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Proposition 4.16. Let \( R \) be a UFD, and let \( K \) be its field of fractions. Let \( f \in \) \( R\left\lbrack x\right\rbrack \) be a nonconstant, irreducible polynomial. Then \( f \) is irreducible as an element of \( K\left\lbrack x\right\rbrack \) . | Proof. First note that \( f \) is primitive: otherwise we could factor out its content, and \( f \) would not be irreducible.\n\nNext, assume \( f = {gh} \), with \( g, h \in K\left\lbrack x\right\rbrack \) ; we have to prove that either \( g \) or \( h \) is a unit in \( K\left\lbrack x\right\rbrack \) . Let \( c, d \... | Yes |
Corollary 4.17. Let \( R \) be a UFD and \( K \) the field of fractions of \( R \) . Let \( f \in R\left\lbrack x\right\rbrack \) be a nonconstant polynomial. Then \( f \) is irreducible in \( R\left\lbrack x\right\rbrack \) if and only if it is irreducible in \( K\left\lbrack x\right\rbrack \) and primitive. | The proof amounts to tying up loose ends, and I leave it to the reader (Exercise 4.21). | No |
Lemma 5.1. Let \( R \) be an integral domain, and let \( f \in R\left\lbrack x\right\rbrack \) be a polynomial of degree \( n \) . Then the number of roots of \( f \), counted with multiplicity, is at most \( n \) . | Proof. The number of roots of \( f \) in \( R \) is less than or equal to the number of roots of \( f \) viewed as a polynomial over the field of fractions \( K \) of \( R \) ; so we may replace \( R \) by \( K \) .\n\nNow, \( K\left\lbrack x\right\rbrack \) is a UFD, and the roots of \( f \) correspond to the irreduci... | Yes |
Corollary 5.2. Let \( R \) be an infinite integral domain, and let \( f, g \in R\left\lbrack x\right\rbrack \) be polynomials. Then \( f = g \) if and only if the evaluation functions \( r \mapsto f\left( r\right), r \mapsto g\left( r\right) \) agree. | Proof. Indeed, the two functions agree if and only if every \( a \in R \) is a root of \( f - g \) ; but a nonzero polynomial over \( R \) cannot have infinitely many roots, by Lemma 5.1 | Yes |
Proposition 5.3. Let \( k \) be a field. A polynomial \( f \in k\left\lbrack x\right\rbrack \) of degree 2 or 3 is irreducible if and only if it has no roots. | Proof. Exercise 5.5. | No |
Let \( {\mathbb{F}}_{2} \) be the field \( \mathbb{Z}/2\mathbb{Z} \). The polynomial \( f\left( t\right) = {t}^{2} + t + 1 \in {\mathbb{F}}_{2}\left\lbrack t\right\rbrack \) is irreducible, since it has no roots: \( f\left( 0\right) = f\left( 1\right) = 1 \) . Therefore the ideal \( \left( {{t}^{2} + t + 1}\right) \) i... | \[ \frac{{\mathbb{F}}_{2}\left\lbrack t\right\rbrack }{\left( {t}^{2} + t + 1\right) }.\] | Yes |
Proposition 5.5. Let \( R \) be a UFD, and let \( K \) be its field of fractions. Let\n\n\[ f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \in R\left\lbrack x\right\rbrack ,\]\n\nand let \( c = \frac{p}{q} \in K \) be a root of \( f \), with \( p, q \in R,\gcd \left( {p, q}\right) = 1 \) . Then \( p \m... | Proof. By hypothesis,\n\n\[ {a}_{0} + {a}_{1}\frac{p}{q} + \cdots + {a}_{n}\frac{{p}^{n}}{{q}^{n}} = 0 \]\n\nthat is,\n\n\[ {a}_{0}{q}^{n} + {a}_{1}p{q}^{n - 1} + \cdots + {a}_{n}{p}^{n} = 0. \]\n\nTherefore\n\n\[ {a}_{0}{q}^{n} = - p\left( {{a}_{1}{q}^{n - 1} + \cdots + {a}_{n}{p}^{n - 1}}\right) ,\]\n\nproving that \... | Yes |
Looking for rational roots of the polynomial\n\n\[ 3 - {2x} + 3{x}^{2} - 2{x}^{3} + 3{x}^{4} - 2{x}^{5} \] | is therefore reduced to trying fractions \( \frac{p}{q} \) with \( q = \pm 1, \pm 2, p = \pm 1, \pm 3 \) . As it happens, \( \frac{3}{2} \) is the only root found among these possibilities, and it follows that it is the only rational root of the polynomial. | Yes |
Let \( k \) be a field, and let \( f\left( t\right) \in k\left\lbrack t\right\rbrack \) be a nonzero irreducible polynomial. Then \[ F \mathrel{\text{:=}} \frac{k\left\lbrack t\right\rbrack }{\left( f\left( t\right) \right) } \] is a field, endowed with a natural homomorphism \( i : k \hookrightarrow F \) (obtained as ... | Proof. Since \( k \) is a field, \( k\left\lbrack t\right\rbrack \) is a PID; hence \( \left( {f\left( t\right) }\right) \) is a maximal ideal of \( k\left\lbrack t\right\rbrack \), by Proposition III 4.13 Therefore \( F \) is indeed a field. Denoting cosets in \( k\left\lbrack t\right\rbrack /\left( {f\left( t\right) ... | Yes |
For \( k = \mathbb{R} \) and \( f\left( x\right) = {x}^{2} + 1 \), the field constructed in Proposition 5.7 is (isomorphic to) \( \mathbb{C} \) | this was checked carefully in Example III 4.8 | No |
Proposition 5.11. Let \( k \) be an algebraically closed field. Then \( k \) is infinite. | Proof. By contradiction, assume that \( k \) is algebraically closed and finite; let the elements of \( k \) be \( {c}_{1},\ldots ,{c}_{N} \). Then there are exactly \( N \) irreducible monic polynomials in \( k\left\lbrack x\right\rbrack \), namely \( (x - \) \( \left. {c}_{1}\right) ,\ldots ,\left( {x - {c}_{N}}\righ... | Yes |
Theorem 5.12. \( \mathbb{C} \) is algebraically closed. | Gauss is credited with providing the first proof 21 of this fundamental theorem (which is indeed known as the fundamental theorem of algebra.) 'Algebraic' proofs of the fundamental theorem of algebra require more than we know at this point (we will encounter one in SVII 7.1 after we have seen a little Galois theory); c... | No |
Proposition 5.13. Every polynomial \( f \in \mathbb{R}\left\lbrack x\right\rbrack \) of degree \( \geq 3 \) is reducible. | Proof. Let \( f \in \mathbb{R}\left\lbrack x\right\rbrack \) be a nonconstant polynomial:\n\n\[ f = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n}, \]\n\nwith all \( {a}_{i} \in \mathbb{R} \). By Theorem 5.12 f has a complex root \( z \):\n\n\[ {a}_{0} + {a}_{1}z + \cdots + {a}_{n}{z}^{n} = 0. \]\n\nApplying complex conj... | Yes |
Proposition 5.15. Let \( f \in \mathbb{Z}\left\lbrack x\right\rbrack \) be a primitive polynomial, and let \( p \) be a prime integer. Assume \( f{\;\operatorname{mod}\;p} \) has the same degree as \( f \) and is irreducible in \( \mathbb{Z}/p\mathbb{Z}\left\lbrack x\right\rbrack \) . Then \( f \) is irreducible in \( ... | Proof. Argue contrapositively: if \( f \) is primitive and reducible in \( \mathbb{Z}\left\lbrack x\right\rbrack \) and \( \deg f = \) \( n \), then \( f = {gh} \) with \( \deg g = d,\deg h = e, d + e = n \), and both \( d, e \), positive. But then the same can be said of \( f{\;\operatorname{mod}\;p} \), so \( f{\;\op... | Yes |
