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Lemma 7.11. Let \( \mathrm{A} \) be an abelian category, and let\n\n(*) \n\n\[ \n0 \rightarrow {L}^{ \bullet } \rightarrow {M}^{ \bullet } \rightarrow {P}^{ \bullet } \rightarrow 0 \n\] \n\nbe an exact sequence of complexes in \( \mathrm{A} \), where \( {P}^{i} \) is projective for all \( i \) . Let \( \mathcal{F} \) :... | Proof. Since \( {P}^{i} \) is projective, the sequence \n\n\[ \n0 \rightarrow {L}^{i} \rightarrow {M}^{i} \rightarrow {P}^{i} \rightarrow 0 \n\] \n\nsplits (see the end of SVIII 6.1). It follows that \n\n\[ \n0 \rightarrow \mathcal{F}\left( {L}^{i}\right) \rightarrow \mathcal{F}\left( {M}^{i}\right) \rightarrow \mathca... | Yes |
Theorem 7.12. Let \( \mathcal{F} : \mathrm{A} \rightarrow \mathrm{B} \) be an additive functor of abelian categories, and assume A has enough projectives. Every exact sequence\n\n\[ 0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \]\n\nin \( \mathrm{A} \) induces a long exact sequence ![cc115a52-9d62-431d-bb8... | Proof. By Lemma 7.8, the given exact sequence is induced by an exact sequence of projective resolutions\n\n\[ 0 \rightarrow {P}_{L}^{ \bullet } \rightarrow {P}_{M}^{ \bullet } \rightarrow {P}_{N}^{ \bullet } \rightarrow 0. \]\n\nBy Lemma 7.11, the corresponding sequence\n\n\[ 0 \rightarrow \mathcal{F}\left( {P}_{L}^{ \... | Yes |
Proposition 7.13. Let \( \mathcal{F} : \mathrm{A} \rightarrow \mathrm{B} \) be a right-exact additive functor. Then \( {\mathrm{L}}_{i}\mathcal{F} = \) 0 for \( i < 0 \), and \( {\mathrm{L}}_{0}\mathcal{F} \) is naturally isomorphic to \( \mathcal{F} \) . | Proof. Projective resolutions \( {P}^{ \bullet } \) of an object \( M \) of \( \mathrm{A} \) are in \( {\mathrm{C}}^{ \leq 0}\left( \mathrm{\;A}\right) \) : it follows that \( \mathrm{C}\left( \mathcal{F}\right) \left( {P}^{ \bullet }\right) \) is 0 in positive degree, hence so is its cohomology. Since \( {H}_{i} = {H}... | Yes |
Take \( G = \mathbb{Z} \). Then \( \mathbb{Z}\left\lbrack G\right\rbrack \) is the ring \( \mathbb{Z}\left\lbrack {x,{x}^{-1}}\right\rbrack \) of Laurent polynomials. As \( \mathbb{Z}\left\lbrack {x,{x}^{-1}}\right\rbrack /\left( {1 - x}\right) \cong \mathbb{Z} \), with the trivial action (check this!), the complex\n\n... | Applying the (contravariant) \( {\operatorname{Hom}}_{\mathbb{Z}\left\lbrack {x,{x}^{-1}}\right\rbrack }\left( {\_, M}\right) \), we see that \( {H}^{ \bullet }\left( {\mathbb{Z}, M}\right) \) is computed by the cohomology of\n\n\[ \n\cdots \rightarrow 0 \rightarrow M\xrightarrow[]{\; \cdot \left( {1 - x}\right) \;}M \... | No |
Let \( G = {C}_{m} \) be a cyclic group of order \( m \) ; then \( \mathbb{Z}\left\lbrack G\right\rbrack \cong \mathbb{Z}\left\lbrack x\right\rbrack /\left( {{x}^{m} - 1}\right) \) . | Again it is not difficult to produce a projective resolution of \( \mathbb{Z} \) (with trivial action) in the category of \( \mathbb{Z}\left\lbrack {C}_{m}\right\rbrack \) -modules: letting \( N = 1 + x + \cdots + {x}^{m - 1} = \left( {1 - {x}^{m}}\right) /\left( {1 - x}\right) \), the reader will verify that the compl... | No |
Proposition 7.16. Let \( G \) be a finite group, and let \( M \) be a \( G \) -module. Then the group cohomology \( {H}^{i}\left( {G, M}\right) \) is the cohomology of the cochain complex | \[ 0 \rightarrow {C}^{0}\left( {G, M}\right) \overset{{d}_{G}^{0}}{ \rightarrow }{C}^{1}\left( {G, M}\right) \overset{{d}_{G}^{1}}{ \rightarrow }{C}^{2}\left( {G, M}\right) \overset{{d}_{G}^{2}}{ \rightarrow }\cdots \] induced by \( \left( \dagger \right) \). Tracing definitions, we see that for \( a \in {C}^{0}\left( ... | Yes |
Claim 7.18. \( {H}^{1}\left( {G,{F}^{ * }}\right) = 0 \) . | Indeed, with notation as above we have \( {H}^{1}\left( {G,{F}^{ * }}\right) \cong \ker {d}_{G}^{1}/\operatorname{im}{d}_{G}^{0} \), and we can compute this quotient explicitly. Let \( \alpha \in {C}^{1}\left( {G,{F}^{ * }}\right) \) ; denote by \( {\alpha }_{g} \) the image of \( g \) in \( {F}^{ * } \) by \( \alpha \... | Yes |
Let \( R \) be a commutative ring, and let\n\n\[ \n{F}_{ \bullet } : \;\cdots \rightarrow {F}_{2} \rightarrow {F}_{1} \rightarrow {F}_{0} \rightarrow 0 \]\n\nbe a resolution of an \( R \) -module \( M \) by flat \( R \) -modules. Then for every \( R \) -module \( N \) ,\n\n\[ \n{\operatorname{Tor}}_{i}^{R}\left( {M, N}... | Indeed, flat modules are acyclic with respect to \( \_ \otimes N \) (Example 8.2), so this is now a consequence of Theorem 8.3. | No |
Theorem 8.9. Let \( \mathrm{A} \) be an abelian category, and let\n\n(*)\n\n\[ \cdots \rightarrow {M}^{-3, \bullet } \rightarrow {M}^{-2, \bullet } \rightarrow {M}^{-1, \bullet } \rightarrow {M}^{0, \bullet } \rightarrow 0 \rightarrow \cdots \]\n\nbe a complex in \( {\mathrm{C}}^{ \leq 0}\left( {{\mathrm{C}}^{ \leq 0}\... | Proof. It is enough to prove the second statement: the first one follows by flipping the double complex corresponding to (*) (cf. Exercise 8.4).\n\n\( {}^{34}\mathrm{\;A} \) clever way out of the sign quagmire in this computation is to choose another way to get a double complex out of \( {\operatorname{Hom}}_{\mathbf{A... | Yes |
