorption laws. We can define a partial order on \( L \) by \( a \preccurlyeq b \) if \( a \vee b = b \) . Furthermore, \( L \) is a lattice with respect to \( \pr... | Proof. We first show that \( L \) is a poset under \( \preccurlyeq \) . Since \( a \vee a = a, a \preccurlyeq a \) and \( \preccurlyeq \) is reflexive. To show that \( \preccurlyeq \) is antisymmetric, let \( a \preccurlyeq b \) and \( b \preccurlyeq a \) . Then \( a \vee b = b \) and \( b \vee a = a \) . By the commut... | No |
A lattice \( L \) is distributive if and only if\n\n\[ a \vee \left( {b \land c}\right) = \left( {a \vee b}\right) \land \left( {a \vee c}\right) \]\n\nfor all \( a, b, c \in L \) . | Proof. Let us assume that \( L \) is a distributive lattice.\n\n\[ a \vee \left( {b \land c}\right) = \left\lbrack {a \vee \left( {a \land c}\right) }\right\rbrack \vee \left( {b \land c}\right) \]\n\n\[ = a \vee \left\lbrack {\left( {a \land c}\right) \vee \left( {b \land c}\right) }\right\rbrack \]\n\n\[ = a \vee \le... | Yes |
Theorem 19.16. A set \( B \) is a Boolean algebra if and only if there exist binary operations \( \vee \) and \( \land \) on \( B \) satisfying the following axioms.\n\n1. \( a \vee b = b \vee a \) and \( a \land b = b \land a \) for \( a, b \in B \) .\n\n2. \( a \vee \left( {b \vee c}\right) = \left( {a \vee b}\right)... | Proof. Let \( B \) be a set satisfying (1)-(5) in the theorem. One of the idempotent laws is satisfied since\n\n\[ a = a \vee O \]\n\n\[ = a \vee \left( {a \land {a}^{\prime }}\right) \]\n\n\[ = \left( {a \vee a}\right) \land \left( {a \vee {a}^{\prime }}\right) \]\n\n\[ = \left( {a \vee a}\right) \land I \]\n\n\[ = a ... | Yes |
If \( a \vee b = a \vee c \) and \( a \land b = a \land c \) for \( a, b, c \in B \), then \( b = c \) . | For \( a \vee b = a \vee c \) and \( a \land b = a \land c \), we have\n\n\[ b = b \vee \left( {b \land a}\right) \]\n\n\[ = b \vee \left( {a \land b}\right) \]\n\n\[ = b \vee \left( {a \land c}\right) \]\n\n\[ = \left( {b \vee a}\right) \land \left( {b \vee c}\right) \]\n\n\[ = \left( {a \vee b}\right) \land \left( {b... | Yes |
Lemma 19.18. Let \( B \) be a finite Boolean algebra. If \( b \) is a nonzero element of \( B \), then there is an atom \( a \) in \( B \) such that \( a \preccurlyeq b \) . | Proof. If \( b \) is an atom, let \( a = b \) . Otherwise, choose an element \( {b}_{1} \), not equal to \( O \) or \( b \) , such that \( {b}_{1} \preccurlyeq b \) . We are guaranteed that this is possible since \( b \) is not an atom. If \( {b}_{1} \) is an atom, then we are done. If not, choose \( {b}_{2} \), not eq... | Yes |
Lemma 19.19. Let \( a \) and \( b \) be atoms in a finite Boolean algebra \( B \) such that \( a \neq b \) . Then \( a \land b = O \) . | Proof. Since \( a \land b \) is the greatest lower bound of \( a \) and \( b \), we know that \( a \land b \preccurlyeq a \) . Hence, either \( a \land b = a \) or \( a \land b = O \) . However, if \( a \land b = a \), then either \( a \preccurlyeq b \) or \( a = O \) . In either case we have a contradiction because \(... | Yes |
Lemma 19.20. Let \( B \) be a Boolean algebra and \( a, b \in B \) . The following statements are equivalent.\n\n1. \( a \preccurlyeq b \) .\n\n2. \( a \land {b}^{\prime } = O \) .\n\n3. \( {a}^{\prime } \vee b = I \) . | Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) . If \( a \preccurlyeq b \), then \( a \vee b = b \) . Therefore,\n\n\[ a \land {b}^{\prime } = a \land {\left( a \vee b\right) }^{\prime } \]\n\n\[ = a \land \left( {{a}^{\prime } \land {b}^{\prime }}\right) \]\n\n\[ = \left( {a \land {a}^{\prime }}\right) \land... | Yes |