Corollary 5.16. There are irreducible polynomials in \( \mathbb{Z}\left\lbrack x\right\rbrack \) and \( \mathbb{Q}\left\lbrack x\right\rbrack \) of arbitrarily large degree. | Proof. By Proposition 4.16, the statement for \( \mathbb{Z}\left\lbrack x\right\rbrack \) implies the one for \( \mathbb{Q}\left\lbrack x\right\rbrack \) . By Proposition 5.15, it suffices to verify that there are irreducible polynomials in \( \mathbb{Z}/p\mathbb{Z}\left\lbrack x\right\rbrack \) of arbitrarily large de... | No |
Proposition 5.17. Let \( R \) be a (commutative) ring, and let \( \mathfrak{p} \) be a prime ideal of \( R \) . Let\n\n\[ f = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \in R\left\lbrack x\right\rbrack \]\n\nbe a polynomial, and assume that\n\n- \( {a}_{n} \notin \mathfrak{p} \) ;\n\n- \( {a}_{i} \in \mathfrak{p} \) ... | Proof. Argue by contradiction. Assume \( f = {gh} \) in \( R\left\lbrack x\right\rbrack \), with both \( d = \deg g \) and \( e = \deg h \) less than \( n = \deg f \) ; write\n\n\[ g = {b}_{0} + {b}_{1}x + \cdots + {b}_{d}{x}^{d},\;h = {c}_{0} + {c}_{1}x + \cdots + {c}_{e}{x}^{e}, \]\n\nand note that necessarily \( d >... | Yes |
For all \( n \) and all primes \( p \), the polynomial \( {x}^{n} - p \) is irreducible in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . | This follows immediately from Eisenstein’s criterion and gives an alternative proof of Corollary 5.16. | No |
Example 5.19. This is probably the most famous application of Eisenstein's criterion. Let \( p \) be a prime integer, and let\n\n\[ f\left( x\right) = 1 + x + {x}^{2} + \cdots + {x}^{p - 1} \in \mathbb{Z}\left\lbrack x\right\rbrack . \]\n\nThese polynomials are called cyclotomic; we will encounter them again in SVII 5.... | \[ f\left( {x + 1}\right) = \frac{{\left( x + 1\right) }^{p} - 1}{\left( {x + 1}\right) - 1} = {x}^{p - 1} + \left( \begin{matrix} p \\ p - 1 \end{matrix}\right) {x}^{p - 2} + \cdots + \left( \begin{array}{l} p \\ 3 \end{array}\right) {x}^{2} + \left( \begin{array}{l} p \\ 2 \end{array}\right) x + \left( \begin{array}{... | Yes |
Theorem 6.1. Let \( {I}_{1},\ldots ,{I}_{k} \) be ideals of \( R \) such that \( {I}_{i} + {I}_{j} = \left( 1\right) \) for all \( i \neq j \) . Then the natural homomorphism\n\n\[ \varphi : R \rightarrow \frac{R}{{I}_{1}} \times \cdots \times \frac{R}{{I}_{k}} \]\n\nis surjective and induces an isomorphism\n\n\[ \wide... | Theorem 6.1 is proven by an induction relying on the following lemma.\n\nLemma 6.2. Let \( {I}_{ | No |
Lemma 6.2. Let \( {I}_{1},\ldots ,{I}_{k} \) be ideals of \( R \) such that \( {I}_{i} + {I}_{k} = \left( 1\right) \) for all \( i = \) \( 1,\ldots, k - 1 \) . Then \( \left( {{I}_{1}\cdots {I}_{k - 1}}\right) + {I}_{k} = \left( 1\right) \) . | Proof. By hypothesis, for \( i = 1,\ldots, k - 1 \) there exists \( {a}_{i} \in {I}_{k} \) such that \( 1 - {a}_{i} \in {I}_{i} \) . Then \[ \left( {1 - {a}_{1}}\right) \cdots \left( {1 - {a}_{k - 1}}\right) \in {I}_{1}\cdots {I}_{k - 1} \] and \[ 1 - \left( {1 - {a}_{1}}\right) \cdots \left( {1 - {a}_{k - 1}}\right) \... | Yes |
Lemma 6.3. Let \( {I}_{1},\ldots ,{I}_{k} \) be ideals of \( R \) such that \( {I}_{i} + {I}_{j} = \left( 1\right) \) for all \( i \neq j \) . Then \( {I}_{1}\cdots {I}_{k} = {I}_{1} \cap \cdots \cap {I}_{k} \) . | Proof. By Lemma 6.2, under the stated hypotheses we have that \( {I}_{1}\cdots {I}_{k - 1} + {I}_{k} = \) (1) for \( k \geq 3 \) . Thus, the general statement is reduced by induction to the case \( k = 2 \) . (By the way, this case is Exercise III 4.5) Assume \( I \) and \( J \) are ideals of \( R \) such that \( I + J... | No |
Corollary 6.4. Let \( R \) be a PID, and let \( {a}_{1},\ldots ,{a}_{k} \in R \) be elements such that \( \gcd \left( {{a}_{i},{a}_{j}}\right) = 1 \) for all \( i \neq j \) . Let \( a = {a}_{1}\cdots {a}_{k} \) . Then the function \[ \varphi : \frac{R}{\left( a\right) } \rightarrow \frac{R}{\left( {a}_{1}\right) } \tim... | This is an immediate consequence of Theorem 6.1, since (in a PID!) \( \gcd \left( {a, b}\right) = \) 1 if and only if \( \left( {a, b}\right) = \left( 1\right) \) as ideals. | Yes |
Lemma 6.5. The function \( N \) is a Euclidean valuation on \( \mathbb{Z}\left\lbrack i\right\rbrack \) ; further, \( N \) is multiplicative in the sense that \( \forall z, w \in \mathbb{Z}\left\lbrack i\right\rbrack \)\n\n\[N\left( {zw}\right) = N\left( z\right) N\left( w\right)\] | Proof. The multiplicativity is an immediate consequence of the elementary properties of complex conjugation:\n\n\[N\left( {zw}\right) = \left( {zw}\right) \left( \overline{zw}\right) = \left( {z\bar{z}}\right) \left( {w\bar{w}}\right) = N\left( z\right) N\left( w\right) .\n\nTo see that \( N \) is a Euclidean valuation... | Yes |
Lemma 6.6. The units of \( \mathbb{Z}\left\lbrack i\right\rbrack \) are \( \pm 1, \pm i \) . | Proof. If \( u \) is a unit in \( \mathbb{Z}\left\lbrack i\right\rbrack \), then there exists \( v \in \mathbb{Z}\left\lbrack i\right\rbrack \) such that \( {uv} = 1 \) . But then \( N\left( u\right) N\left( v\right) = N\left( {uv}\right) = N\left( 1\right) = 1 \) by multiplicativity, so \( N\left( u\right) \) is a uni... | Yes |
Lemma 6.7. Let \( q \in \mathbb{Z}\left\lbrack i\right\rbrack \) be a prime element. Then there is a prime integer \( p \in \mathbb{Z} \) such that \( N\left( q\right) = p \) or \( N\left( q\right) = {p}^{2} \) . | Proof. Since \( q \) is not a unit, \( N\left( q\right) \neq 1 \) (by Lemma 6.6). Thus \( N\left( q\right) \) is a nontrivial product of (integer) primes, and since \( q \) is prime in \( \mathbb{Z}\left\lbrack i\right\rbrack \supseteq \mathbb{Z}, q \) must divide one of the prime integer factors of \( N\left( q\right)... | Yes |
The prime integer 3 is a prime element of \( \mathbb{Z}\left\lbrack i\right\rbrack \) ; this can be verified by proving that 3 is irreducible in \( \mathbb{Z}\left\lbrack i\right\rbrack \) (since \( \mathbb{Z}\left\lbrack i\right\rbrack \) is a UFD). | For this purpose, note that since \( N\left( 3\right) = 9 \), the norm of a factor of 3 would have to be a divisor of 9, that is, 1, 3, or 9. Gaussian integers with norm 1 are units, and those with norm 9 are associates of 3 (Exercise 6.10); thus a nontrivial factor of 3 would necessarily have norm equal to 3 . But the... | No |