Example 8.10. Here is a taste of how convenient Theorem 8.9 is. Let \( {P}^{ \bullet } \) be a complex in \( {\mathrm{C}}^{ \leq 0}\left( \mathrm{\;A}\right) \), where each \( {P}^{i} \) is projective, and let \( {L}^{ \bullet } \) be an exact complex in \( {\mathrm{C}}^{ \geq 0}\left( \mathrm{\;A}\right) \). Since \( ... | According to Theorem 8.9, the corresponding total complex is exact; as seen in Example 8.8 (cf. Exercise 8.5), this says that \( {\operatorname{Hom}}_{\mathrm{K}\left( \mathrm{A}\right) }\left( {{P}^{ \bullet }, L{\left\lbrack i\right\rbrack }^{ \bullet }}\right) = 0 \) for all \( i \). In other words, every cochain mo... | Yes |
Theorem 8.12. Let \( \mathrm{A} \) be an abelian category, and denote by \( {\mathrm{A}}^{\prime } \) the category \( {\mathrm{C}}^{ \leq 0}\left( \mathrm{\;A}\right) \) . Let \( {N}^{ \bullet } \) be an object of \( {\mathrm{A}}^{\prime } \), and let\n\n(*) \n\n\[ \cdots \rightarrow {M}^{-3, \bullet } \rightarrow {M}^... | Proof. The second statement follows from the first, by flipping the corresponding double complex about the main diagonal.\n\nThe first statement follows from Theorem 8.9 and Claim 8.11 Indeed, let \( {M}_{N}^{\bullet , \bullet } \) be the exact complex\n\n\[ \cdots \rightarrow {M}^{-2, \bullet } \rightarrow {M}^{-1, \b... | Yes |
Theorem 8.13. Let \( M, N \) be modules over a commutative ring \( R \), and let \( {P}_{M}^{ \bullet } \) , resp., \( {P}_{N}^{ \bullet } \), be projective resolutions of \( M \), resp., \( N \) . Then\n\n\[ \n{H}^{i}\left( {{P}_{M}^{ \bullet }{ \otimes }_{R}N}\right) \cong {H}^{i}\left( {M{ \otimes }_{R}{P}_{N}^{ \bu... | Proof. Apply Theorem 8.12 to the complex\n\n(*)\n\n\[ \n\cdots \rightarrow {P}_{M}^{-2}{ \otimes }_{R}{P}_{N}^{ \bullet } \rightarrow {P}_{M}^{-1}{ \otimes }_{R}{P}_{N}^{ \bullet } \rightarrow {P}_{M}^{-0}{ \otimes }_{R}{P}_{N}^{ \bullet } \rightarrow 0 \rightarrow \cdots .\n\]\n\nSince each \( {P}_{N}^{j} \) is projec... | Yes |
Theorem 8.14. Let \( M, N \) be modules over a commutative ring \( R \), and let \( {P}_{M}^{ \bullet } \) , resp., \( {Q}_{N}^{ \bullet } \), be a projective resolution of \( M \), resp., an injective resolution of \( N \) . Then\n\n\[ \n{H}^{i}\left( {{\operatorname{Hom}}_{R}\left( {{P}_{M}^{ \bullet }, N}\right) }\r... | The proof of Theorem 8.14 is left to the reader (Exercise 8.9): Example 8.8 and the strategy extensively discussed above will hopefully make this a very easy task. | No |
Proposition 1. Any two left cosets of \( H \) in \( G \) have the same cardinality \( {}^{3} \) and are disjoint if they do not coincide. In particular, \( G \) is the disjoint union of all left cosets of \( H \) in \( G \) . | Proof. For each \( a \in G \), the left translation \( H \rightarrow {aH}, h \mapsto {ah} \), is bijective. Therefore, all left cosets of \( H \) in \( G \) have the same cardinality. The second assertion is a consequence of the following lemma: | Yes |
Lemma 2. Let \( {aH} \) and \( {bH} \) be left cosets of \( H \) in \( G \) . Then the following conditions are equivalent:\n\n(i) \( {aH} = {bH} \) .\n\n(ii) \( {aH} \cap {bH} \neq \varnothing \) .\n\n(iii) \( a \in {bH} \) .\n\n(iv) \( {b}^{-1}a \in H \) . | Proof. The implication (i) \( \Rightarrow \) (ii) is trivial, since \( H \neq \varnothing \) . Next, assume (ii). There exists an element \( c \in {aH} \cap {bH} \), say \( c = a{h}_{1} = b{h}_{2} \), where \( {h}_{1},{h}_{2} \in H \) . This means that \( a = b{h}_{2}{h}_{1}^{-1} \in {bH} \), and we see that (iii) hold... | Yes |
Proposition 6 (Fundamental theorem on homomorphisms). Let \( \varphi : G \rightarrow {G}^{\prime } \) be a group homomorphism and \( N \subset G \) a normal subgroup such that \( N \subset \ker \varphi \) . Then there exists a unique group homomorphism \( \bar{\varphi } : G/N \rightarrow {G}^{\prime } \) satisfying \( ... | Proof. If \( \bar{\varphi } \) exists, then \[ \bar{\varphi }\left( {aN}\right) = \bar{\varphi }\left( {\pi \left( a\right) }\right) = \varphi \left( a\right) \] for \( a \in G \) and we see that \( \bar{\varphi } \) is unique. On the other hand, we can try to set \( \bar{\varphi }\left( {aN}\right) = \varphi \left( a\... | Yes |
Proposition 8 (First isomorphism theorem). Let \( G \) be a group, \( H \subset G \) a subgroup, and \( N \subset G \) a normal subgroup of \( G \) . Then \( {HN} \) is a subgroup of \( G \) admitting \( N \) as a normal subgroup, and \( H \cap N \) is a normal subgroup of \( H \) . The canonical homomorphism\n\n\[ H/H... | Proof. Using the fact that \( N \) is normal in \( G \), one easily shows that \( {HN} \) is a subgroup of \( G \) . Furthermore, \( N \) is normal in \( {HN} \), since it is normal in \( G \) . Now consider the composition of homomorphisms\n\n\[ H \hookrightarrow {HN}\overset{\pi }{ \rightarrow }{HN}/N \]\n\nwhere \( ... | Yes |