Lemma 19.21. Let \( B \) be a Boolean algebra and \( b \) and \( c \) be elements in \( B \) such that \( b \npreceq c \) . Then there exists an atom \( a \in B \) such that \( a \preccurlyeq b \) and \( a \npreceq c \) . | Proof. By Lemma 19.20, \( b \land {c}^{\prime } \neq O \) . Hence, there exists an atom \( a \) such that \( a \preccurlyeq b \land {c}^{\prime } \) . Consequently, \( a \preccurlyeq b \) and \( a \npreceq c \) . | Yes |
Lemma 19.22. Let \( b \in B \) and \( {a}_{1},\ldots ,{a}_{n} \) be the atoms of \( B \) such that \( {a}_{i} \preccurlyeq b \) . Then \( b = {a}_{1} \vee \cdots \vee {a}_{n} \) . Furthermore, if \( a,{a}_{1},\ldots ,{a}_{n} \) are atoms of \( B \) such that \( a \preccurlyeq b,{a}_{i} \preccurlyeq b \), and \( b = a \... | Proof. Let \( {b}_{1} = {a}_{1} \vee \cdots \vee {a}_{n} \) . Since \( {a}_{i} \preccurlyeq b \) for each \( i \), we know that \( {b}_{1} \preccurlyeq b \) . If we can show that \( b \preccurlyeq {b}_{1} \), then the lemma is true by antisymmetry. Assume \( b \npreceq {b}_{1} \) . Then there exists an atom \( a \) suc... | Yes |
Theorem 19.23. Let \( B \) be a finite Boolean algebra. Then there exists a set \( X \) such that \( B \) is isomorphic to \( \mathcal{P}\left( X\right) \) . | Proof. We will show that \( B \) is isomorphic to \( \mathcal{P}\left( X\right) \), where \( X \) is the set of atoms of \( B \) . Let \( a \in B \) . By Lemma 19.22, we can write \( a \) uniquely as \( a = {a}_{1} \vee \cdots \vee {a}_{n} \) for \( {a}_{1},\ldots ,{a}_{n} \in X \) . Consequently, we can define a map \... | Yes |
Theorem 19.30. The set of all circuits is a Boolean algebra. | We leave as an exercise the proof of this theorem for the Boolean algebra axioms not yet verified. | No |
Given a complex circuit, we can now apply the techniques of Boolean algebra to reduce it to a simpler one. Consider the circuit in Figure 19.32. Since\n\n\[ \left( {a \vee b}\right) \land \left( {a \vee {b}^{\prime }}\right) \land \left( {a \vee b}\right) = \left( {a \vee b}\right) \land \left( {a \vee b}\right) \land ... | \[ \left( {a \vee b}\right) \land \left( {a \vee {b}^{\prime }}\right) \land \left( {a \vee b}\right) = \left( {a \vee b}\right) \land \left( {a \vee b}\right) \land \left( {a \vee {b}^{\prime }}\right) \]\n\n\[ = \left( {a \vee b}\right) \land \left( {a \vee {b}^{\prime }}\right) \]\n\n\[ = a \vee \left( {b \land {b}^... | Yes |
If \( F \) is a field, then \( F\left\lbrack x\right\rbrack \) is a vector space over \( F \). | The vectors in \( F\left\lbrack x\right\rbrack \) are simply polynomials, and vector addition is just polynomial addition. If \( \alpha \in F \) and \( p\left( x\right) \in F\left\lbrack x\right\rbrack \), then scalar multiplication is defined by \( {\alpha p}\left( x\right) \). | Yes |
The set of all continuous real-valued functions on a closed interval \( \left\lbrack {a, b}\right\rbrack \) is a vector space over \( \mathbb{R} \) . If \( f\left( x\right) \) and \( g\left( x\right) \) are continuous on \( \left\lbrack {a, b}\right\rbrack \), then \( \left( {f + g}\right) \left( x\right) \) is defined... | For example, if \( f\left( x\right) = \sin x \) and \( g\left( x\right) = {x}^{2} \), then \( \left( {{2f} + {5g}}\right) \left( x\right) = 2\sin x + 5{x}^{2} \) . | No |
Example 20.4. Let \( V = \mathbb{Q}\left( \sqrt{2}\right) = \{ a + b\sqrt{2} : a, b \in \mathbb{Q}\} \) . Then \( V \) is a vector space over Q. If \( u = a + b\sqrt{2} \) and \( v = c + d\sqrt{2} \), then \( u + v = \left( {a + c}\right) + \left( {b + d}\right) \sqrt{2} \) is again in \( V \) . Also, for \( \alpha \in... | We will leave it as an exercise to verify that all of the vector space axioms hold for \( V \) . | No |