A positive integer prime \( p \in \mathbb{Z} \) splits in \( \mathbb{Z}\left\lbrack i\right\rbrack \) if and only if it is the sum of two squares in \( \mathbb{Z} \) . | Proof. First assume that \( p = {a}^{2} + {b}^{2} \), with \( a, b \in \mathbb{Z} \) . Then\n\n\[ p = \left( {a + {bi}}\right) \left( {a - {bi}}\right) \]\n\nin \( \mathbb{Z}\left\lbrack i\right\rbrack \), and \( N\left( {a \pm {bi}}\right) = {a}^{2} + {b}^{2} = p \neq 1 \), so neither of the two factors is a unit in \... | Yes |
Lemma 1.2. Let \( M \) be an \( R \) -module, and let \( S \subseteq M \) be a linearly independent subset. Then there exists a maximal linearly independent subset of \( M \) containing \( S \) . | Proof. Consider the family \( \mathcal{S} \) of linearly independent subsets of \( M \) containing \( S \) , ordered by inclusion. Since \( S \) is linearly independent, \( \mathcal{S} \neq \varnothing \) . By Zorn’s lemma, it suffices to verify that every chain in \( \mathcal{S} \) has an upper bound. Indeed, the unio... | Yes |
Lemma 1.5. An R-module \( M \) is free if and only if it admits a basis. In fact, \( B \subseteq M \) is a basis if and only if the natural homomorphism \( {R}^{\oplus B} \rightarrow M \) is an isomorphism. | Proof. This is immediate from Definition 1.1 if \( B \subseteq M \) is linearly independent and generates \( M \), then the corresponding homomorphism \( {R}^{\oplus B} \rightarrow M \) is injective and surjective. Conversely, if \( \varphi : {R}^{\oplus B} \rightarrow M \) is an isomorphism, then \( B \) is identified... | Yes |
Lemma 1.6. Let \( R = k \) be a field, and let \( V \) be a \( k \) -vector space. Let \( B \) be a maximal linearly independent subset of \( V \) ; then \( B \) is a basis of \( V \). | Proof. Let \( v \in V, v \notin B \) . Then \( B \cup \{ v\} \) is not linearly independent, by the maximality of \( B \) ; therefore, there exist \( {c}_{0},\ldots ,{c}_{t} \in k \) and (distinct) \( {b}_{1},\ldots ,{b}_{t} \in B \) such that\n\n\[ \n{c}_{0}v + {c}_{1}{b}_{1} + \cdots + {c}_{t}{b}_{t} = 0, \n\]\n\nwit... | Yes |
Proposition 1.7. Let \( R = k \) be a field, and let \( V \) be a \( k \) -vector space. Let \( S \) be a linearly independent set of vectors of \( V \) . Then there exists a basis \( B \) of \( V \) containing \( S \) . | Proof. Put Lemma 1.2, Lemma 1.5, and Lemma 1.6 together. | No |
Lemma 1.8. Let \( R = k \) be a field, and let \( V \) be a \( k \) -vector space. Let \( B \) be a minimal generating set for \( V \) ; then \( B \) is a basis of \( V \) . | ## Proof. Exercise 1.6. | No |
Proposition 1.9. Let \( R \) be an integral domain, and let \( M \) be a free \( R \) -module. Let \( B \) be a maximal linearly independent subset of \( M \), and let \( S \) be a linearly independent subset. Then 4 \( \left| S\right| \leq \left| B\right| \) . | Proof. By taking fields of fractions, the general case over an integral domain is easily reduced to the case of vector spaces over a field; see Exercise 1.7 We may then assume that \( R = k \) is a field and \( M = V \) is a \( k \) -vector space.\n\nWe have to prove that there is an injective map \( j : S \hookrightar... | No |
An uncountable subset of \( \mathbb{C}\left\lbrack x\right\rbrack \) is necessarily linearly dependent. | Indeed, \( \mathbb{C}\left\lbrack x\right\rbrack \) has a countable basis over \( \mathbb{C} \) : for example, \( \left\{ {1, x,{x}^{2},{x}^{3},\ldots }\right\} \). | Yes |
Corollary 1.11. Let \( R \) be an integral domain, and let \( A, B \) be sets. Then\n\n\[
{F}^{R}\left( A\right) \cong {F}^{R}\left( B\right) \Leftrightarrow \text{ there is a bijection }A \cong B.\n\] | Proof. Exercise 1.8. | No |
Proposition 1.15. Let \( R \) be an integral domain, and let \( M \) be a free \( R \) -module; assume that \( M \) is generated by \( S : M = \langle S\rangle \) . Then \( S \) contains a maximal linearly independent subset of \( M \) . | Proof. By Exercise 1.7 we may assume that \( R \) is a field and \( M = V \) is a vector space. Use Zorn’s lemma to obtain a linearly independent subset \( B \subseteq S \) which is maximal among subsets of \( S \) . Arguing as in the proof of Lemma 1.6 shows that \( S \) is in the span of \( B \), and it follows that ... | No |
Lemma 2.1. For all \( m \times n \) matrices \( A \) with entries in \( R \) :\n\n- The function \( \varphi : {R}^{n} \rightarrow {R}^{m} \) defined by \( \varphi \left( \mathbf{v}\right) = A \cdot \mathbf{v} \) is a homomorphism of \( R \) -modules.\n\n- Every \( R \) -module homomorphism \( {R}^{n} \rightarrow {R}^{m... | Proof. The first point follows immediately from the elementary properties of matrix multiplication recalled above: \( \forall r, s \in R,\forall \mathbf{v},\mathbf{w} \in {R}^{n} \)\n\n\[ \varphi \left( {r\mathbf{v} + s\mathbf{w}}\right) = A \cdot \left( {r\mathbf{v} + s\mathbf{w}}\right) = {rA} \cdot \mathbf{v} + {sA}... | Yes |
Corollary 2.2. The correspondence introduced in Lemma 2.1 gives an isomorphism of \( R \) -modules\n\n\[ \n{\mathcal{M}}_{m, n}\left( R\right) \cong {\operatorname{Hom}}_{R}\left( {{R}^{n},{R}^{m}}\right) .\n\] | Proof. The reader will check that the correspondence is a bijective homomorphism of \( R \) -modules; this is enough, by Exercise III 5.12 | No |
Lemma 2.3. This diagram commutes. That is, the matrix corresponding to a composition \( \varphi \circ \psi \) is the product of the matrices corresponding to \( \varphi \) and \( \psi \) . | Proof. This follows immediately from the associativity of matrix multiplication: for \( \mathbf{v} \in {R}^{n} \) and \( A \in {\mathcal{M}}_{m, p}\left( R\right), B \in {\mathcal{M}}_{p, n}\left( R\right) \) ,\n\n\[ A \cdot \left( {B \cdot \mathbf{v}}\right) = \left( {A \cdot B}\right) \cdot \mathbf{v} \]\n\nthat is, ... | Yes |
Proposition 2.7. Two matrices \( P, Q \in {\mathcal{M}}_{m, n}\left( R\right) \) are equivalent if \( Q \) may be obtained from \( P \) by a sequence of elementary operations. | Proof. To see that elementary operations produce equivalent matrices, it suffices (by Proposition 2.5) to express them as multiplications on the left or right \( {}^{12} \) by invertible matrices. Indeed, these operations may be performed by suitably multiplying by the matrices obtained from the identity matrix by perf... | No |
Proposition 2.9. Let \( R = k \) be a field, and let \( n \geq 0 \) be an integer. Then \( {\mathrm{{GL}}}_{n}\left( k\right) \) is generated by elementary matrices. | Proof. Let \( A = \left( {a}_{ij}\right) \) be an \( n \times n \) invertible matrix. In particular, some entry in the first column of \( A \) is nonzero; by performing a row switch if necessary, we may assume that \( {a}_{11} \) is nonzero. Multiplying the first row by \( {a}_{11}^{-1} \), we may assume that \( {a}_{1... | Yes |