Proposition 9 (Second isomorphism theorem). Let \( G \) be a group and let \( N, H \) be normal subgroups of \( G \) satisfying \( N \subset H \subset G \) . Then \( N \) is normal in \( H \) as well, and one can view \( H/N \) as a normal subgroup of \( G/N \) . Furthermore, the canonical group homomorphism\n\n\[ \n\l... | Proof. To begin with, let us explain how to view \( H/N \) as a subgroup of \( G/N \) . Look at the group homomorphism \n\n\[ \nH \hookrightarrow G\overset{\pi }{ \rightarrow }G/N \n\] \n\nwhere \( \pi \) is the canonical projection. Since this homomorphism admits \( N \) as kernel, it induces by Proposition 6 a monomo... | Yes |
Lemma 4. Let \( H \subset \mathbb{Z} \) be a subgroup. Then there exists an integer \( m \in \mathbb{Z} \) such that \( H = m\mathbb{Z} \). In particular, every subgroup of \( \mathbb{Z} \) is cyclic. | Proof. We may assume \( H \neq 0 \), i.e., that \( H \) is different from the zero subgroup of \( \mathbb{Z} \) given by the zero element. Then \( H \) must contain positive integers; let \( m \) be the smallest among these. We claim that \( H = m\mathbb{Z} \), where clearly, \( m\mathbb{Z} \subset H \). To show the re... | Yes |
Proposition 5. (i) Every subgroup \( H \) of a cyclic group \( G \) is itself cyclic. | Proof. It follows immediately from the definition of cyclic groups that the image of a cyclic group under a group homomorphism \( \varphi : G \rightarrow {G}^{\prime } \) is cyclic. Since \( \ker \varphi \) is a subgroup of \( G \), it remains to verify assertion (i). Therefore, let \( G \) be cyclic and let \( H \subs... | No |
Proposition 6 (Fermat’s little theorem). Let \( G \) be a finite group, \( a \in G \) . Then ord \( a \) divides ord \( G \) and we have \( {a}^{\operatorname{ord}G} = 1 \) . | Proof. Apply the theorem of Lagrange \( {1.2}/3 \) to the cyclic subgroup of \( G \) that is generated by \( a \) . | No |
Corollary 7. Let \( G \) be a finite group such that \( p \mathrel{\text{:=}} \operatorname{ord}G \) is prime. Then \( G \) is cyclic, \( G \simeq \mathbb{Z}/p\mathbb{Z} \), and every element \( a \in G, a \neq 1 \), is of order \( p \) . In particular, every such element a generates \( G \) . | Proof. For each element \( a \in G, a \neq 1 \), consider the cyclic subgroup \( H \subset G \) generated by \( a \) . Then ord \( a = \operatorname{ord}H \) is different from 1 and, according to Proposition 6, a divisor of \( p = \operatorname{ord}G \) . Since \( p \) is prime, we get ord \( a = \operatorname{ord}H = ... | Yes |
Proposition 4. Let \( R \) be a ring and \( g = \mathop{\sum }\limits_{{i = 0}}^{d}{a}_{i}{X}^{i} \in R\left\lbrack X\right\rbrack \) a polynomial whose leading coefficient \( {a}_{d} \) is a unit in \( R \) . Then, for each \( f \in R\left\lbrack X\right\rbrack \), there exist unique polynomials \( q, r \in R\left\lbr... | Proof. First observe that we have \( \deg \left( {qg}\right) = \deg q + \deg g \) for arbitrary polynomials \( q \in R\left\lbrack X\right\rbrack \), even if \( R \) is not an integral domain. Indeed, the leading coefficient \( {a}_{d} \) of \( g \) is a unit. So if \( q \) is of some degree \( n \geq 0 \) with leading... | Yes |
Corollary 5. If \( \varphi : R \rightarrow {R}^{\prime } \) is a surjective ring homomorphism, then \( {R}^{\prime } \) is canonically isomorphic to \( R/\ker \varphi \) . | To do the proof of Proposition 4 we apply \( {1.2}/6 \) to the additive group of \( R \) . Then it remains only to check that the group homomorphism \( \bar{\varphi } : R/\mathfrak{a} \rightarrow {R}^{\prime } \) obtained from \( {1.2}/6 \) is in fact a ring homomorphism. However, since \( \bar{\varphi } \) is characte... | No |
Proposition 6. The following conditions are equivalent for \( m \in \mathbb{Z}, m > 0 \) :\n\n(i) \( m \) is a prime number.\n\n(ii) \( \mathbb{Z}/m\mathbb{Z} \) is an integral domain.\n\n(iii) \( \mathbb{Z}/m\mathbb{Z} \) is a field. | Proof. For any \( x \in \mathbb{Z} \), let us denote by \( \bar{x} \in \mathbb{Z}/m\mathbb{Z} \) the attached residue class modulo \( m\mathbb{Z} \) . To begin with, assume condition (i), i.e., that \( m \) is a prime number. Then \( m > 1 \) and \( \mathbb{Z}/m\mathbb{Z} \) is nonzero. Now if \( \bar{a} \cdot \bar{b} ... | Yes |
Proposition 8. Let \( R \) be a ring.\n\n(i) An ideal \( \mathfrak{p} \subset R \) is prime if and only if \( R/\mathfrak{p} \) is an integral domain. | Proof. First of all, note that \( \mathfrak{p} \) is a proper ideal in \( R \) if and only if the residue class ring \( R/\mathfrak{p} \) is nonzero, similarly for \( \mathfrak{m} \). Now assertion (i) is easy to verify. Look at residue classes \( \bar{a},\bar{b} \in R/\mathfrak{p} \) of elements \( a, b \in R \). Then... | Yes |
Lemma 10. The zero ideal \( 0 \subset R \) of a ring \( R \) is maximal if and only if \( R \) is a field. | Proof of Lemma 10. Assume that the zero ideal \( 0 \subset R \) is maximal and consider an element \( a \in R - \{ 0\} \) . Then \( {aR} = R \), and there exists an element \( b \in R \) such that \( {ab} = 1 \) . Thus \( {R}^{ * } = R - \{ 0\} \), and \( R \) is a field. Conversely, that the zero ideal of a field is m... | Yes |
