Q stringlengths 27 2.03k | A stringlengths 4 2.98k | Result stringclasses 2
values |
|---|---|---|
Let \( k \) be a field and let \( k\left( \left( x\right) \right) \) be the set of all formal generalized power series in \( x \) with coefficients in \( k \) ; that is, the elements of \( k\left( \left( x\right) \right) \) are formal infinite sums \( \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \) wit... | A straightforward calculation shows that \( k\left( \left( x\right) \right) \) is a commutative ring with identity. Moreover, we can show that \( k\left( \left( x\right) \right) \) is a field. If \( f = \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \) is a nonzero element of \( k\left( \left( x\right) \... | Yes |
Example 1.2 Let \( k \) be a field and let \( x \) be a variable. The rational function field \( k\\left( x\\right) \) is the quotient field of the polynomial ring \( k\\left\\lbrack x\\right\\rbrack \) ; that is, \( k\\left( x\\right) \) consists of all quotients \( f\\left( x\\right) /g\\left( x\\right) \) of polynom... | Null | No |
Example 1.3 Let \( k \) be a field and let \( k\left( \left( x\right) \right) \) be the set of all formal generalized power series in \( x \) with coefficients in \( k \) ; that is, the elements of \( k\left( \left( x\right) \right) \) are formal infinite sums \( \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x... | A straightforward calculation shows that \( k\left( \left( x\right) \right) \) is a commutative ring with identity. Moreover, we can show that \( k\left( \left( x\right) \right) \) is a field. If \( f = \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \) is a nonzero element of \( k\left( \left( x\right) \... | Yes |
Example 1.4 The extension \( \mathbb{C}/\mathbb{R} \) is a finite extension since \( \left\lbrack {\mathbb{C} : \mathbb{R}}\right\rbrack = 2 \) . A basis for \( \mathbb{C} \) as an \( \mathbb{R} \) -vector space is \( \{ 1, i\} \) . As an extension of \( \mathbb{Q} \), both \( \mathbb{C} \) and \( \mathbb{R} \) are inf... | \[ \mathbb{Q}\left( a\right) = \left\{ {\frac{\mathop{\sum }\limits_{i}{\alpha }_{i}{a}^{i}}{\mathop{\sum }\limits_{i}{\beta }_{i}{a}^{i}} : {\alpha }_{i},{\beta }_{i} \in \mathbb{Q},\mathop{\sum }\limits_{i}{\beta }_{i}{a}^{i} \neq 0}\right\} . \] We shall see in Proposition 1.8 that \( \mathbb{Q}\left( a\right) \) is... | Yes |
Example 1.5 If \( k \) is a field, let \( K = k\\left( t\\right) \) be the field of rational functions in \( t \) over \( k \) . If \( f \) is a nonzero element of \( K \), then we can use the construction of \( \\mathbb{Q}\\left( a\\right) \) in the previous example. Let \( F = k\\left( f\\right) \) be the set of all ... | In Example 1.17, we shall see that \( K/F \) is a finite extension provided that \( f \) is not a constant, and in Chapter V we shall prove L\\\ | No |
Example 1.6 Let \( p\left( t\right) = {t}^{3} - 2 \in \mathbb{Q}\left\lbrack t\right\rbrack \) . Then \( p\left( t\right) \) is irreducible over \( \mathbb{Q} \) by the rational root test. Then the ideal \( \left( {p\left( t\right) }\right) \) generated by \( p\left( t\right) \) in \( \mathbb{Q}\left\lbrack t\right\rbr... | \[ {a}^{3} - 2 = {t}^{3} + \left( {p\left( t\right) }\right) - (2 + \left( {p\left( t\right) }\right) = {t}^{3} - 2 + \left( {p\left( t\right) }\right) = 0. \] The element \( a \) is then a root of \( {x}^{3} - 2 \) in \( K \) . Note that we used the identification of \( \mathbb{Q} \) as a subfield in this calculation. | Yes |
Proposition 1.8 Let \( K \) be a field extension of \( F \) and let \( a \in K \) . Then\n\n\[ F\left\lbrack a\right\rbrack = \{ f\left( a\right) : f\left( x\right) \in F\left\lbrack x\right\rbrack \} \]\n\nand\n\n\[ F\left( a\right) = \{ f\left( a\right) /g\left( a\right) : f, g \in F\left\lbrack x\right\rbrack, g\lef... | Proof. The evaluation map \( {\operatorname{ev}}_{a} : F\left\lbrack x\right\rbrack \rightarrow K \) has image \( \{ f\left( a\right) : f \in F\left\lbrack x\right\rbrack \} \) , so this set is a subring of \( K \) . If \( R \) is a subring of \( K \) that contains \( F \) and \( a \), then \( f\left( a\right) \in R \)... | Yes |
Proposition 1.9 Let \( K \) be a field extension of \( F \) and let \( {a}_{1},\ldots ,{a}_{n} \in K \) . Then\n\n\[ F\left\lbrack {{a}_{1},\ldots ,{a}_{n}}\right\rbrack = \left\{ {f\left( {{a}_{1},\ldots ,{a}_{n}}\right) : f \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }\right\} \]\n\nand\n\n\[ F\left( {{a... | Null | No |
Proposition 1.10 Let \( K \) be a field extension of \( F \) and let \( X \) be a subset of \( K \) . If \( \alpha \in F\left( X\right) \), then \( \alpha \in F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) for some \( {a}_{1},\ldots ,{a}_{n} \in X \) . Therefore,\n\n\[ F\left( X\right) = \bigcup \left\{ {F\left( {{a}_{1},... | Proof. Each field \( F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) with the \( {a}_{i} \in X \) is contained in \( F\left( X\right) \) ; hence, \( \bigcup \left\{ {F\left( {{a}_{1},\ldots ,{a}_{n}}\right) : {a}_{i} \in X}\right\} \subseteq F\left( X\right) \) . This union contains \( F \) and \( X \), so if it is a field... | Yes |
Example 1.13 The complex number \( i = \sqrt{-1} \) is algebraic over \( \mathbb{Q} \), since \( {i}^{2} + 1 = 0 \) . If \( r \in \mathbb{Q} \), then \( a = \sqrt[n]{r} \) is algebraic over \( \mathbb{Q} \), since \( a \) is a root of \( {x}^{n} - r \) . If \( \omega = {e}^{{2\pi i}/n} = \cos \left( {{2\pi }/n}\right) ... | Null | No |
Proposition 1.15 Let \( K \) be a field extension of \( F \) and let \( \alpha \in K \) be algebraic over \( F \) . 1. The polynomial \( \min \left( {F,\alpha }\right) \) is irreducible over \( F \) . 2. If \( g\left( x\right) \in F\left\lbrack x\right\rbrack \), then \( g\left( \alpha \right) = 0 \) if and only if \( ... | Proof. If \( p\left( x\right) = \min \left( {F,\alpha }\right) \), then \( F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \cong F\left\lbrack \alpha \right\rbrack \) is an integral domain. Therefore, \( \left( {p\left( x\right) }\right) \) is a prime ideal, so \( p\left( x\right) \) is irreducible. To ... | Yes |
Proposition 1.15 Let \( K \) be a field extension of \( F \) and let \( \alpha \in K \) be algebraic over \( F \). 1. The polynomial \( \min \left( {F,\alpha }\right) \) is irreducible over \( F \). 2. If \( g\left( x\right) \in F\left\lbrack x\right\rbrack \), then \( g\left( \alpha \right) = 0 \) if and only if \( \m... | Proof. If \( p\left( x\right) = \min \left( {F,\alpha }\right) \), then \( F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \cong F\left\lbrack \alpha \right\rbrack \) is an integral domain. Therefore, \( \left( {p\left( x\right) }\right) \) is a prime ideal, so \( p\left( x\right) \) is irreducible. To ... | Yes |