Over a field, every \( m \times n \) matrix is equivalent to a matrix of the form\n\n\[ \left( \begin{matrix} {I}_{r} & 0 \\ 0 & 0 \end{matrix}\right) \]\n\n(where \( r \leq \min \left( {m, n}\right) \) and ’ 0 ’ stands for null matrices of appropriate sizes). | Different matrices of the type displayed in Proposition 2.10 are inequivalent (for example by rank considerations; cf. (3.3). Thus, Proposition 2.10 describes all equivalence classes of matrices over a field and shows that for any given \( m, n \) there are in fact only finitely many such classes (over a field!). | No |
Let \( A \) be a square matrix with entries in an integral domain \( R \). Let \( {A}^{\prime } \) be obtained from \( A \) by switching two rows or two columns. Then \( \det \left( {A}^{\prime }\right) = - \det \left( A\right) . | Proof. These are all essentially immediate from Definition 3.1 For example, switching two columns amounts to correcting each \( \sigma \) in the definition by a fixed transposition, changing the sign of all contributions to the \( \sum \) in the definition. The third point is immediate from distributivity. Combining th... | No |
A square matrix \( A \in {\mathcal{M}}_{n}\left( R\right) \) is invertible if and only if \( \det \left( A\right) \) is a unit in \( R \) . | Proof for \( R = \mathbf{a} \) field. If \( R = k \) is a field, we can use the considerations immediately preceding the statement. The first point is reduced to the case of a block matrix\n\n\[ \left( \begin{matrix} {I}_{r} & 0 \\ 0 & 0 \end{matrix}\right) \]\n\nfor which it is immediate. In fact, this shows that \( \... | No |
Lemma 3.4. With notation as above,\n\n\[ \n\\text{- for all}i = 1,\\ldots, n,\\det \\left( A\\right) = \\mathop{\\sum }\\limits_{{j = 1}}^{n}{a}_{ij}{A}^{\\left( ij\\right) }\\text{,}\n\]\n\n\[ \n\\text{- for all}j = 1,\\ldots, n,\\det \\left( A\\right) = \\mathop{\\sum }\\limits_{{i = 1}}^{n}{a}_{ij}{A}^{\\left( ij\\r... | Proof. This is a simple (if slightly messy) induction on \( n \), which I leave to the diligent reader. | No |
Corollary 3.5. Let \( R \) be a commutative ring and \( A \in {\mathcal{M}}_{n}\left( R\right) \) . Then \[ A \cdot \left( \begin{matrix} {A}^{\left( {11}\right) } & \cdots & {A}^{\left( n1\right) } \\ \vdots & \ddots & \vdots \\ {A}^{\left( 1n\right) } & \cdots & {A}^{\left( nn\right) } \end{matrix}\right) = \left( \b... | Proof. Along the diagonal of the right-hand side, this is a restatement of Lemma 3.4, Off the diagonal, one is evaluating (for example) \[ \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{{i}^{\prime }j}{A}^{\left( ij\right) } \] for \( {i}^{\prime } \neq i \) . By Lemma 3.4 this is the same as the determinant of the matrix obt... | Yes |
Proposition 3.6 (Cramer’s rule). Assume \( \det \left( A\right) \) is a unit, and let \( {A}^{\left( j\right) } \) be the matrix obtained by replacing the \( j \) -th column of \( A \) by the column vector \( b \) . Then\n\n\[ \n{x}_{j} = \det {\left( A\right) }^{-1}\det \left( {A}^{\left( j\right) }\right) .\n\] | Proof. Using Lemma 3.4, expand \( \det \left( {A}^{\left( j\right) }\right) \) with respect to the \( j \) -th column:\n\n\[ \n\det \left( {A}^{\left( j\right) }\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{A}^{\left( ij\right) }{b}_{i} \n\]\n\nTherefore\n\n\[ \n\left( \begin{matrix} {x}_{1} \\ \vdots \\ {x}_{n} \end{m... | Yes |
Proposition 3.7. The row rank of a matrix over a field \( k \) equals its column rank. | Proof. Equivalent matrices have the same ranks. Indeed, let \( P \in {\mathcal{M}}_{m, n}\left( k\right) \) ; the row space of \( P \) consists of all row vectors\n\n\[ \left( \begin{array}{lll} {a}_{1} & \cdots & {a}_{m} \end{array}\right) = \left( \begin{array}{lll} {v}_{1} & \cdots & {v}_{m} \end{array}\right) \cdot... | No |
Lemma 4.2. Submodules and direct sums of torsion-free modules are torsion-free. Free modules over an integral domain are torsion-free. | Proof. The first statement is immediate; the second follows from the first, since an integral domain is torsion-free as a module over itself. | No |
Let \( R = \mathbb{Z}\left\lbrack x\right\rbrack \), and let \( I = \left( {2, x}\right) \) . Then \( I \) is not a free \( R \) -module. More generally, let \( I \) be any nonprincipal ideal of an integral domain \( R \) ; then \( I \) is a torsion-free module which is not free. | Indeed, if \( I \) were free, then its rank would have to be 1 at most, by Proposition 1.9 (a basis for \( I \) would be a linearly independent subset of \( R \), and \( R \) has rank 1 over itself); thus one element would suffice to generate \( I \), and \( I \) would be principal. | Yes |
Lemma 4.5. Let \( R \) be an integral domain. Assume that every cyclic \( R \) -module is torsion-free. Then \( R \) is a field. | Proof. Let \( c \in R, c \neq 0 \) ; then \( M = R/\left( c\right) \) is a cyclic module. Note that \( \operatorname{Tor}\left( M\right) = \) \( M \) : indeed, the class of 1 generates \( R/\left( c\right) \) and belongs to \( \operatorname{Tor}\left( M\right) \) since \( c \cdot 1 \) is 0 \( {\;\operatorname{mod}\;\le... | Yes |
Lemma 4.8. If \( R \) is a Noetherian ring, then every finitely generated \( R \) -module is finitely presented. | Proof. If \( M \) is a finitely generated module, there is an exact sequence\n\n\[ \n{R}^{m}\overset{\pi }{ \rightarrow }M \rightarrow 0 \n\]\n\nfor some \( m \) . Since \( R \) is Noetherian, \( {R}^{m} \) is Noetherian as an \( R \) -module (Corollary III 6.8). Thus \( \ker \pi \) is finitely generated; that is, ther... | Yes |
Proposition 4.10. Let \( R \) be an integral domain. Then \( R \) is a field if and only if every finitely generated \( R \) -module is free. | Proof. If \( R \) is a field, then every \( R \) -module is free, by Proposition 1.7. For the converse, assume that every finitely generated \( R \) -module is free; in particular, every cyclic module is free; in particular, every cyclic module is torsion-free. But then \( R \) is a field, by Lemma 4.5 | Yes |
Lemma 4.12. Let \( A, B \) be matrices with entries in an integral domain \( R \), and let \( M, N \) denote the corresponding \( R \)-modules. Then \( M \oplus N \) corresponds to the block matrix \[ \left( \begin{matrix} A & 0 \\ 0 & B \end{matrix}\right) \] | Proof. This follows immediately from Exercise 4.16 | No |