Corollary 11. An ideal in \( \mathbb{Z} \) is prime if and only if it is of type \( p\mathbb{Z} \) for a prime number \( p \) or for \( p = 0 \) . An ideal in \( \mathbb{Z} \) is maximal if and only if it is a nonzero prime ideal. | Just use the fact that \( \mathbb{Z} \) is a principal ideal domain by \( {2.2}/3 \) and that the zero ideal of an integral domain is prime. | No |
Proposition 12. Let \( R \) be a ring and \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{n} \subset R \) pairwise coprime ideals, i.e., assume that \( {\mathfrak{a}}_{i} + {\mathfrak{a}}_{j} = R \) for \( i \neq j \) . Then, writing \( {\pi }_{i} : R \rightarrow R/{\mathfrak{a}}_{i} \) for the canonical projections, the... | Proof. To begin with, let us show for \( j = 1,\ldots, n \) that the ideals \( {\mathfrak{a}}_{j} \) and \( \mathop{\bigcap }\limits_{{i \neq j}}{\mathfrak{a}}_{i} \) are coprime in the sense that their sum yields \( R \) . To do this, fix an index \( j \) . Since \( {\mathfrak{a}}_{j} \) and \( {\mathfrak{a}}_{i} \) a... | Yes |
Corollary 13. Let \( {a}_{1},\ldots ,{a}_{n} \in \mathbb{Z} \) be integers that are pairwise relatively prime. Then the system of simultaneous congruences \( x \equiv {x}_{i}{\;\operatorname{mod}\;{a}_{i}}, i = 1,\ldots, n \) , is solvable for arbitrary integers \( {x}_{1},\ldots ,{x}_{n} \in \mathbb{Z} \), and the sol... | Of course, it has to be checked that relatively prime integers \( a,{a}^{\prime } \in \mathbb{Z} \) satisfy the equations\n\n\[ \left( {a,{a}^{\prime }}\right) = \left( 1\right) \;\text{ and }\;\left( {a \cdot {a}^{\prime }}\right) = \left( a\right) \cap \left( {a}^{\prime }\right) \]\n\nfor details see \( {2.4}/{13} \... | Yes |
Proposition 2. Every Euclidean domain is a principal ideal domain. | Proof. Proceeding as in \( {1.3}/4 \), let \( \mathfrak{a} \subset R \) be an ideal, where we may assume \( \mathfrak{a} \neq 0 \) . Choose an element \( a \) of \( \mathfrak{a} - \{ 0\} \) such that \( \delta \left( a\right) \) is minimal with respect to the Euclidean function \( \delta \) considered on \( R \) . We c... | Yes |
Proposition 6. Let \( R \) be a principal ideal domain and \( p \in R \) a nonzero nonunit. The following conditions are equivalent:\n\n(i) \( p \) is irreducible.\n\n(ii) \( p \) is a prime element.\n\n(iii) \( \left( p\right) \) is a maximal ideal in \( R \) . | Proof. In view of Remark 5, it remains to show that (i) implies (iii). Therefore, assume that \( p \) is irreducible and let \( \mathfrak{a} \) be an ideal in \( R \) satisfying \( \left( p\right) \subset \mathfrak{a} \subset R \) , say \( \mathfrak{a} = \left( a\right) \), since \( R \) is a principal ideal domain. Th... | Yes |
Proposition 7. Let \( R \) be a principal ideal domain. Then every nonzero nonunit \( a \in R \) is a product of prime elements. | Proof. Fix an element \( a \in R - \left( {{R}^{ * }\cup \{ 0\} }\right) \) . If \( a \) is irreducible (and thereby prime), nothing has to be shown. Otherwise, decompose \( a \) into the product \( {bc} \) of two nonunits in \( R \) . If one of the factors \( b \) and \( c \) is not yet irreducible, we can further dec... | No |
Lemma 8. Every principal ideal domain \( R \) is Noetherian, i.e., every ascending chain of ideals \( {\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \subset \ldots \subset R \) becomes stationary in the sense that there is some \( n \in \mathbb{N} \) such that \( {\mathfrak{a}}_{i} = {\mathfrak{a}}_{n} \) for all \( i \... | The assertion is easy to verify. Since the union of an ascending chain of ideals is itself an ideal, we can consider \( \mathfrak{a} = \mathop{\bigcup }\limits_{{i > 1}}{\mathfrak{a}}_{i} \) as an ideal in \( R \) ; it is a principal ideal, say \( \mathfrak{a} = \left( a\right) \) . However, since \( a \in \overline{\m... | Yes |
Lemma 9. Let \( R \) be an integral domain. For an element \( a \in R \), consider factorizations\n\n\[ a = {p}_{1}\ldots {p}_{r} = {q}_{1}\ldots {q}_{s} \]\n\ninto prime elements \( {p}_{i} \) and irreducible elements \( {q}_{j} \) . Then \( r = s \), and one can renumber the \( {q}_{j} \) in such a way that \( {p}_{i... | Proof. Since \( {p}_{1} \mid {q}_{1}\ldots {q}_{s} \) and \( {p}_{1} \) is prime, there exists an index \( j \) such that \( {p}_{1} \mid {q}_{j} \) . Renumbering the \( {q}_{j} \), we may assume \( j = 1 \) . Hence, using the fact that \( {q}_{1} \) is irreducible, there is an equation \( {q}_{1} = {\varepsilon }_{1}{... | Yes |
Proposition 13. Let \( {x}_{1},\ldots ,{x}_{n} \) be elements of an integral domain \( R \). (i) If \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \), the ideal generated by the \( {x}_{i} \) in \( R \), is principal, say generated by an element \( d \in R \), then \( d = \gcd \left( {{x}_{1},\ldots ,{x}_{n}}\right) \). | Proof. (i) Assume \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) = \left( d\right) \). Then we get \( {x}_{i} \in \left( d\right) \) and therefore \( d \mid {x}_{i} \) for all \( i \). Moreover, due to \( d \in \left( {{x}_{1},\ldots ,{x}_{n}}\right) \), there is an equation \( d = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i... | Yes |