Proposition 20.5. Let \( V \) be a vector space over \( F \). Then each of the following statements is true.\n\n1. \( {0v} = \mathbf{0} \) for all \( v \in V \). | Proof. To prove (1), observe that\n\n\[ \n{0v} = \left( {0 + 0}\right) v = {0v} + {0v} \n\]\n\nconsequently, \( \mathbf{0} + {0v} = {0v} + {0v} \). Since \( V \) is an abelian group, \( \mathbf{0} = {0v} \). | Yes |
Let \( W \) be the subspace of \( {\mathbb{R}}^{3} \) defined by \( W = \left\{ \left( {{x}_{1},2{x}_{1} + {x}_{2},{x}_{1} - {x}_{2}}\right) \right. \) : \( \left. {{x}_{1},{x}_{2} \in \mathbb{R}}\right\} \) . We claim that \( W \) is a subspace of \( {\mathbb{R}}^{3} \) . | Since\n\n\[ \alpha \left( {{x}_{1},2{x}_{1} + {x}_{2},{x}_{1} - {x}_{2}}\right) = \left( {\alpha {x}_{1},\alpha \left( {2{x}_{1} + {x}_{2}}\right) ,\alpha \left( {{x}_{1} - {x}_{2}}\right) }\right) \]\n\n\[ = \left( {\alpha {x}_{1},2\left( {\alpha {x}_{1}}\right) + \alpha {x}_{2},\alpha {x}_{1} - \alpha {x}_{2}}\right)... | Yes |
Proposition 20.8. Let \( S = \left\{ {{v}_{1},{v}_{2},\ldots ,{v}_{n}}\right\} \) be vectors in a vector space \( V \) . Then the span of \( S \) is a subspace of \( V \) . | Proof. Let \( u \) and \( v \) be in \( S \) . We can write both of these vectors as linear combinations of the \( {v}_{i} \) ’s:\n\n\[ u = {\alpha }_{1}{v}_{1} + {\alpha }_{2}{v}_{2} + \cdots + {\alpha }_{n}{v}_{n} \]\n\n\[ v = {\beta }_{1}{v}_{1} + {\beta }_{2}{v}_{2} + \cdots + {\beta }_{n}{v}_{n} \]\n\nThen\n\n\[ u... | No |
Proposition 20.9. Let \( \\left\\{ {{v}_{1},{v}_{2},\\ldots ,{v}_{n}}\\right\\} \) be a set of linearly independent vectors in a vector space. Suppose that\n\n\[ v = {\\alpha }_{1}{v}_{1} + {\\alpha }_{2}{v}_{2} + \\cdots + {\\alpha }_{n}{v}_{n} = {\\beta }_{1}{v}_{1} + {\\beta }_{2}{v}_{2} + \\cdots + {\\beta }_{n}{v}... | Proof. If\n\n\[ v = {\\alpha }_{1}{v}_{1} + {\\alpha }_{2}{v}_{2} + \\cdots + {\\alpha }_{n}{v}_{n} = {\\beta }_{1}{v}_{1} + {\\beta }_{2}{v}_{2} + \\cdots + {\\beta }_{n}{v}_{n}, \]\n\nthen\n\n\[ \\left( {{\\alpha }_{1} - {\\beta }_{1}}\\right) {v}_{1} + \\left( {{\\alpha }_{2} - {\\beta }_{2}}\\right) {v}_{2} + \\cdo... | Yes |
Proposition 20.10. A set \( \left\{ {{v}_{1},{v}_{2},\ldots ,{v}_{n}}\right\} \) of vectors in a vector space \( V \) is linearly dependent if and only if one of the \( {v}_{i} \) ’s is a linear combination of the rest. | Proof. Suppose that \( \left\{ {{v}_{1},{v}_{2},\ldots ,{v}_{n}}\right\} \) is a set of linearly dependent vectors. Then there exist scalars \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) such that\n\n\[{\alpha }_{1}{v}_{1} + {\alpha }_{2}{v}_{2} + \cdots + {\alpha }_{n}{v}_{n} = \mathbf{0},\]\n\nwith at least one of the \(... | Yes |
The vectors \( {e}_{1} = \left( {1,0,0}\right) ,{e}_{2} = \left( {0,1,0}\right) \), and \( {e}_{3} = \left( {0,0,1}\right) \) form a basis for \( {\mathbb{R}}^{3} \). | The set certainly spans \( {\mathbb{R}}^{3} \), since any arbitrary vector \( \left( {{x}_{1},{x}_{2},{x}_{3}}\right) \) in \( {\mathbb{R}}^{3} \) can be written as \( {x}_{1}{e}_{1} + {x}_{2}{e}_{2} + {x}_{3}{e}_{3} \). Also, none of the vectors \( {e}_{1},{e}_{2},{e}_{3} \) can be written as a linear combination of t... | Yes |