Proposition 4.13. Let \( A \) be a matrix with entries in an integral domain \( R \), and let \( B \) be obtained from \( A \) by any sequence of the following operations:\n\n- switch two rows or two columns;\n\n- add to one row (resp., column) a multiple of another row (resp., column);\n\n- multiply all entries in one... | Proof. The first three operations are the 'elementary operations' of [2.3] and they transform a matrix into an equivalent one (by Proposition 2.7); as observed above, this does not affect the corresponding module, up to isomorphism.\n\nAs for the fourth operation, if \( u \) is a unit and the only nonzero entry in (say... | Yes |
Example 4.14. The matrix with integer entries\n\n\\[ \n\\left( \\begin{array}{ll} 1 & 3 \\\\ 2 & 3 \\\\ 5 & 9 \\end{array}\\right) \n\\]\n\ndetermines an abelian group \( G \) . | Subtract three times the first column from the second column, obtaining\n\n\\[ \n\\left( \\begin{matrix} 1 & 0 \\\\ 2 & - 3 \\\\ 5 & - 6 \\end{matrix}\\right) \n\\]\n\nthe \( \\left( {1,1}\\right) \) entry is a unit and the only nonzero entry in the first row, so we can remove the first row and column:\n\n\\[ \n\\left(... | Yes |
Proposition 5.1. Let \( R \) be a PID, let \( F \) be a finitely generated free module over \( R \) , and let \( M \subseteq F \) be a submodule. Then \( M \) is free. | We will actually prove a more precise result, in view of the full statement of the classification theorem: we will show that there is a basis \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) of \( F \) and elements \( {a}_{1},\ldots ,{a}_{m} \) of \( R \) (with \( m \leq n \) ) such that\n\n\[ \n{y}_{1} = {a}_{1}{x}_{1},\... | Yes |
Lemma 5.2. Let \( R \) be a PID, let \( F \) be a finitely generated free module over \( R \), and let \( M \subseteq F \) be a nonzero submodule. Then there exist \( a \in R, x \in F, y \in M \), and submodules \( {F}^{\prime } \subseteq F \) and \( {M}^{\prime } \subseteq M \), such that \( y = {ax} \neq 0,{M}^{\prim... | Proof. For all \( \varphi \in {\operatorname{Hom}}_{R}\left( {F, R}\right) ,\varphi \left( M\right) \) is a submodule of \( R \), that is, an ideal. The family of all these ideals is nonempty, and PIDs are Noetherian; therefore (by Proposition VI1.1 there exists a maximal element in the family, say \( \alpha \left( M\r... | Yes |
Corollary 5.3. Let \( R \) be a PID, let \( F \) be a finitely generated free module over \( R \) , and let \( M \subseteq F \) be a submodule. Then there exist a basis \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) of \( F \) and nonzero elements \( {a}_{1},\ldots ,{a}_{m} \) of \( R\left( {m \leq n}\right) \) such tha... | Proof. Now that we know that submodules of a free module are free, we see that the submodule \( {F}^{\prime } \subseteq F \) produced in Lemma 5.2 is free. The first part of the statement then follows from Lemma 5.2, by an inductive argument analogous to the proof of Proposition 5.1, and is left to the reader.\n\nThe m... | No |
Proposition 5.4. Let \( R \) be an integral domain. Then \( R \) is a PID if and only if for every finitely generated \( R \) -module \( M \) and every epimorphism \[ {R}^{{m}_{0}}\overset{{\pi }_{0}}{ \rightarrow }M \rightarrow 0 \] there exist a free \( R \) -module \( {R}^{{m}_{1}} \) and a homomorphism \( {\pi }_{1... | Proof. The fact that the stated condition implies that \( R \) is a PID was proved in Claim 4.11. For the converse, let \( {\pi }_{0} : {R}^{{m}_{0}} \rightarrow M \) be an epimorphism; then \( \ker {\pi }_{0} \) is free by Proposition 5.1 the result follows by choosing any isomorphism \( {\pi }_{1} : {R}^{{m}_{1}} \ri... | Yes |
Lemma 5.7. Let \( M \) be a torsion module, expressed as in Theorem 5.6 (with \( \operatorname{rk}M = \) \( 0) \) . Then \( \operatorname{Ann}\left( M\right) = \left( {a}_{m}\right) \) . Further, the prime ideals \( \left( {q}_{i}\right) \) are precisely the prime ideals of \( R \) containing \( \operatorname{Ann}\left... | Proof. By hypothesis\n\n\[ M \cong \frac{R}{\left( {a}_{1}\right) } \oplus \cdots \oplus \frac{R}{\left( {a}_{m}\right) } \]\n\nwith \( {a}_{1}\left| \cdots \right| {a}_{m} \) . If \( r \in \operatorname{Ann}\left( M\right) \), then\n\n\[ 0 = r\left( {1,\ldots ,1}\right) = \left( {r,\ldots, r}\right) . \]\n\nIn particu... | Yes |
Proposition 6.2. Two matrices \( A, B \in {\mathcal{M}}_{n}\left( R\right) \) are similar if and only if there exists an invertible matrix \( P \) such that\n\n\[ B = {PA}{P}^{-1}\text{.} \] | The reader who has really understood Proposition 2.5 will not need any detailed proof of this statement: it should be apparent from staring at the butterfly diagram\n\n\n\nwhich I am essentially copying from [2.2] Th... | No |
Proposition 6.5. Let \( \alpha \) be a linear transformation of a free \( R \) -module \( F \cong {R}^{n} \) . Then \( \det \left( \alpha \right) \neq 0 \) if and only if \( \alpha \) is injective. | Proof. Embed \( R \) in its field of fractions \( K \), and view \( \alpha \) as a linear transformation of \( {K}^{n} \) ; note that the determinant of \( \alpha \) is the same whether it is computed over \( R \) or over \( K \) . Then \( \alpha \) is injective as a linear transformation \( {R}^{n} \rightarrow {R}^{n}... | Yes |
Lemma 6.7. Let \( A, B \in {\mathcal{M}}_{n}\left( R\right) \) . Then \( \operatorname{tr}\left( {AB}\right) = \operatorname{tr}\left( {BA}\right) \) . | Proof. Let \( A = \left( {a}_{ij}\right), B = \left( {b}_{ij}\right) \) . Then \( {AB} = \left( {\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik}{b}_{kj}}\right) \) ; hence\n\n\[ \operatorname{tr}\left( {AB}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik}{b}_{ki} \]\n\nThis expression... | Yes |
Proposition 6.9. Let \( F \) be a free \( R \) -module of rank \( n \), and let \( \alpha \in {\operatorname{End}}_{R}\left( F\right) \) .\n\n- The characteristic polynomial \( {P}_{\alpha }\left( t\right) \) is a monic polynomial of degree \( n \).\n\n- The coefficient of \( {t}^{n - 1} \) in \( {P}_{\alpha }\left( t\... | Proof. The first point is immediate, and the third is checked by setting \( t = 0 \) . To verify the second assertion, let \( A = \left( {a}_{ij}\right) \) be a matrix representing \( \alpha \) with respect to any basis for \( F \), so that\n\n\[{P}_{\alpha }\left( t\right) = \det \left( \begin{matrix} t - {a}_{11} & -... | Yes |