Corollary 14. Let \( R \) be a principal ideal domain and \( a = \varepsilon {p}_{1}^{{\nu }_{1}}\ldots {p}_{r}^{{\nu }_{r}} \) a prime factorization of some element \( a \in R \), where \( \varepsilon \) is a unit and the prime elements \( {p}_{i} \) are pairwise nonassociated. Then the Chinese remainder theorem \( {2... | Proof. Using Proposition 13 in conjunction with Proposition 12, the ideals \( \left( {p}_{1}^{{\nu }_{1}}\right) ,\ldots ,\left( {p}_{r}^{{\nu }_{r}}\right) \) are pairwise coprime in \( R \), since \( \gcd \left( {{p}_{i}^{{\nu }_{i}},{p}_{j}^{{\nu }_{j}}}\right) = 1 \) for \( i \neq j \) . Likewise, we have \( \left(... | Yes |
Proposition 15 (Euclidean algorithm). Let \( R \) be a Euclidean domain. For two elements \( x, y \in R - \{ 0\} \) consider the sequence \( {z}_{0},{z}_{1},\ldots \in R \), which is inductively given by\n\n\[ \n{z}_{0} = x \n\]\n\n\[ \n{z}_{1} = y \n\]\n\n\[ \n{z}_{i + 1} = \left\{ \begin{array}{l} \text{ the remainde... | Proof. Let \( \delta : R - \{ 0\} \rightarrow \mathbb{N} \) be the Euclidean function considered on \( R \) and fix an index \( i > 0 \) such that \( {z}_{i} \neq 0 \) . According to the definition of the sequence \( {z}_{0},{z}_{1},\ldots \), there is an equation of type\n\n\[ \n{z}_{i - 1} = {q}_{i}{z}_{i} + {z}_{i +... | Yes |
Proposition 1. Let \( \varphi : R \rightarrow {R}^{\prime } \) be a ring homomorphism and \( \sigma : M \rightarrow {R}^{\prime } \) a monoid homomorphism, where we view \( {R}^{\prime } \) as a monoid with respect to the ring multiplication. Then there exists a unique ring homomorphism \( \Phi : R\left\lbrack M\right\... | Proof. To verify the uniqueness assertion, consider an element \( \mathop{\sum }\limits_{{\mu \in M}}{a}_{\mu }{X}^{\mu } \) in \( R\left\lbrack M\right\rbrack \) . If there exists a homomorphism \( \Phi \) satisfying the stated conditions, we must have\n\n\[ \Phi \left( {\sum {a}_{\mu }{X}^{\mu }}\right) = \sum \Phi \... | Yes |
Proposition 2. If \( R \) is an integral domain, then for finitely many variables \( {X}_{1},\ldots ,{X}_{n} \), the polynomial ring \( R\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is also an integral domain. | Proof. We have already seen in \( {2.1}/3 \) that the proposition is true in the case of one variable. But then, using the isomorphism\n\n\[ R\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \simeq \left( {R\left\lbrack {{X}_{1},\ldots ,{X}_{n - 1}}\right\rbrack }\right) \left\lbrack {X}_{n}\right\rbrack ,\]\n\nthe ... | Yes |
Proposition 3. Let \( f, g \in R\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) be polynomials with coefficients in a ring \( R \) . Then\n\n\[ \deg \left( {f + g}\right) \leq \max \left( {\deg f,\deg g}\right) \]\n\n\[ \deg \left( {f \cdot g}\right) \leq \deg f + \deg g \]\n\nand \( \deg \left( {f \cdot g}\righ... | Proof. The estimate for \( \deg \left( {f + g}\right) \) becomes clear if we decompose polynomials in \( R\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) into the sums of their homogeneous parts. Furthermore, if \( \deg f = r \) and \( \deg g = s \), and if \( f = \mathop{\sum }\limits_{{i = 0}}^{r}{f}_{i}, g = ... | Yes |
Proposition 5. Let \( \varphi : R \rightarrow {R}^{\prime } \) be a ring homomorphism and consider finitely many elements \( {x}_{1},\ldots ,{x}_{n} \in {R}^{\prime } \) . Then there exists a unique ring homomorphism \( \Phi : R\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \rightarrow {R}^{\prime } \) satisfying ... | Writing \( x = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) and \( {x}^{\mu } = {x}_{1}^{{\mu }_{1}}\ldots {x}_{n}^{{\mu }_{n}} \) for \( \mu \in {\mathbb{N}}^{n} \) in the situation of the preceding proposition, we can describe the homomorphism \( \Phi \) by\n\n\[ \Phi : R\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbr... | Yes |
Proposition 2. Let \( f \in K\left\lbrack X\right\rbrack, f \neq 0 \), be a polynomial with coefficients in a field \( K \) . A zero \( \alpha \in K \) of \( f \) is a multiple zero (i.e., a zero of multiplicity \( \geq 2 \) ) if and only if \( \left( {f}^{\prime }\right) \left( \alpha \right) = 0 \) . | Proof. If \( \alpha \) is a zero of \( f \) of multiplicity \( r \geq 1 \), then there is a factorization of type \( f = {\left( X - \alpha \right) }^{r}g \) for some \( g \in K\left\lbrack X\right\rbrack \) satisfying \( g\left( \alpha \right) \neq 0 \) . Since\n\n\[ \n{f}^{\prime } = {\left( X - \alpha \right) }^{r}{... | Yes |
Corollary 3. An element \( \alpha \in K \) is a multiple zero of a nonzero polynomial \( f \in K\left\lbrack X\right\rbrack \) if and only if \( \alpha \) is a zero of \( \gcd \left( {f,{f}^{\prime }}\right) \) . | For example, if \( p \) is a prime number, the polynomial \( f = {X}^{p} - X \in {\mathbb{F}}_{p}\left\lbrack X\right\rbrack \) does not admit multiple zeros. Indeed, we have \( {f}^{\prime } = - 1 \), since the \( p \) -fold sum \( p \cdot 1 \) of the unit element \( 1 \in {\mathbb{F}}_{p} = \mathbb{Z}/p\mathbb{Z} \) ... | No |