The element \( \sqrt[3]{2} \) satisfies the polynomial \( {x}^{3} - 2 \) over \( \mathbb{Q} \) , which is irreducible by the Eisenstein criterion, so \( {x}^{3} - 2 \) is the minimal polynomial of \( \sqrt[3]{2} \) over \( \mathbb{Q} \) . Thus, \( \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rb... | If \( p \) is a prime, then \( {x}^{n} - p \) is irreducible over \( \mathbb{Q} \), again by Eisenstein, so \( \left\lbrack {\mathbb{Q}\left( \sqrt[n]{p}\right) : \mathbb{Q}}\right\rbrack = n \) . | Yes |
Example 1.17 Here is a very nice, nontrivial example of a finite field extension. Let \( k \) be a field and let \( K = k\\left( t\\right) \) be the field of rational functions in \( t \) over \( k \) . Let \( u \\in K \) with \( u \\notin k \) . Write \( u = f\\left( t\\right) /g\\left( t\\right) \) with \( f, g \\in ... | To see this, first note that \( K = F\\left( t\\right) \) . By using Proposition 1.15, we need to determine the minimal polynomial of \( t \) over \( F \) to determine \( \\left\\lbrack {K : F}\\right\\rbrack \) . Consider the polynomial \( p\\left( x\\right) = {ug}\\left( x\\right) - f\\left( x\\right) \\in F\\left\\l... | Yes |
Lemma 1.19 If \( K \) is a finite extension of \( F \), then \( K \) is algebraic and finitely generated over \( F \) . | Proof. Suppose that \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) is a basis for \( K \) over \( F \) . Then every element of \( K \) is of the form \( \mathop{\sum }\limits_{i}{a}_{i}{\alpha }_{i} \) with \( {a}_{i} \in F \), so certainly we have \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) ; thus, \( K ... | Yes |
Lemma 1.19 If \( K \) is a finite extension of \( F \), then \( K \) is algebraic and finitely generated over \( F \) . | Proof. Suppose that \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) is a basis for \( K \) over \( F \) . Then every element of \( K \) is of the form \( \mathop{\sum }\limits_{i}{a}_{i}{\alpha }_{i} \) with \( {a}_{i} \in F \), so certainly we have \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) ; thus, \( K ... | Yes |
Proposition 1.20 Let \( F \subseteq L \subseteq K \) be fields. Then \[ \left\lbrack {K : F}\right\rbrack = \left\lbrack {K : L}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack \] | Proof. Let \( \left\{ {{a}_{i} : i \in I}\right\} \) be a basis for \( L/F \), and let \( \left\{ {{b}_{j} : j \in J}\right\} \) be a basis for \( K/L \) . Consider the set \( \left\{ {{a}_{i}{b}_{j} : i \in I, j \in J}\right\} \) . We will show that this set is a basis for \( K/F \) . If \( x \in K \), then \( x = \ma... | Yes |
Proposition 1.21 Let \( K \) be a field extension of \( F \) . If each \( {\alpha }_{i} \in K \) is algebraic over \( F \), then \( F\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\rbrack \) is a finite dimensional field extension of \( F \) with\n\n\[ \left\lbrack {F\left\lbrack {{\alpha }_{1},\ldots ,{\alpha... | Proof. We prove this by induction on \( n \) ; the case \( n = 1 \) follows from Proposition 1.15. If we set \( L = F\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n - 1}}\right\rbrack \), then by induction \( L \) is a field and \( \left\lbrack {L : F}\right\rbrack \leq \mathop{\prod }\limits_{{i = 1}}^{{n - 1}}\left\... | Yes |
Corollary 1.22 If \( K \) is a field extension of \( F \), then \( \alpha \in K \) is algebraic over \( F \) if and only if \( \left\lbrack {F\left( \alpha \right) : F}\right\rbrack < \infty \) . Moreover, \( K \) is algebraic over \( F \) if \( \left\lbrack {K : F}\right\rbrack < \infty \) . | Null | No |
Proposition 1.23 Let \( K \) be a field extension of \( F \), and let \( X \) be a subset of \( K \) such that each element of \( X \) is algebraic over \( F \) . Then \( F\\left( X\\right) \) is algebraic over \( F \) . If \( \\left| X\\right| < \\infty \), then \( \\left\\lbrack {F\\left( X\\right) : F}\\right\\rbrac... | Proof. Let \( a \in F\\left( X\\right) \) . By Proposition 1.10, there are \( {\\alpha }_{1},\\ldots ,{\\alpha }_{n} \in X \) with \( a \in F\\left( {{\\alpha }_{1},\\ldots ,{\\alpha }_{n}}\\right) \) . By Proposition 1.21, \( F\\left( {{\\alpha }_{1},\\ldots ,{\\alpha }_{n}}\\right) \) is algebraic over \( F \) . Thus... | Yes |
Theorem 1.24 Let \( F \subseteq L \subseteq K \) be fields. If \( L/F \) and \( K/L \) are algebraic, then \( K/F \) is algebraic. | Proof. Let \( \alpha \in K \), and let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {x}^{n} \) be the minimal polynomial of \( \alpha \) over \( L \) . Since \( L/F \) is algebraic, the field \( {L}_{0} = F\left( {{a}_{0},\ldots ,{a}_{n - 1}}\right) \) is a finite extension of \( F \) by Corollary 1.22. Now \( f... | Yes |
Corollary 1.26 Let \( K \) be a field extension of \( F \), and let \( L \) be the algebraic closure of \( F \) in \( K \). Then \( L \) is a field, and therefore is the largest algebraic extension of \( F \) contained in \( K \). | Proof. Let \( a, b \in L \). Then \( F\left( {a, b}\right) \) is algebraic over \( F \) by Proposition 1.23, so \( F\left( {a, b}\right) \subseteq L \), and since \( a \pm b,{ab}, a/b \in F\left( {a, b}\right) \subseteq L \), the set \( L \) is closed under the field operations, so it is a subfield of \( K \). Each ele... | Yes |
Example 1.27 Let \( F = \mathbb{Q} \), and view all fields in this example as subfields of \( \mathbb{C} \) . Let \( \omega = {e}^{{2\pi i}/3} \), so \( {\omega }^{3} = 1 \) and \( \omega \neq 1 \) . The composite of \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) and \( \mathbb{Q}\left( {\omega \sqrt[3]{2}}\right) \) is \( ... | To see that this is the composite, note that both \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) and \( \mathbb{Q}\left( {\omega \sqrt[3]{2}}\right) \) are contained in \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \), so their composite is also contained in \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) . However, ... | Yes |
Example 1.28 The composite of \( \mathbb{Q}\left( \sqrt{2}\right) \) and \( \mathbb{Q}\left( \sqrt{3}\right) \) is the field \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \) . This composite can be generated by a single element over \( \mathbb{Q} \) . In fact, \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) = \mathbb{Q... | To see this, the inclusion \( \supseteq \) is clear. For the reverse inclusion, let \( a = \sqrt{2} + \sqrt{3} \) . Then \( {\left( a - \sqrt{2}\right) }^{2} = 3 \) . Multiplying this and rearranging gives \( 2\sqrt{2}a = {a}^{2} - 1 \), so\n\n\[ \sqrt{2} = \frac{{a}^{2} - 1}{2a} \in \mathbb{Q}\left( a\right) \]\n\nSim... | Yes |