Lemma 6.10. If \( \alpha \) and \( \beta \) are similar, then \( {\mathcal{I}}_{\alpha } = {\mathcal{I}}_{\beta } \) . | Proof. By hypothesis there exists an invertible \( \pi \) such that \( \beta = \pi \circ \alpha \circ {\pi }^{-1} \) . As\n\n\[ \n{\beta }^{k} = {\left( \pi \circ \alpha \circ {\pi }^{-1}\right) }^{k} = \left( {\pi \circ \alpha \circ {\pi }^{-1}}\right) \circ \left( {\pi \circ \alpha \circ {\pi }^{-1}}\right) \circ \cd... | Yes |
Theorem 6.11 (Cayley-Hamilton). Let \( {P}_{\alpha }\left( t\right) \) be the characteristic polynomial of the linear transformation \( \alpha \in {\operatorname{End}}_{R}\left( F\right) \) . Then\n\n\[ \n{P}_{\alpha }\left( \alpha \right) = 0 \n\] | This beautiful observation can be proved directly by judicious use of Cramer's rule \( {30} \), in the form of Corollary 3.5, cf. Exercise 6.9. In any case, the Cayley-Hamilton theorem will become essentially evident once we connect these linear algebra considerations with the classification theorem for finitely genera... | No |
Lemma 6.14. Let \( F \) be a finitely generated \( R \) -module, and let \( \alpha \in {\operatorname{End}}_{R}\left( F\right) \) . Then the set of eigenvalues of \( \alpha \) is precisely the set of roots in \( R \) of the characteristic polynomial \( {P}_{\alpha }\left( t\right) \) . | Proof. This is a straightforward consequence of Proposition 6.5\n\n\[ \lambda \text{is an eigenvalue for}\alpha \Leftrightarrow \exists \mathbf{v} \neq 0\text{such that}\alpha \left( \mathbf{v}\right) = {\lambda I}\left( \mathbf{v}\right) \]\n\n\[ \Leftrightarrow \exists \mathbf{v} \neq 0\text{such that}\left( {{\lambd... | Yes |
The matrix \[ \left( \begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}\right) \] has no eigenvalues over \( \mathbb{R} \), while it has eigenvalues over \( \mathbb{C} \). | Indeed, the characteristic polynomial \( {t}^{2} + 1 \) has no real roots and two complex roots. The reader should observe that, as a linear transformation of the real plane \( {\mathbb{R}}^{2} \), this matrix corresponds to a \( {90}^{ \circ } \) counterclockwise rotation; the reason why this transformation has no (re... | No |
Corollary 6.18. The number of eigenvalues of a linear transformation of \( {R}^{n} \) is at most \( n \) . If the base ring \( R \) is an algebraically closed field, then every linear transformation has exactly \( n \) eigenvalues (counted with algebraic multiplicity). | Proof. Immediate from Lemmas 6.14, V15.1, and V15.10. | No |
Lemma 7.2. Let \( \alpha ,\beta \) be linear transformations of a free \( R \) -module \( F \) . Then the corresponding \( R\left\lbrack t\right\rbrack \) -module structures on \( F \) are isomorphic if and only if \( \alpha \) and \( \beta \) are similar. | Proof. Denote by \( {F}_{\alpha },{F}_{\beta } \) the two \( R\left\lbrack t\right\rbrack \) -modules defined on \( F \) by \( \alpha ,\beta \) as per Claim 7.1\n\nAssume first that \( \alpha \) and \( \beta \) are similar. Then there exists an invertible \( R \) -linear transformation \( \pi : F \rightarrow F \) such ... | No |
Theorem 7.5. Let \( k \) be a field, and let \( V \) be a finite-dimensional vector space. Let \( \alpha \) be a linear transformation on \( V \), and endow \( V \) with the corresponding \( k\left\lbrack t\right\rbrack \) -module structure, as in Claim 7.1. Then the following hold:\n\n- There exist distinct monic irre... | Proof. Since \( \dim V \) is finite, \( V \) is finitely generated as a \( k \) -module and a fortiori as a \( k\left\lbrack t\right\rbrack \) -module. The two isomorphisms are then obtained by applying Theorem 5.6 All the relevant polynomials may be chosen to be monic since every polynomial over a field is the associa... | No |
Proposition 7.9. Let \( {f}_{1}\left( t\right) \left| \cdots \right| {f}_{m}\left( t\right) \) be the invariant factors of a linear transformation \( \alpha \) on a vector space \( V \) . Then the minimal polynomial \( {m}_{\alpha }\left( t\right) \) equals \( {f}_{m}\left( t\right) \) , and the characteristic polynomi... | Proof. Tracing definitions, \( \left( {{m}_{\alpha }\left( t\right) }\right) \) is the annihilator ideal of \( V \) when this is viewed as a \( k\left\lbrack t\right\rbrack \) -module via \( \alpha \) (as in Claim 7.1). Therefore the equality of \( {m}_{\alpha }\left( t\right) \) and \( {f}_{m}\left( t\right) \) is a r... | No |
Corollary 7.10 (Cayley-Hamilton). The minimal polynomial of a linear transformation divides its characteristic polynomial. | Proof. This has now become evident, as promised in [6.2, | No |
Proposition 7.11. Let \( A \in {\mathcal{M}}_{n}\left( k\right) \) be a square matrix. Then \( A \) is similar to its transpose. | Proof. If \( B \) is similar to \( A \) and we can prove that \( B \) is similar to its transpose \( {B}^{t} \) , then \( A \) is similar to its transpose \( {A}^{t} \) : because \( B = {PA}{P}^{-1},{B}^{t} = {QB}{Q}^{-1} \) give\n\n\[ \n{A}^{t} = \left( {{P}^{t}{QP}}\right) A{\left( {P}^{t}QP\right) }^{-1}.\n\] \n\nTh... | Yes |
Lemma 7.12. Assume that the characteristic polynomial \( {P}_{\alpha }\left( t\right) \) factors completely; that is,\n\n\[ \n{P}_{\alpha }\left( t\right) = \mathop{\prod }\limits_{{i = 1}}^{s}{\left( t - {\lambda }_{i}\right) }^{{m}_{i}} \n\] \n\nwhere \( {\lambda }_{i}, i = 1,\ldots, s \), are the distinct eigenvalue... | Proof. The first statement follows from uniqueness of factorizations. The statement about the minimal polynomial is immediate from Proposition 7.9 and the bookkeeping giving the equivalence of the two formulations in Theorem 7.5 | Yes |
One use of the Jordan canonical form is the enumeration of all possible similarity classes of transformations with given eigenvalues. For example, there are 5 similarity classes of linear transformations with a single eigenvalue \( \lambda \) with algebraic multiplicity 4, over a 4-dimensional vector space: indeed, the... | \[ \left( \begin{matrix} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{matrix}\right) ,\;\left( \begin{matrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{matrix}\right) ,\;\left( \begin{matrix} \lambda & 1 & 0 & 0 \\ 0 & ... | Yes |