Lemma 5 (Gauss). Let \( R \) be a unique factorization domain and \( p \in R \) a prime element. Then \( {\nu }_{p}\left( \cdot \right) \) satisfies the following relation for elements \( f, g \in Q\left( R\right) \left\lbrack X\right\rbrack \) :\n\n\[{\nu }_{p}\left( {fg}\right) = {\nu }_{p}\left( f\right) + {\nu }_{p... | Proof. As mentioned above, the stated relation holds for constant polynomials, i.e., for \( f, g \in Q\left( R\right) \) and hence also for \( f \in Q\left( R\right) \) and arbitrary polynomials \( g \in Q\left( R\right) \left\lbrack X\right\rbrack \) .\n\nTo deal with the general case we may assume \( f, g \neq 0 \) .... | Yes |
Corollary 6. Let \( R \) be a unique factorization domain and \( h \in R\left\lbrack X\right\rbrack \) a monic polynomial. Assume that there is a factorization \( h = f \cdot g \) into monic polynomials \( f, g \in Q\left( R\right) \left\lbrack X\right\rbrack \) . Then necessarily \( f, g \in R\left\lbrack X\right\rbra... | Proof. We have \( {\nu }_{p}\left( h\right) = 0 \), as well as \( {\nu }_{p}\left( f\right) ,{\nu }_{p}\left( g\right) \leq 0 \) for every prime element \( p \in R \), due to the fact that \( h, f \), and \( g \) are monic. Furthermore, Gauss’s lemma yields\n\n\[{\nu }_{p}\left( f\right) + {\nu }_{p}\left( g\right) = {... | Yes |
Proposition 7 (Gauss). Let \( R \) be a unique factorization domain. Then the polynomial ring \( R\left\lbrack X\right\rbrack \) is a unique factorization domain as well. A polynomial \( q \in R\left\lbrack X\right\rbrack \) is prime if and only if:\n\n(i) \( q \) is prime in \( R \), or\n\n(ii) \( q \) is primitive in... | Proof. Let \( q \) be a prime element in \( R \) . Then \( R/{qR} \) is an integral domain and the same is true for \( R\left\lbrack X\right\rbrack /{qR}\left\lbrack X\right\rbrack \simeq \left( {R/{qR}}\right) \left\lbrack X\right\rbrack \) . From this we conclude that \( q \) is prime also in \( R\left\lbrack X\right... | Yes |
Proposition 1 (Eisenstein's criterion). Let \( R \) be a unique factorization domain and \( f = {a}_{n}{X}^{n} + \ldots + {a}_{0} \in R\left\lbrack X\right\rbrack \) a primitive polynomial of degree \( > 0 \) . Assume there is a prime element \( p \in R \) such that\n\n\[ p \nmid {a}_{n},\;p \mid {a}_{i}\text{ for }i <... | Proof. Suppose \( f \) is reducible in \( R\left\lbrack X\right\rbrack \) . Then there is a factorization\n\n\[ f = {gh},\;\text{ say }\;g = \mathop{\sum }\limits_{{i = 0}}^{r}{b}_{i}{X}^{i}, h = \mathop{\sum }\limits_{{i = 0}}^{s}{c}_{i}{X}^{i}, \]\n\nwhere \( r + s = n \), and \( r > 0, s > 0 \) . Furthermore, from o... | Yes |
Proposition 2. Let \( R \) be a unique factorization domain, \( p \in R \) a prime element, and \( f \in R\left\lbrack X\right\rbrack \) a polynomial of degree \( > 0 \) whose leading coefficient is not divisible by \( p \) . Furthermore, let \( \Phi : R\left\lbrack X\right\rbrack \rightarrow R/\left( p\right) \left\lb... | Proof. First, assume that \( f \in R\left\lbrack X\right\rbrack \) is primitive. Then if \( f \) is reducible, there is a factorization \( f = {gh} \) in \( R\left\lbrack X\right\rbrack \), where \( \deg g > 0 \) and \( \deg h > 0 \) . Furthermore, \( p \) cannot divide the leading coefficient of \( g \) or \( h \), si... | Yes |
Lemma 1. (i) Let \( A \) be a principal ideal domain and \( a \in A \) an element with prime factorization \( a = {p}_{1}\ldots {p}_{r} \) . Then \( {l}_{A}\left( {A/{aA}}\right) = r \) . | Now assertion (i) is easy to justify. Renumbering the \( {p}_{i} \), we can look at a prime factorization of type \( a = \varepsilon {p}_{1}^{{\nu }_{1}}\ldots {p}_{s}^{{\nu }_{s}} \) for a unit \( \varepsilon \) and pairwise nonassociated prime elements \( {p}_{1},\ldots ,{p}_{s} \), where \( r = {\nu }_{1} + \ldots +... | Yes |
Lemma 5. Let \( A \) be a principal ideal domain and let \( Q \simeq {\bigoplus }_{i = 1}^{n}A/{\alpha }_{i}A \) be an \( A \) -module, where \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in A - \{ 0\} \) are nonunits such that \( {\alpha }_{i} \mid {\alpha }_{i + 1} \) for \( 1 \leq i < n \) . Then the elements \( {\alpha }... | Proof. For technical reasons we invert the numbering of the elements \( {\alpha }_{i} \) and consider two decompositions\n\n\[ Q \simeq {\bigoplus }_{i = 1}^{n}A/{\alpha }_{i}A \simeq {\bigoplus }_{j = 1}^{m}A/{\beta }_{j}A \]\n\nsuch that \( {\alpha }_{i + 1}\left| {{\alpha }_{i}\text{for}1 \leq i < n\text{, as well a... | Yes |
Proposition 6. Let \( A \) be a principal ideal domain, \( F \) a finite free \( A \) -module with basis \( {x}_{1},\ldots ,{x}_{r} \), as well as \( M \subset F \) a submodule of rank \( n \) with corresponding elementary divisors \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) . Furthermore, let \( {z}_{1},\ldots ,{z}_{m} ... | Proof. To start with, let us verify the assertion for \( t = 1 \) . Note that \( \left( {\alpha }_{1}\right) \subset A \) is the ideal generated by all elements of type \( \varphi \left( z\right) \) for \( z \in M \) and \( \varphi \in {F}^{ * } \) ; this can be read from the assertion of Theorem 2 or from its proof. I... | Yes |