Lemma 2.2 Let \( K = F\left( X\right) \) be a field extension of \( F \) that is generated by a subset \( X \) of \( K \) . If \( \sigma ,\tau \in \operatorname{Gal}\left( {K/F}\right) \) with \( {\left. {\left. \sigma \right| }_{X} = \tau \right| }_{X} \), then \( \sigma = \tau \) . Therefore, \( F \) -automorphisms o... | Proof. Let \( a \in K \) . Then there is a finite subset \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \subseteq X \) with \( a \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) . This means there are polynomials \( f, g \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) with \( a = f\left... | Yes |
Lemma 2.3 Let \( \tau : K \rightarrow L \) be an \( F \) -homomorphism and let \( \alpha \in K \) be algebraic over \( F \) . If \( f\left( x\right) \) is a polynomial over \( F \) with \( f\left( \alpha \right) = 0 \) , then \( f\left( {\tau \left( \alpha \right) }\right) = 0 \) . Therefore, \( \tau \) permutes the ro... | Proof. Let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \) . Then\n\n\[ 0 = \tau \left( 0\right) = \tau \left( {f\left( \alpha \right) }\right) = \mathop{\sum }\limits_{i}\tau \left( {a}_{i}\right) \tau {\left( \alpha \right) }^{i}. \]\n\nBut, since each \( {a}_{i} \in F \), we have \( \tau \left(... | Yes |
Corollary 2.4 If \( \left\lbrack {K : F}\right\rbrack < \infty \), then \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| < \infty \) . | Proof. We can write \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) for some \( {\alpha }_{i} \in K \) . Any \( F \) - automorphism of \( K \) is determined by what it does to the \( {\alpha }_{i} \) . By Lemma 2.3, there are only finitely many possibilities for the image of any \( {\alpha }_{i} \) ; hen... | Yes |
Consider the extension \( \mathbb{C}/\mathbb{R} \). We claim that \( \operatorname{Gal}\left( {\mathbb{C}/\mathbb{R}}\right) = \) \( \{ \mathrm{{id}},\sigma \} \), where \( \sigma \) is complex conjugation. | Both of these functions are \( \mathbb{R} \) - automorphisms of \( \mathbb{C} \), so they are contained in \( \operatorname{Gal}\left( {\mathbb{C}/\mathbb{R}}\right) \). To see that there is no other automorphism of \( \mathbb{C}/\mathbb{R} \), note that an element of \( \operatorname{Gal}\left( {\mathbb{C}/\mathbb{R}}... | Yes |
The Galois group of \( \mathbb{Q}\left( \sqrt[3]{2}\right) /\mathbb{Q} \) is \( \langle \mathrm{{id}}\rangle \). | To see this, if \( \sigma \) is a \( \mathbb{Q} \) -automorphism of \( \mathbb{Q}\left( \sqrt[3]{2}\right) \), then \( \sigma \left( \sqrt[3]{2}\right) \) is a root of \( \min \left( {\mathbb{Q},\sqrt[3]{2}}\right) = {x}^{3} - 2 \). If \( \omega = {e}^{{2\pi i}/3} \), then the roots of this polynomial are \( \sqrt[3]{2... | Yes |
Example 2.7 Let \( K = {\mathbb{F}}_{2}\left( t\right) \) be the rational function field in one variable over \( {\mathbb{F}}_{2} \), and let \( F = {\mathbb{F}}_{2}\left( {t}^{2}\right) \) . Then \( \left\lbrack {K : F}\right\rbrack = 2 \) . | The element \( t \) satisfies the polynomial \( {x}^{2} - {t}^{2} \in F\left\lbrack x\right\rbrack \), which has only \( t \) as a root, since \( {x}^{2} - {t}^{2} = {\left( x - t\right) }^{2} \) in \( K\left\lbrack x\right\rbrack \) . Consequently, if \( \sigma \) is an \( F \) -automorphism of \( K \), then \( \sigma... | Yes |
Example 2.8 Let \( F = {\mathbb{F}}_{2} \) . The polynomial \( 1 + x + {x}^{2} \) is irreducible over \( F \) , since it has no roots in \( F \) . In fact, this is the only irreducible quadratic over \( F \) ; the three other quadratics factor over \( F \) . Let \( K = F\left\lbrack x\right\rbrack /\left( {1 + x + {x}^... | To see that \( \operatorname{Gal}\left( {K/F}\right) \) has exactly two elements, we need to check that there is indeed an automorphism \( \sigma \) with \( \sigma \left( \alpha \right) = \alpha + 1 \) . If \( \sigma \) does exist, then \( \sigma \left( {a + {b\alpha }}\right) = a + b\left( {\alpha + 1}\right) = \left(... | No |
Lemma 2.9 Let \( K \) be a field.\n\n1. If \( {L}_{1} \subseteq {L}_{2} \) are subfields of \( K \), then \( \operatorname{Gal}\left( {K/{L}_{2}}\right) \subseteq \operatorname{Gal}\left( {K/{L}_{1}}\right) \) . | Proof. The first four parts are simple consequences of the definitions. We leave the proofs of parts \( 2,3 \), and 4 to the reader and prove part 1 for the sake of illustration. If \( \sigma \in \operatorname{Gal}\left( {K/{L}_{2}}\right) \), then \( \sigma \left( a\right) = a \) for all \( a \in {L}_{2} \) . Thus, \(... | No |
Corollary 2.10 If \( K \) is a field extension of \( F \), then there is \( 1 - 1 \) inclusion reversing correspondence between the set of subgroups of \( \operatorname{Gal}\left( {K/F}\right) \) of the form \( \operatorname{Gal}\left( {K/L}\right) \) for some subfield \( L \) of \( K \) containing \( F \) and the set ... | Proof. This follows immediately from the lemma. If \( \mathcal{G} \) and \( \mathcal{F} \) are respectively the set of groups and fields in question, then the map that sends a subfield \( L \) of \( K \) to the subgroup \( \operatorname{Gal}\left( {K/L}\right) \) of \( \operatorname{Aut}\left( K\right) \) sends \( \mat... | Yes |
Lemma 2.12 (Dedekind’s Lemma) Let \( {\tau }_{1},\ldots ,{\tau }_{n} \) be distinct characters from \( G \) to \( {K}^{ * } \) . Then the \( {\tau }_{i} \) are linearly independent over \( K \) ; that is, if \( \mathop{\sum }\limits_{i}{c}_{i}{\tau }_{i}\left( g\right) = 0 \) for all \( g \in G \), where the \( {c}_{i}... | Proof. Suppose that the lemma is false. Choose \( k \) minimal (relabeling the \( {\tau }_{i} \) if necessary) so that there are \( {c}_{i} \in K \) with \( \mathop{\sum }\limits_{i}{c}_{i}{\tau }_{i}\left( g\right) = 0 \) for all \( g \in G \) . Then all \( {c}_{i} \neq 0 \) . Since \( {\tau }_{1} \neq {\tau }_{2} \),... | Yes |
Proposition 2.13 If \( K \) is a finite field extension of \( F \), then \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| \leq \) \( \left\lbrack {K : F}\right\rbrack \) . | Proof. The group \( \operatorname{Gal}\left( {K/F}\right) \) is finite by Corollary 2.4. Let \( \operatorname{Gal}\left( {K/F}\right) = \) \( \left\{ {{\tau }_{1},\ldots ,{\tau }_{n}}\right\} \), and suppose that \( \left\lbrack {K : F}\right\rbrack < n \) . Let \( {\alpha }_{1},\ldots ,{\alpha }_{m} \) be a basis for ... | Yes |