Proposition 7.16. The geometric multiplicity of \( \lambda \) as an eigenvalue of \( \alpha \) equals the number of Jordan blocks corresponding to \( \lambda \) in the Jordan canonical form of \( \alpha \) . | Proof. As the geometric multiplicity is clearly additive in direct sums, it suffices to show that the geometric multiplicity of \( \lambda \) for the transformation corresponding to a single Jordan block\n\n\[ J = \left( \begin{matrix} \lambda & 1 & \ldots & 0 & 0 \\ 0 & \lambda & \ldots & 0 & 0 \\ \vdots & \vdots & \d... | Yes |
Proposition 7.18. Assume the characteristic polynomial of \( \alpha \in {\operatorname{End}}_{k}\left( V\right) \) factors completely over \( k \) . Then \( \alpha \) is diagonalizable if and only if the minimal polynomial of \( \alpha \) has no multiple roots. | Proof. Again, diagonalizability is equivalent to having all Jordan blocks of size 1 in the Jordan canonical form of \( \alpha \) . Therefore, if the characteristic polynomial of \( \alpha \) factors completely, then \( \alpha \) is diagonalizable if and only if all exponents \( {r}_{ij} \) appearing in Theorem 7.5 equa... | Yes |
Proposition 1.3. Let \( k \subseteq k\left( \alpha \right) \) be a simple extension. Consider the evaluation map \( \epsilon : k\left\lbrack t\right\rbrack \rightarrow k\left( \alpha \right) \), defined by \( f\left( t\right) \mapsto f\left( \alpha \right) \). Then we have the following:\n\n- \( \epsilon \) is injectiv... | Proof. Let \( F = k\left( \alpha \right) \). By the ’first isomorphism theorem’, the image of \( \epsilon : k\left\lbrack t\right\rbrack \rightarrow F \) is isomorphic to \( k\left\lbrack t\right\rbrack /\ker \left( \epsilon \right) \). Since \( F \) is an integral domain, so is \( k\left\lbrack t\right\rbrack /\ker \l... | Yes |
Consider the extension \( \mathbb{Q} \subseteq \mathbb{R} \). The polynomial \( {x}^{2} - 2 \in \mathbb{Q}\left\lbrack x\right\rbrack \) has roots in \( \mathbb{R} \): therefore, by Proposition V15.7 there exists a homomorphism (hence a field extension) | \[ \bar{\epsilon } : \;\frac{\mathbb{Q}\left\lbrack t\right\rbrack }{\left( {t}^{2} - 2\right) } \hookrightarrow \mathbb{R} \] such that the image of (the coset of) \( t \) is a root \( \alpha \) of \( {x}^{2} - 2 \). Proposition 1.3 simply identifies the image of this homomorphism with \( \mathbb{Q}\left( \alpha \righ... | No |
Proposition 1.5. Let \( {k}_{1} \subseteq {F}_{1} = {k}_{1}\left( {\alpha }_{1}\right) ,{k}_{2} \subseteq {F}_{2} = {k}_{2}\left( {\alpha }_{2}\right) \) be two finite simple extensions. Let \( {p}_{1}\left( t\right) \in {k}_{1}\left\lbrack t\right\rbrack \), resp., \( {p}_{2}\left( t\right) \in {k}_{2}\left\lbrack t\r... | Proof. Since every element of \( {k}_{1}\left( {\alpha }_{1}\right) \) is a linear combination of powers of \( {\alpha }_{1} \) with coefficients in \( {k}_{1}, j \) is determined by its action on \( {k}_{1} \) (which agrees with \( i \) ) and by \( j\left( {\alpha }_{1}\right) \), which is prescribed to be \( {\alpha ... | Yes |
Corollary 1.7. Let \( k \subseteq F = k\\left( \\alpha \\right) \) be a simple finite extension, and let \( p\\left( x\\right) \) be the minimal polynomial of \( \\alpha \) over \( k \) . Then \( \\left| {{\\operatorname{Aut}}_{k}\\left( F\\right) }\\right| \) equals the number of distinct roots of \( p\\left( x\\right... | Proof. Let \( j \\in {\\operatorname{Aut}}_{k}\\left( F\\right) \) . Since every element of \( F \) is a polynomial expression in \( \\alpha \) with coefficients in \( k \), and \( j \) extends the identity on \( k, j \) is determined by \( j\\left( \\alpha \\right) \) . Now\n\n\\[ \np\\left( {j\\left( \\alpha \\right)... | Yes |
Lemma 1.9. Let \( k \subseteq F \) be a finite extension. Then every \( \alpha \in F \) is algebraic over \( k \), of degree \( \leq \left\lbrack {F : k}\right\rbrack \) . | Proof. Since \( k \subseteq k\left( \alpha \right) \subseteq F \), the dimension of \( k\left( \alpha \right) \) as a \( k \) -vector space is bounded by \( {\dim }_{k}F = \left\lbrack {F : k}\right\rbrack \). Concretely, if \( k \subseteq F \) is finite and \( \alpha \in F \), then the powers \( 1,\alpha ,{\alpha }^{2... | Yes |
Proposition 1.10. Let \( k \subseteq E \subseteq F \) be field extensions. Then \( k \subseteq F \) is finite if and only if both \( k \subseteq E \) and \( E \subseteq F \) are finite. In this case,\n\n\[ \left\lbrack {F : k}\right\rbrack = \left\lbrack {F : E}\right\rbrack \left\lbrack {E : k}\right\rbrack . \] | Proof. If \( F \) is finite-dimensional as a vector space over \( k \), then so is its subspace \( E \) ; and any linear dependence relation of elements of \( F \) over the field \( k \) gives one over the larger field \( E \) . It follows that if \( k \subseteq F \) is finite, then so are \( k \subseteq E \) and \( E ... | Yes |
Let \( k \subseteq F \) be a field extension, and let \( \alpha \in F \) be an algebraic element over \( k \), of \( {odd} \) degree. Then I claim that \( \alpha \) may be written as a polynomial in \( {\alpha }^{2} \), with coefficients in \( k \) . | Indeed, \( k\left( {\alpha }^{2}\right) \) is intermediate between \( k \) and \( k\left( \alpha \right) \) :\n\n\[ k \subseteq k\left( {\alpha }^{2}\right) \subseteq k\left( \alpha \right) \]\n\nwhat can we say about the degree \( d \) of \( k\left( \alpha \right) \) over \( k\left( {\alpha }^{2}\right) \) ? Since \( ... | Yes |
Proposition 1.15. Let \( k \subseteq F = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) be a finitely generated field extension. Then the following are equivalent:\n\n(i) \( k \subseteq F \) is a finite extension.\n\n(ii) \( k \subseteq F \) is an algebraic extension.\n\n(iii) Each \( {\alpha }_{i} \) is algeb... | Proof. Lemma 1.9 shows that (i) \( \Rightarrow \) (ii); (ii) \( \Rightarrow \) (iii) trivially. Thus, we only need to prove that (iii) \( \Rightarrow \) (i), and to bound the degree of \( F \) over \( k \) in the process.\n\nAssume that each \( {\alpha }_{i} \) is algebraic over \( k \), and let \( {d}_{i} \) be the de... | No |
Let \( \overline{\mathbb{Q}} \subseteq \mathbb{C} \) be the set of complex numbers that are algebraic over \( \mathbb{Q} \); then \( \overline{\mathbb{Q}} \) is a field, by Corollary 1.16, and the extension \( \mathbb{Q} \subseteq \overline{\mathbb{Q}} \) is (tautologically) algebraic. | Note that \( \mathbb{Q} \subseteq \overline{\mathbb{Q}} \) is not a finite extension, because in it there are elements of arbitrarily high degree over \( \mathbb{Q} \): indeed, there exist irreducible polynomials in \( \mathbb{Q}\left\lbrack x\right\rbrack \) of arbitrarily high degree, as we observed in Corollary V15.... | Yes |