Proposition 20.14. Let \( \\left\\{ {{e}_{1},{e}_{2},\\ldots ,{e}_{m}}\\right\\} \) and \( \\left\\{ {{f}_{1},{f}_{2},\\ldots ,{f}_{n}}\\right\\} \) be two bases for a vector space V. Then \( m = n \) . | Proof. Since \( \\left\\{ {{e}_{1},{e}_{2},\\ldots ,{e}_{m}}\\right\\} \) is a basis, it is a linearly independent set. By Proposition \( {20.11}, n \\leq m \) . Similarly, \( \\left\\{ {{f}_{1},{f}_{2},\\ldots ,{f}_{n}}\\right\\} \) is a linearly independent set, and the last proposition implies that \( m \\leq n \) .... | Yes |
We claim that \( E \) is an extension field of \( F \). To see this, we need only show that \( \sqrt{2} \) is in \( E \). | Since \( \sqrt{2} + \sqrt{3} \) is in \( E \), \( 1/\left( {\sqrt{2} + \sqrt{3}}\right) = \sqrt{3} - \sqrt{2} \) must also be in \( E \). Taking linear combinations of \( \sqrt{2} + \sqrt{3} \) and \( \sqrt{3} - \sqrt{2} \), we find that \( \sqrt{2} \) and \( \sqrt{3} \) must both be in \( E \). | Yes |
Let \( p\left( x\right) = {x}^{2} + x + 1 \in {\mathbb{Z}}_{2}\left\lbrack x\right\rbrack \) . Since neither 0 nor 1 is a root of this polynomial, we know that \( p\left( x\right) \) is irreducible over \( {\mathbb{Z}}_{2} \) . We will construct a field extension of \( {\mathbb{Z}}_{2} \) containing an element \( \alph... | By Theorem 17.22, the ideal \( \langle p\left( x\right) \rangle \) generated by \( p\left( x\right) \) is maximal; hence, \( {\mathbb{Z}}_{2}\left\lbrack x\right\rbrack /\langle p\left( x\right) \rangle \) is a field. Let \( f\left( x\right) + \langle p\left( x\right) \rangle \) be an arbitrary element of \( {\mathbb{Z... | Yes |
Let \( p\left( x\right) = {x}^{5} + {x}^{4} + 1 \in {\mathbb{Z}}_{2}\left\lbrack x\right\rbrack \) . Then \( p\left( x\right) \) has irreducible factors \( {x}^{2} + x + 1 \) and \( {x}^{3} + x + 1 \) . | For a field extension \( E \) of \( {\mathbb{Z}}_{2} \) such that \( p\left( x\right) \) has a root in \( E \), we can let \( E \) be either \( {\mathbb{Z}}_{2}\left\lbrack x\right\rbrack /\left\langle {{x}^{2} + x + 1}\right\rangle \) or \( {\mathbb{Z}}_{2}\left\lbrack x\right\rbrack /\left\langle {{x}^{3} + x + 1}\ri... | No |
We will show that \( \sqrt{2 + \sqrt{3}} \) is algebraic over \( \mathbb{Q} \) . | If \( \alpha = \sqrt{2 + \sqrt{3}} \), then \( {\alpha }^{2} = 2 + \sqrt{3} \) . Hence, \( {\alpha }^{2} - 2 = \sqrt{3} \) and \( {\left( {\alpha }^{2} - 2\right) }^{2} = 3 \) . Since \( {\alpha }^{4} - 4{\alpha }^{2} + 1 = 0 \), it must be true that \( \alpha \) is a zero of the polynomial \( {x}^{4} - 4{x}^{2} + 1 \i... | Yes |
Theorem 21.9. Let \( E \) be an extension field of \( F \) and \( \alpha \in E \) . Then \( \alpha \) is transcendental over \( F \) if and only if \( F\left( \alpha \right) \) is isomorphic to \( F\left( x\right) \), the field of fractions of \( F\left\lbrack x\right\rbrack \) . | Proof. Let \( {\phi }_{\alpha } : F\left\lbrack x\right\rbrack \rightarrow E \) be the evaluation homomorphism for \( \alpha \) . Then \( \alpha \) is transcendental over \( F \) if and only if \( {\phi }_{\alpha }\left( {p\left( x\right) }\right) = p\left( \alpha \right) \neq 0 \) for all nonconstant polynomials \( p\... | Yes |
Theorem 21.10. Let \( E \) be an extension field of a field \( F \) and \( \alpha \in E \) with \( \alpha \) algebraic over \( F \) . Then there is a unique irreducible monic polynomial \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) of smallest degree such that \( p\left( \alpha \right) = 0 \) . If \( f\left( ... | Proof. Let \( {\phi }_{\alpha } : F\left\lbrack x\right\rbrack \rightarrow E \) be the evaluation homomorphism. The kernel of \( {\phi }_{\alpha } \) is a principal ideal generated by some \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) with \( \deg p\left( x\right) \geq 1 \) . We know that such a polynomial ex... | Yes |