Proposition 2.14 Let \( G \) be a finite group of automorphisms of \( K \) with \( F = \mathcal{F}\left( G\right) \) . Then \( \left| G\right| = \left\lbrack {K : F}\right\rbrack \), and so \( G = \operatorname{Gal}\left( {K/F}\right) \) . | Proof. By the previous proposition, \( \left| G\right| \leq \left\lbrack {K : F}\right\rbrack \) since \( G \subseteq \operatorname{Gal}\left( {K/F}\right) \) . Suppose that \( \left| G\right| < \left\lbrack {K : F}\right\rbrack \) . Let \( n = \left| G\right| \), and take \( {\alpha }_{1},\ldots ,{\alpha }_{n + 1} \in... | Yes |
Corollary 2.16 Let \( K \) be a finite extension of \( F \) . Then \( K/F \) is Galois if and only if \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| = \left\lbrack {K : F}\right\rbrack \) . | Proof. If \( K/F \) is a Galois extension, then \( F = \mathcal{F}\left( {\operatorname{Gal}\left( {K/F}\right) }\right) \), so by Proposition 2.14, \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| = \left\lbrack {K : F}\right\rbrack \) . Conversely, if \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right|... | Yes |
Corollary 2.17 Let \( K \) be a field extension of \( F \), and let \( a \in K \) be algebraic over \( F \) . Then \( \left| {\operatorname{Gal}\left( {F\left( a\right) /F}\right) }\right| \) is equal to the number of distinct roots of \( \min \left( {F, a}\right) \) in \( F\left( a\right) \) . Therefore, \( F\left( a\... | Proof. If \( \tau \in \operatorname{Gal}\left( {F\left( a\right) /F}\right) \), we have seen that \( \tau \left( a\right) \) is a root of \( \min \left( {F, a}\right) \) . Moreover, if \( \sigma ,\tau \in \operatorname{Gal}\left( {F\left( a\right) /F}\right) \) with \( \sigma \neq \tau \), then \( \sigma \left( a\right... | Yes |
The extension \( \mathbb{Q}\left( \sqrt[3]{2}\right) /\mathbb{Q} \) is not Galois, for we have seen that \( \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rbrack = 3 \) but \( \left| {\operatorname{Gal}\left( {\mathbb{Q}\left( \sqrt[3]{2}\right) /\mathbb{Q}}\right) }\right| = 1 \) . The polynomia... | Null | No |
Example 2.19 Let \( k \) be a field of characteristic \( p > 0 \), and let \( k\left( t\right) \) be the rational function field in one variable over \( k \) . Consider the field extension \( k\left( t\right) /k\left( {t}^{p}\right) \) . Then \( t \) satisfies the polynomial \( {x}^{p} - {t}^{p} \in k\left( {t}^{p}\rig... | Null | No |
Example 2.20 Let \( F \) be a field of characteristic not 2, and let \( a \in F \) be an element that is not the square of any element in \( F \) . Let \( K = F\left\lbrack x\right\rbrack /\left( {{x}^{2} - a}\right) \) , a field since \( {x}^{2} - a \) is irreducible over \( F \) . We view \( F \) as a subfield of \( ... | \[ \sigma \left( {\alpha + {\beta u}}\right) = \alpha - {\beta u} \] then \( \sigma \) is an automorphism of \( K \) since \( u \) and \( - u \) are roots in \( K \) of \( {x}^{2} - a \) . Thus, \( \mathrm{{id}},\sigma \in \mathrm{{Gal}}\left( {K/F}\right) \), so \( \left| {\mathrm{{Gal}}\left( {K/F}\right) }\right| = ... | Yes |
Example 2.21 The extension \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) /\mathbb{Q} \) is Galois, where \( \omega = {e}^{{2\pi i}/3} \). | In fact, the field \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) is the field generated over \( \mathbb{Q} \) by the three roots \( \sqrt[3]{2} \) , \( \omega \sqrt[3]{2} \), and \( {\omega }^{2}\sqrt[3]{2} \), of \( {x}^{3} - 2 \), and since \( \omega \) satisfies \( {x}^{2} + x + 1 \) over \( \mathbb{Q} \) and ... | Yes |
This example shows us that any finite group can occur as the Galois group of a Galois extension. We will use this example a number of times in later sections. Let \( k \) be a field and let \( K = k\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) be the field of rational functions in \( n \) variables over \( k \) . ... | The straightforward but somewhat messy calculation that this does define a field automorphism on \( K \) is left to Problem 5. We can then view \( {S}_{n} \subseteq \) \( \operatorname{Aut}\left( K\right) \) . Let \( F = \mathcal{F}\left( {S}_{n}\right) \) . By Proposition 2.14, \( K/F \) is a Galois extension with \( ... | No |
Lemma 3.1 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) and \( \alpha \in F \) . Then \( \alpha \) is a root of \( f \) if and only if \( x - \alpha \) divides \( f \) . Furthermore, \( f \) has at most \( \deg \left( f\right) \) roots in any extension field of \( F \) . | Proof. By the division algorithm, \( f\left( x\right) = q\left( x\right) \cdot \left( {x - \alpha }\right) + r\left( x\right) \) for some \( q\left( x\right) \) and \( r\left( x\right) \) with \( r\left( x\right) = 0 \) or \( \deg \left( r\right) < \deg \left( {x - \alpha }\right) \) . In either case, we see that \( r\... | Yes |
Theorem 3.3 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) have degree \( n \) . There is an extension field \( K \) of \( F \) with \( \left\lbrack {K : F}\right\rbrack \leq n \) such that \( K \) contains a root of \( f \) . In addition, there is a field \( L \) containing \( F \) with \( \left\lbrack {L... | Proof. Let \( p\left( x\right) \) be an irreducible factor of \( f\left( x\right) \) in \( F\left\lbrack x\right\rbrack \), and let \( K \) be the field \( F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \) . Then \( F \) is isomorphic to a subfield of \( K \) ; namely the map \( \varphi : F \rightarrow... | Yes |
Corollary 3.5 If \( {f}_{1}\left( x\right) ,\ldots ,{f}_{n}\left( x\right) \in F\left\lbrack x\right\rbrack \), then there is a splitting field for \( \left\{ {{f}_{1},\ldots ,{f}_{n}}\right\} \) over \( F \) . | Proof. Suppose that \( {f}_{1},\ldots ,{f}_{n} \in F\left\lbrack x\right\rbrack \) . Note that a splitting field of \( \left\{ {{f}_{1},\ldots ,{f}_{n}}\right\} \) is the same as a splitting field of the product \( {f}_{1}\cdots {f}_{n} \) . If \( f = {f}_{1}\cdots {f}_{n} \), then by Theorem 3.3, there is a field \( L... | Yes |
Corollary 3.8 Let \( F \) be a field and let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be a polynomial of degree \( n \) . If \( K \) is a splitting field of \( f \) over \( F \), then \( \left\lbrack {K : F}\right\rbrack \leq n! \) . | Proof. We prove this by induction on \( n = \deg \left( f\right) \) . If \( n = 1 \), then the result is clear. Suppose that \( n > 1 \) and that the result is true for polynomials of degree \( n - 1 \) . Let \( K \) be a splitting field of \( f \) over \( F \), and let \( a \) be a root of \( f \) in \( K \) . Then \(... | Yes |
Example 3.7 Let \( F = {\mathbb{F}}_{2} \) and \( K = F\left\lbrack x\right\rbrack /\left( {1 + x + {x}^{2}}\right) \cong F\left( \alpha \right) \), where \( \alpha \) is a root of \( 1 + x + {x}^{2} \) . Then \( 1 + x + {x}^{2} \) factors as \( \left( {x - \alpha }\right) \left( {x - \left( {\alpha + 1}\right) }\right... | Null | No |