Proposition 2. Let \( K \) be a field and \( P \subset K \) its prime subfield. Then:\n\n(i) char \( K = p > 0 \Leftrightarrow P \simeq {\mathbb{F}}_{p} \) for \( p \) prime.\n\n(ii) char \( K = 0 \Leftrightarrow P \simeq \mathbb{Q} \) . | Proof. We have \( \operatorname{char}{\mathbb{F}}_{p} = p \), as well as \( \operatorname{char}\mathbb{Q} = 0 \) . Since \( \operatorname{char}P = \operatorname{char}K \), we get char \( K = p \) from \( P \simeq {\mathbb{F}}_{p} \), and \( \operatorname{char}K = 0 \) from \( P \simeq \mathbb{Q} \) . This justifies in ... | No |
Proposition 6. Let \( K \subset L \) be a field extension and \( \alpha \in L \) algebraic over \( K \) with minimal polynomial \( f \in K\left\lbrack X\right\rbrack \) . Writing \( K\left\lbrack \alpha \right\rbrack \) for the subring of \( L \) that is generated by \( \alpha \) and \( K \), i.e., for the image of the... | Proof. We have \( K\left\lbrack \alpha \right\rbrack = \operatorname{im}\varphi \simeq K\left\lbrack X\right\rbrack /\left( f\right) \), due to the fundamental theorem on homomorphisms. Since \( \ker \varphi = \left( f\right) \) is a nonzero prime ideal in \( K\left\lbrack X\right\rbrack \), we conclude from \( {2.4}/6... | Yes |
Proposition 7. Every finite field extension \( K \subset L \) is algebraic. | Proof. Let \( \left\lbrack {L : K}\right\rbrack = n \) and consider an element \( \alpha \in L \) . Then the powers \( {\alpha }^{0},\ldots ,{\alpha }^{n} \) give rise to a system of length \( n + 1 \) and hence to a system in \( L \) that is linearly dependent over \( K \) . It follows that there exists a nontrivial e... | Yes |
Proposition 9. Let \( L = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) be a finitely generated field extension of \( K \) . Assume that \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are algebraic over \( K \) . Then:\n\n(i) \( L = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = K\left\lbrack {{\alpha }_{1... | Proof. We conclude by induction on \( n \) . The case \( n = 1 \) was already dealt with in Proposition 6. Therefore, let \( n > 1 \) . We may assume by the induction hypothesis that \( K\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n - 1}}\right\rbrack \) is a finite field extension of \( K \) . Furthermore, it follo... | Yes |
Proposition 12. Let \( K \subset L \subset M \) be field extensions. If \( \alpha \in M \) is algebraic over \( L \) and if \( L/K \) is algebraic, then \( \alpha \) is algebraic over \( K \) as well. In particular, the field extension \( M/K \) is algebraic if and only if \( M/L \) and \( L/K \) are algebraic. | Proof. Let \( f = {X}^{n} + {c}_{1}{X}^{n - 1} + \ldots + {c}_{n} \in L\left\lbrack X\right\rbrack \) be the minimal polynomial of \( \alpha \) over \( L \) . Then \( \alpha \) is algebraic over the subfield \( K\left( {{c}_{1},\ldots ,{c}_{n}}\right) \) of \( L \), and we can conclude from Proposition 6 that\n\n\[ \le... | Yes |
Corollary 3. Every finite ring homomorphism \( A \rightarrow B \) is integral. | Proof. Use condition (iii) of Lemma 1 for \( M = B \) to see that \( A \rightarrow B \) is integral. | No |
Corollary 4. Let \( \varphi : A \rightarrow B \) be a ring homomorphism of finite type and assume \( B = A\left\lbrack {{b}_{1},\ldots ,{b}_{r}}\right\rbrack \) for elements \( {b}_{1},\ldots ,{b}_{r} \in B \) that are integral over \( A \) . Then \( A \rightarrow B \) is finite and, in particular, integral. | Proof. Consider the chain of \ | No |
Corollary 5. Let \( A \rightarrow B \) and \( B \rightarrow C \) be two finite (resp. integral) ring homomorphisms. Then their composition \( A \rightarrow C \) is also finite (resp. integral). | Proof. To settle the assertion for finite homomorphisms, we use the same argument as the one applied in the induction step of the proof of Corollary 4. Hence, it remains to consider the case of integral homomorphisms. Therefore, assume that \( A \rightarrow B \) and \( B \rightarrow C \) are integral and consider an el... | Yes |
Lemma 7. Let \( A \hookrightarrow B \) be an integral extension of integral domains. If one of the rings \( A \) and \( B \) is a field, then the same is true for the other as well. | Proof. Let \( A \) be a field and \( b \neq 0 \) an element of \( B \) . Then \( b \) satisfies an integral equation over \( A \), say\n\n\[ \n{b}^{n} + {a}_{1}{b}^{n - 1} + \ldots + {a}_{n} = 0,\;{a}_{1},\ldots ,{a}_{n} \in A, \n\]\n\nand we may assume \( {a}_{n} \neq 0 \) . Indeed, pass to the field of fractions of \... | Yes |