Proposition 21.12. Let \( E \) be a field extension of \( F \) and \( \alpha \in E \) be algebraic over \( F \) . Then \( F\left( \alpha \right) \cong F\left\lbrack x\right\rbrack /\langle p\left( x\right) \rangle \), where \( p\left( x\right) \) is the minimal polynomial of \( \alpha \) over \( F \) . | Proof. Let \( {\phi }_{\alpha } : F\left\lbrack x\right\rbrack \rightarrow E \) be the evaluation homomorphism. The kernel of this map is \( \langle p\left( x\right) \rangle \), where \( p\left( x\right) \) is the minimal polynomial of \( \alpha \) . By the First Isomorphism Theorem for rings, the image of \( {\phi }_{... | Yes |
Theorem 21.13. Let \( E = F\\left( \\alpha \\right) \) be a simple extension of \( F \), where \( \\alpha \\in E \) is algebraic over \( F \). Suppose that the degree of \( \\alpha \) over \( F \) is \( n \). Then every element \( \\beta \\in E \) can be expressed uniquely in the form\n\n\[ \n\\beta = {b}_{0} + {b}_{1}... | Proof. Since \( {\\phi }_{\\alpha }\\left( {F\\left\\lbrack x\\right\\rbrack }\\right) \\cong F\\left( \\alpha \\right) \), every element in \( E = F\\left( \\alpha \\right) \) must be of the form \( {\\phi }_{\\alpha }\\left( {f\\left( x\\right) }\\right) = f\\left( \\alpha \\right) \), where \( f\\left( \\alpha \\rig... | Yes |
Example 21.14. Since \( {x}^{2} + 1 \) is irreducible over \( \mathbb{R},\left\langle {{x}^{2} + 1}\right\rangle \) is a maximal ideal in \( \mathbb{R}\left\lbrack x\right\rbrack \) . So \( E = \mathbb{R}\left\lbrack x\right\rbrack /\left\langle {{x}^{2} + 1}\right\rangle \) is a field extension of \( \mathbb{R} \) tha... | We know that \( {\alpha }^{2} = - 1 \) in \( E \), since\n\n\[{\alpha }^{2} + 1 = {\left( x + \left\langle {x}^{2} + 1\right\rangle \right) }^{2} + \left( {1 + \left\langle {{x}^{2} + 1}\right\rangle }\right)\]\n\n\[= \left( {{x}^{2} + 1}\right) + \left\langle {{x}^{2} + 1}\right\rangle\]\n\n\[= 0\text{.}\]\n\nHence, w... | Yes |
Theorem 21.15. Every finite extension field \( E \) of a field \( F \) is an algebraic extension. | Proof. Let \( \alpha \in E \) . Since \( \left\lbrack {E : F}\right\rbrack = n \), the elements\n\n\[ 1,\alpha ,\ldots ,{\alpha }^{n} \]\n\ncannot be linearly independent. Hence, there exist \( {a}_{i} \in F \), not all zero, such that\n\n\[ {a}_{n}{\alpha }^{n} + {a}_{n - 1}{\alpha }^{n - 1} + \cdots + {a}_{1}\alpha +... | Yes |
Theorem 21.17. 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\n\n\[ \left\lbrack {K : F}\right\rbrack = \left\lbrack {K : E}\right\rbrack \left\lbrack {E : F}\right\rbrack . \] | Proof. Let \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \) be a basis for \( E \) as a vector space over \( F \) and \( \left\{ {{\beta }_{1},\ldots ,{\beta }_{m}}\right\} \) be a basis for \( K \) as a vector space over \( E \) . We claim that \( \left\{ {{\alpha }_{i}{\beta }_{j}}\right\} \) is a basis fo... | Yes |
Let \( E \) be an extension field of \( F \) . If \( \alpha \in E \) is algebraic over \( F \) with minimal polynomial \( p\left( x\right) \) and \( \beta \in F\left( \alpha \right) \) with minimal polynomial \( q\left( x\right) \), then \( \deg q\left( x\right) \) divides \( \deg p\left( x\right) \) . | We know that \( \deg p\left( x\right) = \left\lbrack {F\left( \alpha \right) : F}\right\rbrack \) and \( \deg q\left( x\right) = \left\lbrack {F\left( \beta \right) : F}\right\rbrack \) . Since \( F \subset \) \( F\left( \beta \right) \subset F\left( \alpha \right) , \)\n\n\[ \left\lbrack {F\left( \alpha \right) : F}\r... | Yes |