Corollary 3.8 Let \( F \) be a field and let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be a polynomial of degree \( n \) . If \( K \) is a splitting field of \( f \) over \( F \), then \( \left\lbrack {K : F}\right\rbrack \leq n! \) . | Proof. We prove this by induction on \( n = \deg \left( f\right) \) . If \( n = 1 \), then the result is clear. Suppose that \( n > 1 \) and that the result is true for polynomials of degree \( n - 1 \) . Let \( K \) be a splitting field of \( f \) over \( F \), and let \( a \) be a root of \( f \) in \( K \) . Then \(... | Yes |
Example 3.9 Let \( k \) be a field, and let \( K = k\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) be the rational function field in \( n \) variables over \( k \) . We view the symmetric group \( {S}_{n} \) as a subgroup of \( \operatorname{Aut}\left( K\right) \) by defining\n\n\[ \sigma \left( \frac{f\left( {{x}_... | To show this, we use the concept of splitting fields. Let\n\n\[ f\left( t\right) = {t}^{n} - {s}_{1}{t}^{n - 1} + \cdots + {\left( -1\right) }^{n}{s}_{n} \in k\left( {{s}_{1},{s}_{2},\ldots ,{s}_{n}}\right) \left\lbrack t\right\rbrack . \]\n\nThen \( f\left( t\right) = \left( {t - {x}_{1}}\right) \cdots \left( {t - {x}... | Yes |
Lemma 3.10 If \( K \) is a field, then the following statements are equivalent:\n\n1. There are no algebraic extensions of \( K \) other than \( K \) itself.\n\n2. There are no finite extensions of \( K \) other than \( K \) itself.\n\n3. If \( L \) is a field extension of \( K \), then \( K = \{ a \in L : a \) is alge... | Proof. (1) \( \Rightarrow \) (2): This is clear, since any finite extension of \( F \) is an algebraic extension of \( F \) .\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) : Let \( a \in L \) be algebraic over \( K \) . Then \( K\left( a\right) \) is a finite extension of \( K \), so \( K\left( a\right) = K \) .... | Yes |
Example 3.12 The complex field \( \mathbb{C} \) is algebraically closed. | Null | No |
Lemma 3.13 If \( K/F \) is algebraic, then \( \left| K\right| \leq \max \{ \left| F\right| ,\left| \mathbb{N}\right| \} \) . | Proof. In this proof, we require some facts of cardinal arithmetic, facts that can be found in Proposition 2.1 in Appendix B. If \( a \in K \), pick a labeling \( {a}_{1},\ldots ,{a}_{n} \) of the roots of \( \min \left( {F, a}\right) \) in \( K \) . If \( \mathcal{M} \) is the set of all monic polynomials over \( F \)... | Yes |
Theorem 3.14 Let \( F \) be a field. Then \( F \) has an algebraic closure. | Proof. Let \( S \) be a set containing \( F \) with \( \left| S\right| > \max \{ \left| F\right| ,\left| \mathbb{N}\right| \} \) . Let \( \mathcal{A} \) be the set of all algebraic extension fields of \( F \) inside \( S \) . Then \( \mathcal{A} \) is ordered by defining \( K \leq L \) if \( L \) is an extension field ... | Yes |
Corollary 3.15 Let \( S \) be a set of nonconstant polynomials over \( F \) . Then \( S \) has a splitting field over \( F \) . | Proof. Let \( K \) be an algebraic closure of \( F \) . Then each \( f\left( x\right) \in S \) splits over \( K \) . Let \( X \) be the set of roots of all \( f \in S \) . Then \( F\left( X\right) \subseteq K \) is a splitting field for \( S \) over \( F \), since each \( f \) splits over \( F\left( X\right) \) and thi... | Yes |
Corollary 3.16 If \( F \) is a field, then the splitting field of the set of all nonconstant polynomials over \( F \) is an algebraic closure of \( F \) . | Null | No |
Lemma 3.17 Let \( \sigma : F \rightarrow {F}^{\prime } \) be a field isomorphism. Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be irreducible, let \( \alpha \) be a root of \( f \) in some extension field \( K \) of \( F \), and let \( {\alpha }^{\prime } \) be a root of \( \sigma \left( f\right) \) in s... | Proof. Since \( f \) is irreducible and \( f\left( \alpha \right) = 0 \), the minimal polynomial of \( \alpha \) over \( F \) is a constant multiple of \( f \) . Thus, \( f \) and \( \min \left( {F,\alpha }\right) \) generate the same principal ideal in \( F\left\lbrack x\right\rbrack \) . We then have an \( F \) -isom... | Yes |
Lemma 3.18 Let \( \sigma : F \rightarrow {F}^{\prime } \) be a field isomorphism, let \( K \) be a field extension of \( F \), and let \( {K}^{\prime } \) be a field extension of \( {F}^{\prime } \) . Suppose that \( K \) is a splitting field of \( \left\{ {f}_{i}\right\} \) over \( F \) and that \( \tau : K \rightarro... | Proof. Because \( K \) is a splitting field of a set \( \left\{ {f}_{i}\right\} \) of polynomials over \( F \) , given \( {f}_{i} \) there are \( a,{\alpha }_{1},\ldots ,{\alpha }_{n} \in K \) with \( {f}_{i}\left( x\right) = a\mathop{\prod }\limits_{j}\left( {x - {\alpha }_{j}}\right) \) . Therefore, \( \tau \left( {{... | Yes |
Theorem 3.19 Let \( \sigma : F \rightarrow {F}^{\prime } \) be a field isomorphism, let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) , and let \( \sigma \left( f\right) \in {F}^{\prime }\left\lbrack x\right\rbrack \) be the corresponding polynomial over \( {F}^{\prime } \) . Let \( K \) be the splitting fiel... | Proof. We prove this by induction on \( n = \left\lbrack {K : F}\right\rbrack \) . If \( n = 1 \), then \( f \) splits over \( F \), and the result is trivial in this case. So, suppose that \( n > 1 \) and that the result is true for splitting fields of degree less than \( n \) . If \( f \) splits over \( F \), then th... | Yes |
Theorem 3.20 (Isomorphism Extension Theorem) Let \( \sigma : F \rightarrow {F}^{\prime } \) be a field isomorphism. Let \( S = \left\{ {{f}_{i}\left( x\right) }\right\} \) be a set of polynomials over \( F \), and let \( {S}^{\prime } = \left\{ {\sigma \left( {f}_{i}\right) }\right\} \) be the corresponding set over \(... | Proof. We prove this with a Zorn’s lemma argument. Let \( \mathcal{S} \) be the set of all pairs \( \left( {L,\varphi }\right) \) such that \( L \) is a subfield of \( K \) and \( \varphi : L \rightarrow {K}^{\prime } \) is a homomorphism extending \( \sigma \) . This set is nonempty since \( \left( {F,\sigma }\right) ... | Yes |
Corollary 3.21 Let \( F \) be a field, and let \( S \) be a subset of \( F\left\lbrack x\right\rbrack \) . Any two splitting fields of \( S \) over \( F \) are \( F \) -isomorphic. In particular, any two algebraic closures of \( F \) are \( F \) -isomorphic. | Proof. For the proof of the first statement, the isomorphism extension theorem gives an isomorphism extending id on \( F \) between any two splitting fields of \( S \) . The second statement follows from the first, since any algebraic closure of \( F \) is a splitting field of the set of all nonconstant polynomials in ... | Yes |