Corollary 1.18. Let \( k \subseteq E \subseteq F \) be field extensions. Then \( k \subseteq F \) is algebraic if and only if both \( k \subseteq E \) and \( E \subseteq F \) are algebraic. | Proof. If \( k \subseteq F \) is algebraic, then every element of \( F \) is algebraic over \( k \), hence over \( E \), and every element of \( E \) is algebraic over \( k \) ; thus \( E \subseteq F \) and \( k \subseteq E \) are algebraic.\n\nConversely, assume \( k \subseteq E \) and \( E \subseteq F \) are both alg... | Yes |
Consider the extension \( \mathbb{Q} \subseteq \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \). | -By Proposition 1.15 we know that this is a finite (hence algebraic) extension, of degree at most 4 .\n\n- Thus any five elements in \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \) must be linearly dependent over \( \mathbb{Q} \) . We consider powers of \( \sqrt{2} + \sqrt{3} \) :\n\n\[ 1,\;{\left( \sqrt{2} + \sqrt{3... | Yes |
Lemma 2.1. For a field \( K \), the following are equivalent:\n\n- \( K \) is algebraically closed.\n\n- \( K \) has no nontrivial algebraic extensions.\n\n- If \( K \subseteq L \) is any extension and \( \alpha \in L \) is algebraic over \( K \), then \( \alpha \in K \) . | The proof is a straightforward application of the definitions and a good exercise (Exercise 2.1). | No |
Theorem 2.3. Every field \( k \) admits an algebraic closure \( k \subseteq \bar{k} \) ; this extension is unique up to isomorphism. | Concerning existence, the idea is to construct ’by hand’ a huge extension \( K \) of \( k \) where every polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) factors completely. The elements of \( K \) which are algebraic over \( k \) will form an algebraic closure of \( k \) .\n\nThe construction is done... | No |
Lemma 2.4. Let \( k \) be a field. Then there exists an extension \( k \subseteq K \) such that every nonconstant polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) has at least one root in \( K \) . | Proof. (This construction is apparently due to Emil Artin.) Consider a set \( \mathcal{T} = \) \( \left\{ {t}_{f}\right\} \) in bijection with the set of nonconstant monic polynomials \( f\left( x\right) \in k\left\lbrack x\right\rbrack \), and let \( k\left\lbrack \mathcal{T}\right\rbrack \) be the corresponding polyn... | Yes |
Lemma 2.6. Let \( k \subseteq L \) be a field extension, with \( L \) algebraically closed. Let\n\n\[ \bar{k} \mathrel{\text{:=}} \{ \alpha \in L \mid \alpha \text{ is algebraic over }k\} .\n\]\n\nThen \( \bar{k} \) is an algebraic closure of \( k \) . | The construction reviewed above provides us with an algebraically closed field \( L \) containing any given field \( k \), so the lemma is all we need to prove.\n\nBy Corollary 1.16, \( \bar{k} \) is a field, and the extension \( k \subseteq \bar{k} \) is tautologically algebraic. To verify that \( \bar{k} \) is algebr... | Yes |
Lemma 2.8. Let \( k \subseteq L \) be a field extension, with \( L \) algebraically closed. Let \( k \subseteq F \) be any algebraic extension. Then there exists a morphism of extensions \( i : F \rightarrow L \) . | Proof. This argument also relies on Zorn’s lemma. Consider the set \( Z \) of homomorphisms\n\n\[ \n{i}_{K} : K \rightarrow L \n\]\n\nwhere \( K \) is an intermediate field, \( k \subseteq K \subseteq F \), and \( {i}_{K} \) restricts to the identity on \( k \) ; \( Z \) is nonempty, since the extension \( {i}_{k} : k ... | Yes |
Theorem 2.9 (Nullstellensatz). Let \( k \subseteq F \) be a field extension, and assume that \( F \) is a finite-type \( k \) -algebra. Then \( k \subseteq F \) is a finite (hence algebraic) extension. | ## Proof for uncountable fields. Assume that \( k \) is uncountable.\n\nLet \( k \subseteq F \) be a field extension, and assume that \( F \) is finitely generated as an algebra over \( k \) ; in particular, it is finitely generated as a field extension. We have to prove that \( k \subseteq F \) is a finite extension, ... | No |
Let \( K \) be an algebraically closed field, and let \( I \) be an ideal of \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then \( I \) is maximal if and only if\n\n\[ I = \left( {{x}_{1} - {c}_{1},\ldots ,{x}_{n} - {c}_{n}}\right) \]\n\nfor \( {c}_{1},\ldots ,{c}_{n} \in K \) . | Proof. For \( {c}_{1},\ldots ,{c}_{n} \in K \)\n\n\[ \frac{K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }{\left( {x}_{1} - {c}_{1},\ldots ,{x}_{n} - {c}_{n}\right) } \cong K \]\n\n(Exercise III 4.12) is a field; therefore \( \left( {{x}_{1} - {c}_{1},\ldots ,{x}_{n} - {c}_{n}}\right) \) is maximal. Conversely, ... | No |
Lemma 2.15. Let \( K \) be a field, and let \( S \) be a subset of \( {\mathbb{A}}_{K}^{n} \) . Then the ideal \( \mathcal{I}\left( S\right) \) is a radical ideal of \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . | Proof. The inclusion \( \mathcal{I}\left( S\right) \subseteq \sqrt{\mathcal{I}\left( S\right) } \) holds for every ideal, so it is trivially satisfied. To verify the inclusion \( \sqrt{\mathcal{I}\left( S\right) } \subseteq \mathcal{I}\left( S\right) \), let \( f \in \sqrt{\mathcal{I}\left( S\right) } \) . Then there i... | Yes |
Proposition 2.16 (Weak Nullstellensatz). Let \( K \) be an algebraically closed field, and let \( I \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an ideal. Then \( \mathcal{V}\left( I\right) = \varnothing \) if and only if \( I = \left( 1\right) \) . | Proof. If \( I = \left( 1\right) \), then \( \mathcal{V}\left( I\right) = \varnothing \) by definition.\n\nConversely, assume that \( I \neq \left( 1\right) \) . By Proposition V13.5, \( I \) is then contained in a maximal ideal \( \mathfrak{m} \) . Since \( K \) is algebraically closed, by Corollary 2.10 we have\n\n\[... | Yes |
Corollary 2.18. Let \( K \) be an algebraically closed field. Then for any \( n \geq 0 \) the functions\n\n\[ \left\{ {\text{ algebraic subsets of }{\mathbb{A}}_{K}^{n}}\right\} \overset{\mathcal{I}}{ \leftrightarrow }\left\{ {\text{ radical ideals in }K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }\right\} \]\n... | Proof. Proposition 2.17 shows that \( \mathcal{I} \circ \mathcal{V} \) is the identity on radical ideals, so \( \mathcal{V} \) is injective, and \( \mathcal{V} \) is surjective by definition of affine algebraic set. It follows that \( \mathcal{V} \) is a bijection and \( \mathcal{I} \) is its inverse. | Yes |
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