Let us determine an extension field of \( \mathbb{Q} \) containing \( \sqrt{3} + \sqrt{5} \) . | It is easy to determine that the minimal polynomial of \( \sqrt{3} + \sqrt{5} \) is \( {x}^{4} - {16}{x}^{2} + 4 \) . It follows that\n\n\[ \left\lbrack {\mathbb{Q}\left( {\sqrt{3} + \sqrt{5}}\right) : \mathbb{Q}}\right\rbrack = 4 \]\n\nWe know that \( \{ 1,\sqrt{3}\} \) is a basis for \( \mathbb{Q}\left( \sqrt{3}\righ... | Yes |
Let us compute a basis for \( \mathbb{Q}\left( {\sqrt[3]{5},\sqrt{5}i}\right) \), where \( \sqrt{5} \) is the positive square root of 5 and \( \sqrt[3]{5} \) is the real cube root of 5. | We know that \( \sqrt{5}i \notin \mathbb{Q}\left( \sqrt[3]{5}\right) \), so\n\n\[ \left\lbrack {\mathbb{Q}\left( {\sqrt[3]{5},\sqrt{5}i}\right) : \mathbb{Q}\left( \sqrt[3]{5}\right) }\right\rbrack = 2. \]\n\nIt is easy to determine that \( \{ 1,\sqrt{5}i\} \) is a basis for \( \mathbb{Q}\left( {\sqrt[3]{5},\sqrt{5}i}\r... | Yes |
Theorem 21.22. Let \( E \) be a field extension of \( F \) . Then the following statements are equivalent.\n\n1. \( E \) is a finite extension of \( F \) .\n\n2. There exists a finite number of algebraic elements \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in E \) such that \( E = \) \( F\left( {{\alpha }_{1},\ldots ,{\alp... | Proof. \( \;\left( 1\right) \Rightarrow \left( 2\right) \) . 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 \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) in \( E \) such that \( E = F\left( {{\alpha }_{1... | Yes |
Theorem 21.23. Let \( E \) be an extension field of \( F \) . The set of elements in \( E \) that are algebraic over \( F \) form a field. | Proof. Let \( \alpha ,\beta \in E \) be algebraic over \( F \) . Then \( F\left( {\alpha ,\beta }\right) \) is a finite extension of \( F \) . Since every element of \( F\left( {\alpha ,\beta }\right) \) is algebraic over \( F,\alpha \pm \beta ,{\alpha \beta } \), and \( \alpha /\beta \left( {\beta \neq 0}\right) \) ar... | Yes |
Theorem 21.25. A field \( F \) is algebraically closed if and only if every nonconstant polynomial in \( F\left\lbrack x\right\rbrack \) factors into linear factors over \( F\left\lbrack x\right\rbrack \) . | Proof. Let \( F \) be an algebraically closed field. If \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) is a nonconstant polynomial, then \( p\left( x\right) \) has a zero in \( F \), say \( \alpha \) . Therefore, \( x - \alpha \) must be a factor of \( p\left( x\right) \) and so \( p\left( x\right) = \left( {x... | Yes |
Corollary 21.26. An algebraically closed field \( F \) has no proper algebraic extension \( E \) . | Proof. Let \( E \) be an algebraic extension of \( F \) ; then \( F \subset E \) . For \( \alpha \in E \), the minimal polynomial of \( \alpha \) is \( x - \alpha \) . Therefore, \( \alpha \in F \) and \( F = E \) . | Yes |
Theorem 21.27. 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. | No |
Theorem 21.31. Let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be a nonconstant polynomial. Then there exists a splitting field \( E \) for \( p\left( x\right) \) . | Proof. We will use mathematical induction on the degree of \( p\left( x\right) \) . If \( \deg p\left( x\right) = 1 \), then \( p\left( x\right) \) is a linear polynomial and \( E = F \) . Assume that the theorem is true for all polynomials of degree \( k \) with \( 1 \leq k < n \) and let \( \deg p\left( x\right) = n ... | Yes |