Corollary 3.22 Let \( F \) be a field, and let \( N \) be an algebraic closure of \( F \). If \( K \) is an algebraic extension of \( F \), then \( K \) is isomorphic to a subfield of \( N \). | Proof. Let \( M \) be an algebraic closure of \( K \). By Theorem 1.24, \( M \) is algebraic over \( F \); hence, \( M \) is also an algebraic closure of \( F \). Therefore, by the previous corollary, \( M \cong N \). If \( f : M \rightarrow N \) is an \( F \)-isomorphism, then \( f\left( K\right) \) is a subfield of \... | Yes |
Example 3.24 If \( \left\lbrack {K : F}\right\rbrack = 2 \), then \( K \) is normal over \( F \) . | For, if \( a \in K - F \) , then \( K = F\left( a\right) \), since \( \left\lbrack {K : F}\right\rbrack = 2 \) . If \( p\left( x\right) = \min \left( {F, a}\right) \), then \( p \) has one root in \( K \) ; hence, since \( \deg \left( p\right) = 2 \), this polynomial factors over \( K \) . Because \( K \) is generated ... | Yes |
Example 3.25 If \( F \subseteq L \subseteq K \) are fields such that \( K/F \) is normal, then \( K/L \) is normal. | This is true because if \( K \) is the splitting field over \( F \) of a set of polynomials \( S \subseteq F\left\lbrack x\right\rbrack \), then \( K \) is generated over \( F \) by the roots of the polynomials in \( S \) . Consequently, \( K \) is generated by the roots as an extension of \( L \), so \( K \) is a spli... | Yes |
Example 3.26 The field \( \mathbb{Q}\left( {\omega ,\sqrt[3]{2}}\right) \) is normal over \( \mathbb{Q} \), since it is the splitting field of \( {x}^{3} - 2 \) over \( \mathbb{Q} \) . Similarly, if \( i = \sqrt{-1} \), then \( \mathbb{Q}\left( {\sqrt[4]{2}, i}\right) \) is normal over \( \mathbb{Q} \), since it is the... | Null | No |
Example 3.27 Let \( F \) be a field of characteristic \( p > 0 \), and suppose that \( K = F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) with \( {a}_{i}^{p} \in F \) for each \( i \) . Then we show that \( K \) is normal over \( F \) . | The minimal polynomial of \( {a}_{i} \) divides \( {x}^{p} - {a}_{i}^{p} \), which factors completely over \( K \) as \( {x}^{p} - {a}_{i}^{p} = {\left( x - {a}_{i}\right) }^{p} \) ; hence, \( \min \left( {F,{a}_{i}}\right) \) splits over \( K \) . Thus, \( K \) is the splitting field of \( \left\{ {\min \left( {F,{a}_... | Yes |
Proposition 3.28 If \( K \) is algebraic over \( F \), then the following statements are equivalent:\n\n1. The field \( K \) is normal over \( F \) .\n\n2. If \( M \) is an algebraic closure of \( K \) and if \( \tau : K \rightarrow M \) is an \( F \) - homomorphism, then \( \tau \left( K\right) = K \) .\n\n3. If \( F ... | Proof. (1) \( \Rightarrow \) (2): Let \( M \) be an algebraic closure of \( K \), and let \( \tau : K \rightarrow M \) be an \( F \) -homomorphism. If \( K \) is the splitting field for \( S \subseteq F\left\lbrack x\right\rbrack \) over \( F \) , then so is \( \tau \left( K\right) \subseteq M \) by Lemma 3.17. Since \... | Yes |
Lemma 4.3 Let \( f\left( x\right) \) and \( g\left( x\right) \) be polynomials over a field \( F \) . 1. If \( f \) has no repeated roots in any splitting field, then \( f \) is separable over \( F \) . 2. If \( g \) divides \( f \) and if \( f \) is separable over \( F \), then \( g \) is separable over \( F \) . 3. I... | Proof. For property 1, if \( f \) has no repeated roots in any splitting field, then neither does any irreducible factor of \( f \) . Thus, \( f \) is separable over \( F \) . To show property 2, if \( g \) divides \( f \) with \( f \) separable over \( F \), then no irreducible factor of \( f \) has a repeated root. H... | Yes |
If \( f \) has no repeated roots in any splitting field, then \( f \) is separable over \( F \) . | For property 1, if \( f \) has no repeated roots in any splitting field, then neither does any irreducible factor of \( f \) . Thus, \( f \) is separable over \( F \) . | Yes |
Proposition 4.5 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be a nonconstant polynomial. Then \( f \) has no repeated roots in a splitting field if and only if \( \gcd \left( {f,{f}^{\prime }}\right) = 1 \) in \( F\left\lbrack x\right\rbrack \) . | Proof. We first point out that \( f \) and \( {f}^{\prime } \) are relatively prime in \( F\left\lbrack x\right\rbrack \) if and only if they are relatively prime in \( K\left\lbrack x\right\rbrack \) . To prove this, suppose that \( \gcd \left( {f,{f}^{\prime }}\right) = 1 \) in \( F\left\lbrack x\right\rbrack \) . Th... | Yes |
Proposition 4.6 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial.\n\n1. If \( \operatorname{char}\left( F\right) = 0 \), then \( f \) is separable over \( F \) . If \( \operatorname{char}\left( F\right) = p > 0 \), then \( f \) is separable over \( F \) if and only if \( {f}^{\pr... | Proof. If \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) is irreducible over \( F \), then the only possibility for \( \gcd \left( {f,{f}^{\prime }}\right) \) is 1 or \( f \) . If \( \operatorname{char}\left( F\right) = 0 \), then \( \deg \left( {f}^{\prime }\right) = \deg \left( f\right) - 1 \) ; thus, \( f \... | Yes |
Example 4.8 If \( F \) is a field of characteristic 0, then any algebraic extension of \( F \) is separable over \( F \), since every polynomial in \( F\left\lbrack x\right\rbrack \) is separable over \( F \) . If \( k \) is a field of characteristic \( p > 0 \) and if \( k\left( x\right) \) is the rational function fi... | Null | No |
Corollary 4.10 Let \( L \) be a finite extension of \( F \). 1. \( L \) is separable over \( F \) if and only if \( L \) is contained in a Galois extension of \( F \). | Proof. If \( L \subseteq K \) with \( K/F \) Galois, then \( K/F \) is separable by Theorem 4.9. Hence, \( L/F \) is separable. Conversely, suppose that \( L/F \) is separable. Since \( \left\lbrack {L : F}\right\rbrack < \infty \), we may write \( L = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \), and each \... | Yes |
Theorem 4.13 Let \( F \) be a field of characteristic \( p \) . Then \( F \) is perfect if and only if \( {F}^{p} = F \) . | Proof. Suppose that \( F \) is perfect. Let \( a \in F \), and consider the field \( K = \) \( F\left( \alpha \right) \), where \( \alpha \) is a root of \( {x}^{p} - a \) . The minimal polynomial of \( \alpha \) divides \( {x}^{p} - a = {\left( x - \alpha \right) }^{p} \) . However, \( K \) is separable over \( F \) s... | Yes |
Theorem 4.13 Let \( F \) be a field of characteristic \( p \) . Then \( F \) is perfect if and only if \( {F}^{p} = F \) . | Proof. Suppose that \( F \) is perfect. Let \( a \in F \), and consider the field \( K = \) \( F\left( \alpha \right) \), where \( \alpha \) is a root of \( {x}^{p} - a \) . The minimal polynomial of \( \alpha \) divides \( {x}^{p} - a = {\left( x - \alpha \right) }^{p} \) . However, \( K \) is separable over \( F \) s... | Yes |
Any finite field is perfect | The map \( \varphi : F \rightarrow F \) given by \( \varphi \left( a\right) = {a}^{p} \) is a nonzero field homomorphism, so \( \varphi \) is injective. Since \( F \) is finite, \( \varphi \) is also surjective. Thus, \( {F}^{p} = \operatorname{im}\left( \varphi \right) = F \), so \( F \) is perfect by Theorem 4.13. | Yes |
Lemma 4.16 Let \( F \) be a field of characteristic \( p > 0 \) . If \( \alpha \) is algebraic over \( F \), then \( \alpha \) is purely inseparable over \( F \) if and only if \( {\alpha }^{{p}^{n}} \in F \) for some \( n \) . When this happens, \( \min \left( {F,\alpha }\right) = {\left( x - \alpha \right) }^{{p}^{n}... | Proof. If \( {\alpha }^{{p}^{n}} = a \in F \), then \( \alpha \) is a root of the polynomial \( {x}^{{p}^{n}} - a \) . This polynomial factors over \( F\left( \alpha \right) \) as \( {\left( x - \alpha \right) }^{{p}^{n}} \), and \( \min \left( {F,\alpha }\right) \) divides this polynomial, so \( \min \left( {F,\alpha ... | Yes |
1. If \( \alpha \in K \) is separable and purely inseparable over \( F \), then \( \alpha \in F \) . | Proof. Suppose that \( \alpha \in K \) is both separable and purely inseparable over \( F \) . Then \( \min \left( {F,\alpha }\right) \) has only one distinct root, and it also has no repeated roots. Therefore, \( p\left( x\right) = x - \alpha \), so \( \alpha \in F \) . | Yes |
Proposition 4.20 Let \( K \) be a field extension of \( F \). If \( S \) and \( I \) are the separable and purely inseparable closures of \( F \) in \( K \), respectively, then \( S \) and \( I \) are field extensions of \( F \) with \( S/F \) separable, \( I/F \) purely inseparable, and \( S \cap I = F \). If \( K/F \... | Proof. Let \( a, b \in S \) . Then \( F\left( {a, b}\right) \) is a separable extension of \( F \) by Lemma 4.10. Hence, \( a \pm b,{ab} \), and \( a/b \) are separable over \( F \), so they all lie in \( S \) . Thus, \( S \) is a field. For \( I \), if \( c, d \in I \), then there are \( n, m \) with \( {c}^{{p}^{n}} ... | Yes |
Proposition 4.20 Let \( K \) be a field extension of \( F \) . If \( S \) and \( I \) are the separable and purely inseparable closures of \( F \) in \( K \), respectively, then \( S \) and \( I \) are field extensions of \( F \) with \( S/F \) separable, \( I/F \) purely inseparable, and \( S \cap I = F \) . If \( K/F... | Proof. Let \( a, b \in S \) . Then \( F\left( {a, b}\right) \) is a separable extension of \( F \) by Lemma 4.10. Hence, \( a \pm b,{ab} \), and \( a/b \) are separable over \( F \), so they all lie in \( S \) . Thus, \( S \) is a field. For \( I \), if \( c, d \in I \), then there are \( n, m \) with \( {c}^{{p}^{n}} ... | Yes |
Proposition 4.21 If \( F \subseteq L \subseteq K \) are fields such that \( L/F \) and \( K/L \) are separable, then \( K/F \) is separable. | Proof. Let \( S \) be the separable closure of \( F \) in \( K \) . Then \( L \subseteq S \), as \( L/F \) is separable. Also, since \( K/L \) is separable, \( K/S \) is separable. But \( K/S \) is purely inseparable, so \( K = S \) . Thus, \( K \) is separable over \( F \) . | Yes |
Example 4.22 Let \( K \) be a finite extension of \( F \), and suppose that \( \operatorname{char}\left( F\right) \) does not divide \( \left\lbrack {K : F}\right\rbrack \) . We show that \( K/F \) is separable. | If \( \operatorname{char}\left( F\right) = 0 \) , then this is clear, so suppose that \( \operatorname{char}\left( F\right) = p > 0 \) . Let \( S \) be the separable closure of \( F \) in \( K \) . Then \( K/S \) is purely inseparable, so \( \left\lbrack {K : S}\right\rbrack = {p}^{n} \) for some \( n \) by Lemma 4.17.... | Yes |
Theorem 4.23 Let \( K \) be a normal extension of \( F \), and let \( S \) and \( I \) be the separable and purely inseparable closures of \( F \) in \( K \) , respectively. Then \( S/F \) is Galois, \( I = \mathcal{F}\left( {\operatorname{Gal}\left( {K/F}\right) }\right) \), and \( \operatorname{Gal}\left( {S/F}\right... | Proof. Let \( a \in S \), and set \( f\left( x\right) = \min \left( {F, a}\right) \) . Since \( K \) is normal over \( F \) , the polynomial \( f \) splits over \( K \) . Since \( a \) is separable over \( F \), the polynomial \( f \) has no repeated roots, so all its roots are separable over \( S \) . Thus, \( f \) sp... | Yes |
We give an example of a field extension \( K/F \) in which \( K \) is not separable over the purely inseparable closure \( I \) of \( F \) in \( K \) . This is also an example of a nonseparable field extension \( K/F \) in which the purely inseparable closure is \( F \) . Let \( k \) be a field of characteristic 2, let... | To do this, we show that if \( a \in K \) with \( {a}^{2} \in F \), then \( a \in F \) . A basis for \( K/F \) is \( 1, u,\sqrt{uy} \), and \( u\sqrt{uy} \) . Say \( {a}^{2} \in F \) and write \( a = \alpha + {\beta u} + \gamma \sqrt{uy} + {\delta u}\sqrt{uy} \) with \( \alpha ,\beta ,\gamma ,\delta \in F \) . Then\n\n... | Yes |
Theorem 5.1 (Fundamental Theorem of Galois Theory) Let \( K \) be a finite Galois extension of \( F \), and let \( G = \operatorname{Gal}\left( {K/F}\right) \). Then there is a \( 1 - 1 \) inclusion reversing correspondence between intermediate fields of \( K/F \) and subgroups of \( G \), given by \( L \mapsto \operat... | Proof. We have seen in Lemma 2.9 that the maps \( L \mapsto \operatorname{Gal}\left( {K/L}\right) \) and \( H \mapsto \mathcal{F}\left( H\right) \) give injective inclusion reversing correspondences between the set of fixed fields \( L \) with \( F \subseteq L \subseteq K \) and the set of subgroups of \( G \) of the f... | Yes |
Which is the Galois group of the field \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) over \( \mathbb{Q} \)? | The subfield \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) is not Galois over \( \mathbb{Q} \), since the minimal polynomial of \( \sqrt[3]{2} \) does not split over \( \mathbb{Q}\left( \sqrt[3]{2}\right) \). Therefore, the corresponding subgroup is not normal in \( G \). However, every subgroup of an Abelian group is norm... | Yes |