Lemma 21.32. Let \( \\phi : E \\rightarrow F \) be an isomorphism of fields. Let \( K \) be an extension field of \( E \) and \( \\alpha \\in K \) be algebraic over \( E \) with minimal polynomial \( p\\left( x\\right) \) . Suppose that \( L \) is an extension field of \( F \) such that \( \\beta \) is root of the poly... | Proof. If \( p\\left( x\\right) \) has degree \( n \), then by Theorem 21.13 we can write any element in \( E\\left( \\alpha \\right) \) as a linear combination of \( 1,\\alpha ,\\ldots ,{\\alpha }^{n - 1} \) . Therefore, the isomorphism that we are seeking must be\n\n\[ \n\\bar{\\phi }\\left( {{a}_{0} + {a}_{1}\\alpha... | Yes |
Theorem 21.33. Let \( \phi : E \rightarrow F \) be an isomorphism of fields and let \( p\left( x\right) \) be a nonconstant polynomial in \( E\left\lbrack x\right\rbrack \) and \( q\left( x\right) \) the corresponding polynomial in \( F\left\lbrack x\right\rbrack \) under the isomorphism. If \( K \) is a splitting fiel... | Proof. We will use mathematical induction on the degree of \( p\left( x\right) \) . We can assume that \( p\left( x\right) \) is irreducible over \( E \) . Therefore, \( q\left( x\right) \) is also irreducible over \( F \) . If \( \deg p\left( x\right) = 1 \), then by the definition of a splitting field, \( K = E \) an... | Yes |
Theorem 21.35. The set of all constructible real numbers forms a subfield \( F \) of the field of real numbers. | Proof. Let \( \alpha \) and \( \beta \) be constructible numbers. We must show that \( \alpha + \beta ,\alpha - \beta ,{\alpha \beta } \), and \( \alpha /\beta \left( {\beta \neq 0}\right) \) are also constructible numbers. We can assume that both \( \alpha \) and \( \beta \) are positive with \( \alpha > \beta \) . It... | No |
Lemma 21.37. If \( \alpha \) is a constructible number, then \( \sqrt{\alpha } \) is a constructible number. | Proof. In Figure 21.38 the triangles \( \bigtriangleup {ABD},\bigtriangleup {BCD} \), and \( \bigtriangleup {ABC} \) are similar; hence, \( 1/x = x/\alpha \), or \( {x}^{2} = \alpha \). | Yes |
Lemma 21.39. Let \( F \) be a subfield of \( \mathbb{R} \). 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 \( F \). 2. If a circle has a center at a point with coordinates in \( F \) and a radius that is also in \( F \), then i... | Proof. Let \( \left( {{x}_{1},{y}_{1}}\right) \) and \( \left( {{x}_{2},{y}_{2}}\right) \) be points on a line whose coordinates are in \( F \). If \( {x}_{1} = {x}_{2} \), then the equation of the line through the two points is \( x - {x}_{1} = 0 \), which has the form \( {ax} + {by} + c = 0 \). If \( {x}_{1} \neq {x}... | Yes |
Theorem 21.41. A real number \( \alpha \) is a constructible number if and only if there exists a sequence of fields\n\n\[ \n\\mathbb{Q} = {F}_{0} \\subset {F}_{1} \\subset \\cdots \\subset {F}_{k}\n\]\n\nsuch that \( {F}_{i} = {F}_{i - 1}\\left( \\sqrt{{\\alpha }_{i}}\\right) \) with \( {\\alpha }_{i} \\in {F}_{i} \) ... | Proof. The existence of the \( {F}_{i} \) ’s and the \( {\\alpha }_{i} \) ’s is a direct consequence of Lemma 21.40 and of the fact that\n\n\[ \n\\left\\lbrack {{F}_{k} : \\mathbb{Q}}\\right\\rbrack = \\left\\lbrack {{F}_{k} : {F}_{k - 1}}\\right\\rbrack \\left\\lbrack {{F}_{k - 1} : {F}_{k - 2}}\\right\\rbrack \\cdots... | Yes |
Proposition 22.2. If \( F \) is a finite field of characteristic \( p \), then the order of \( F \) is \( {p}^{n} \) for some \( n \in \mathbb{N} \) . | Proof. Let \( \phi : \mathbb{Z} \rightarrow F \) be the ring homomorphism defined by \( \phi \left( n\right) = n \cdot 1 \) . Since the characteristic of \( F \) is \( p \), the kernel of \( \phi \) must be \( p\mathbb{Z} \) and the image of \( \phi \) must be a subfield of \( F \) isomorphic to \( {\mathbb{Z}}_{p} \) ... | Yes |