Example 5.3 Let \( K = \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \). Then \( K \) is the splitting field of \( \left\{ {{x}^{2} - 2,{x}^{2} - 3}\right\} \) over \( \mathbb{Q} \) or, alternatively, the splitting field of \( \left( {{x}^{2} - 2}\right) \left( {{x}^{2} - 3}\right) \) over \( \mathbb{Q} \). The dimension... | \[ \text{id} : \sqrt{2} \rightarrow \sqrt{2},\;\sqrt{3} \rightarrow \sqrt{3}\text{,} \] \[ \sigma : \sqrt{2} \rightarrow - \sqrt{2},\;\sqrt{3} \rightarrow \sqrt{3} \] \[ \tau : \sqrt{2} \rightarrow \sqrt{2},\;\sqrt{3} \rightarrow - \sqrt{3} \] \[ {\sigma \tau } : \sqrt{2} \rightarrow - \sqrt{2},\;\sqrt{3} \rightarrow -... | Yes |
Example 5.4 Let \( F = \mathbb{C}\left( t\right) \) be the rational function field in one variable over \( \mathbb{C} \), and let \( f\left( x\right) = {x}^{n} - t \in F\left\lbrack x\right\rbrack \) . The polynomial \( f \) is irreducible over \( F \) by the Eisenstein criterion, since \( F \) is the quotient field of... | To see this, if \( \omega = \exp \left( {{2\pi i}/n}\right) \), then \( {\omega }^{n} = 1 \), so \( {\omega }^{i}\alpha \) is a root of \( f\left( x\right) \) for each \( i \) . There are exactly \( n \) distinct powers of \( \omega \), so the \( n \) distinct elements \( \alpha ,{\omega \alpha },\ldots ,{\omega }^{n -... | Yes |
Theorem 5.5 (Natural Irrationalities) Let \( K \) be a finite Galois extension of \( F \), and let \( L \) be an arbitrary extension of \( F \) . Then \( {KL}/L \) is Galois and \( \operatorname{Gal}\left( {{KL}/L}\right) \cong \operatorname{Gal}\left( {K/K \cap L}\right) \) . Moreover, \( \left\lbrack {{KL} : L}\right... | Proof. Define \( \theta : \operatorname{Gal}\left( {{KL}/L}\right) \rightarrow \operatorname{Gal}\left( {K/F}\right) \) by \( \theta \left( \sigma \right) = {\left. \sigma \right| }_{K} \) . This map is well defined since \( K \) is normal over \( F \), and \( \theta \) is a group homomorphism. The kernel of \( \theta ... | Yes |
Theorem 5.6 (Primitive Element Theorem) A finite extension \( K/F \) is simple if and only if there are only finitely many fields \( L \) with \( F \subseteq L \subseteq K \) . | Proof. We prove this with the assumption that \( \left| F\right| = \infty \) . The case for finite fields requires a different proof, which we will handle in Section 6. Suppose that there are only finitely many intermediate fields of \( K/F \) . Since \( \left\lbrack {K : F}\right\rbrack < \infty \), we can write \( K ... | Yes |
Corollary 5.7 If \( K/F \) is finite and separable, then \( K = F\left( \alpha \right) \) for some \( \alpha \in K \) . | Proof. If \( K \) is finite and separable over \( F \), then \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) for some \( {\alpha }_{i} \) . Let \( N \) be the splitting field over \( F \) of \( \left\{ {\min \left( {F,{\alpha }_{i}}\right) : 1 \leq i \leq n}\right\} \) . Then \( N/F \) is Galois by Theor... | Yes |
Corollary 5.8 If \( K/F \) is finite and \( F \) has characteristic 0, then \( K = F\left( \alpha \right) \) for some \( \alpha \) . | Proof. This corollary follows immediately from the preceding corollary since any finite extension of a field of characteristic 0 is separable. | Yes |
Proposition 5.9 Let \( K \) be an algebraic extension of \( F \), and let \( N \) be the normal closure of \( K/F \). 1. The field \( N \) is a normal extension of \( F \) containing \( K \). Moreover, if \( M \) is a normal extension of \( F \) with \( K \subseteq M \subseteq N \), then \( M = N \). 2. If \( K = F\lef... | Proof. Since \( N \) is a splitting field over \( F \) of a set of polynomials, \( N \) is normal over \( F \). It is clear that \( N \) contains \( K \). Suppose that \( M \) is a normal extension of \( F \) with \( K \subseteq M \subseteq N \). If \( a \in K \), then \( a \in M \), so by normality \( \min \left( {F, ... | Yes |
Corollary 5.10 Let \( K \) be an algebraic extension of \( F \), and let \( N \) be the normal closure of \( K/F \). If \( {N}^{\prime } \) is any normal extension of \( F \) containing \( K \), then there is an \( F \)-homomorphism from \( N \) to \( {N}^{\prime } \). Consequently, if \( {N}^{\prime } \) does not cont... | Proof. Suppose that \( {N}^{\prime } \) is normal over \( F \) and contains \( K \). Then \( \min \left( {F, a}\right) \) splits over \( {N}^{\prime } \) for each \( a \in K \). By the isomorphism extension theorem, the identity map on \( F \) extends to a homomorphism \( \sigma : N \rightarrow {N}^{\prime } \). Then \... | Yes |
Example 5.11 Let \( F = \mathbb{Q} \) and \( K = \mathbb{Q}\left( \sqrt[3]{2}\right) \) . If \( {\omega }^{3} = 1 \) and \( \omega \neq 1 \), then \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) is the splitting field of \( {x}^{3} - 2 \) over \( \mathbb{Q} \), so it is normal over \( \mathbb{Q} \) . This field is ... | Null | No |
If \( K \) is an extension of \( F \), and if \( a \in K \) has minimal polynomial \( p\left( x\right) \) over \( F \), then the normal closure of \( F\left( a\right) /F \) is the field \( F\left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \), where the \( {a}_{i} \) are the roots of \( p\left( x\right) \) . | Null | No |
Lemma 5.13 Let \( f\left( x\right) \in \mathbb{R}\left\lbrack x\right\rbrack \) .\n\n1. If \( f\left( x\right) = {x}^{2} - a \) for some \( a > 0 \), then \( f \) has a root in \( \mathbb{R} \) . Therefore, every nonnegative real number has a real square root.\n\n2. If \( \deg \left( f\right) \) is odd, then \( f \) ha... | Proof. Suppose that \( f\left( x\right) = {x}^{2} - a \) with \( a > 0 \) . Then \( f\left( 0\right) < 0 \) and \( f\left( u\right) > 0 \) for \( u \) sufficiently large. Therefore, there is a \( c \in \left\lbrack {0, u}\right\rbrack \) with \( f\left( c\right) = 0 \) by the intermediate value theorem. In other words,... | Yes |
Lemma 5.14 Every complex number has a complex square root. Therefore, there is no field extension \( N \) of \( \mathbb{C} \) with \( \left\lbrack {N : \mathbb{C}}\right\rbrack = 2 \) . | Proof. To prove this, we use the polar coordinate representation of complex numbers. Let \( a \in \mathbb{C} \), and set \( a = r{e}^{i\theta } \) with \( r \geq 0 \) . Then \( \sqrt{r} \in \mathbb{R} \) by Lemma 5.13, so \( b = \sqrt{r}{e}^{{i\theta }/2} \in \mathbb{C} \) . We have \( {b}^{2} = r{\left( {e}^{{i\theta ... | Yes |
Lemma 6.1 If \( K \) is a field and \( G \) is a finite subgroup of \( {K}^{ * } \), then \( G \) is cyclic. | Proof. Let \( n = \left| G\right| \) and \( m = \exp \left( G\right) \) . Then \( m \) divides \( n \) by Lagrange’s theorem. If \( g \in G \), then \( {g}^{m} = 1 \), so each element of \( G \) is a root of the polynomial \( {x}^{m} - 1 \) . This polynomial has at most \( m \) roots in the field \( K \) . However, \( ... | Yes |
Corollary 6.2 If \( F \) is a finite field, then \( {F}^{ * } \) is cyclic. | Null | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.