Lemma 22.3 Freshman's Dream. Let \( p \) be prime and \( D \) be an integral domain of characteristic \( p \) . Then\n\n\[ \n{a}^{{p}^{n}} + {b}^{{p}^{n}} = {\left( a + b\right) }^{{p}^{n}} \n\]\n\nfor 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 2.4) to verify the case for \( n = 1 \) ; that is,\n\n\[ \n{\left( a + b\right) }^{p} = \mathop{\sum }\limits_{{k = 0}}^{p}\left( \begin{array}{l} p \\ k \end{array}\right) {a}^{k}{b}^{p - k... | Yes |
The polynomial \( {x}^{2} - 2 \) is separable over \( \mathbb{Q} \) since it factors as \( \left( {x - \sqrt{2}}\right) (x + \) \( \sqrt{2} \) ). In fact, \( \mathbb{Q}\left( \sqrt{2}\right) \) is a separable extension of \( \mathbb{Q} \) . Let \( \alpha = a + b\sqrt{2} \) be any element in \( \mathbb{Q}\left( \sqrt{2}... | \[ {x}^{2} - {2ax} + {a}^{2} - 2{b}^{2} = \left( {x - \left( {a + b\sqrt{2}}\right) }\right) \left( {x - \left( {a - b\sqrt{2}}\right) }\right) . \] | Yes |
Lemma 22.5. Let \( F \) be a field and \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) . Then \( f\left( x\right) \) is separable if and only if \( f\left( x\right) \) and \( {f}^{\prime }\left( x\right) \) are relatively prime. | Proof. Let \( f\left( x\right) \) be separable. Then \( f\left( x\right) \) factors over some extension field of \( F \) as \( f\left( x\right) = \left( {x - {\alpha }_{1}}\right) \left( {x - {\alpha }_{2}}\right) \cdots \left( {x - {\alpha }_{n}}\right) \), where \( {\alpha }_{i} \neq {\alpha }_{j} \) for \( i \neq j ... | Yes |
Theorem 22.6. For every prime \( p \) and every positive integer \( n \), there exists a finite field \( F \) with \( {p}^{n} \) elements. Furthermore, any field of order \( {p}^{n} \) is isomorphic to the splitting field of \( {x}^{{p}^{n}} - x \) over \( {\mathbb{Z}}_{p} \) . | Proof. Let \( f\left( x\right) = {x}^{{p}^{n}} - x \) and let \( F \) be the splitting field of \( f\left( x\right) \) . Then by Lemma 22.5, \( f\left( x\right) \) has \( {p}^{n} \) distinct zeros in \( F \), since \( {f}^{\prime }\left( x\right) = {p}^{n}{x}^{{p}^{n} - 1} - 1 = - 1 \) is relatively prime to \( f\left(... | Yes |
Theorem 22.7. Every subfield of the Galois field \( \operatorname{GF}\left( {p}^{n}\right) \) has \( {p}^{m} \) elements, where \( m \) divides \( n \) . Conversely, if \( m \mid n \) for \( m > 0 \), then there exists a unique subfield of \( \mathrm{{GF}}\left( {p}^{n}\right) \) isomorphic to \( \mathrm{{GF}}\left( {p... | Proof. Let \( F \) be a subfield of \( E = \operatorname{GF}\left( {p}^{n}\right) \) . Then \( F \) must be a field extension of \( K \) that contains \( {p}^{m} \) elements, where \( K \) is isomorphic to \( {\mathbb{Z}}_{p} \) . Then \( m \mid n \), since \( \left\lbrack {E : K}\right\rbrack = \left\lbrack {E : F}\ri... | Yes |
Theorem 22.10. 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 \( {F}^{ * } \) of order \( n \) . By the Fundamental Theorem of Finite Abelian Groups (Theorem 13.4), \[ G \cong {\mathbb{Z}}_{{p}_{1}^{{e}_{1}}} \times \cdots \times {\mathbb{Z}}_{{p}_{k}^{{e}_{k}}} \] where \( n = {p}_{1}^{{e}_{1}}\cdots {p}_{k}^{{e}_{k}} \) and the \( {p}_... | Yes |
Corollary 22.12. Every finite extension \( E \) of a finite field \( F \) is a simple extension of \( F \) . | Proof. Let \( \alpha \) be a generator for the cyclic group \( {E}^{ * } \) of nonzero elements of \( E \) . Then \( E = F\left( \alpha \right) \) . | Yes |
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