Corollary 8. Let \( K \subset L \) be a field extension satisfying \( L = K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) for some elements \( {x}_{1},\ldots ,{x}_{n} \in L \), i.e., assume that \( K \subset L \), as a ring extension, is of finite type. Then the extension \( K \subset L \) is finite. | Proof. Due to Theorem 6 on Noether normalization, there are elements \( {y}_{1},\ldots ,{y}_{r} \) in \( L \) such that the ring extension \( K\left\lbrack {{y}_{1},\ldots ,{y}_{r}}\right\rbrack \hookrightarrow L \) is finite and the elements \( {y}_{1},\ldots ,{y}_{r} \) are algebraically independent over \( K \) . Si... | Yes |
Corollary 9. Let \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) be the polynomial ring in \( n \) variables over a field \( K \) and let \( \mathfrak{m} \subset K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) be a maximal ideal. Then the canonical map \( K \rightarrow K\left\lbrack {{X}_{1},\ldots ,... | Proof. Since \( \mathfrak{m} \) is a maximal ideal in \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \), we see that \( L \) is a field. Furthermore, if \( {x}_{i} \in L \) is the residue class of the variable \( {X}_{i} \) for each \( i \), we get \( L = K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \)... | Yes |
Proposition 1. Let \( K \) be a field and \( f \in K\left\lbrack X\right\rbrack \) a polynomial of degree \( \geq 1 \) . Then there exists a finite algebraic field extension \( K \subset L \) such that \( f \) admits a zero in \( L \) . If \( f \) is irreducible, we can set \( L \mathrel{\text{:=}} K\left\lbrack X\righ... | Proof. We may assume that \( f \) is irreducible; otherwise, decompose \( f \) into its prime factors and replace it by one of these. Then \( \left( f\right) \) is a maximal ideal in\n\n\( K\left\lbrack X\right\rbrack \) by \( {2.4}/6 \), and it follows that \( L \mathrel{\text{:=}} K\left\lbrack X\right\rbrack /\left(... | Yes |
Theorem 4. Every field \( K \) admits an extension field \( L \) that is algebraically closed. | For the proof of the theorem we need to know that every ring \( R \neq 0 \) contains a maximal ideal. The latter result is a consequence of Zorn's lemma, whose assertion we will explain next. | No |
Lemma 5 (Zorn). Let \( M \) be a partially ordered set such that every subset of \( M \) that is totally ordered with respect to the order induced from \( M \) admits an upper bound in \( M \) . Then there exists a maximal element in \( M{.}^{2} \) | For an elementary justification of the above result we refer to [12], Appendix 2, §2. However, it should be pointed out that Zorn's lemma is of axiomatic character. It is equivalent to the so-called axiom of choice, asserting that the Cartesian product of a nonempty family of nonempty sets is nonempty. | No |
Proposition 6. Let \( R \) be a ring and \( \mathfrak{a} \subsetneq R \) a proper ideal. Then \( R \) admits a maximal ideal \( \mathfrak{m} \) containing \( \mathfrak{a} \) . In particular, every ring \( R \neq 0 \) admits a maximal ideal. | Proof. Let \( M \) be the set of all proper ideals \( \mathfrak{b} \subsetneq R \) containing \( \mathfrak{a} \) . Then \( M \) is partially ordered under the inclusion of ideals. Furthermore, since \( \mathfrak{a} \in M \), we see that \( M \neq \varnothing \) . We claim that every totally ordered subset \( N \subset ... | Yes |
Corollary 7. Let \( K \) be a field. Then there exists an algebraically closed field \( \bar{K} \) extending \( K \), where \( \bar{K} \) is algebraic over \( K \) ; such a field \( \bar{K} \) is called an algebraic closure of \( K \) . | Proof. If we look a bit closer at the construction of an algebraically closed field \( L \) extending a field \( K \), as exercised in the proof of Theorem 4 above, we can easily realize that \( L \) is algebraic over \( K \) and therefore admits the properties of an algebraic closure of \( K \) . Indeed, by its constr... | Yes |
Lemma 8. Let \( K \) be a field and \( {K}^{\prime } = K\left( \alpha \right) \) a simple algebraic field extension of \( K \) with attached minimal polynomial \( f \in K\left\lbrack X\right\rbrack \) of \( \alpha \) . Furthermore, let \( \sigma : K \rightarrow L \) be a field homomorphism.\n\n(i) If \( {\sigma }^{\pri... | Proof. For every extension \( {\sigma }^{\prime } : {K}^{\prime } \rightarrow L \) of \( \sigma \) we get from \( f\left( \alpha \right) = 0 \) necessarily \( {f}^{\sigma }\left( {{\sigma }^{\prime }\left( \alpha \right) }\right) = {\sigma }^{\prime }\left( {f\left( \alpha \right) }\right) = 0 \) . Moreover, since \( {... | Yes |
Proposition 9. Let \( K \subset {K}^{\prime } \) be an algebraic field extension and \( \sigma : K \rightarrow L \) a field homomorphism with image in an algebraically closed field \( L \) . Then \( \sigma \) admits an extension \( {\sigma }^{\prime } : {K}^{\prime } \rightarrow L \) . In addition, if \( {K}^{\prime } ... | Proof. The main work was already done in Lemma 8, and it remains only to apply Zorn’s lemma. Let \( M \) be the set of all pairs \( \left( {F,\tau }\right) \) consisting of an intermediate field \( F, K \subset F \subset {K}^{\prime } \), as well as an extension \( \tau : F \rightarrow L \) of \( \sigma \) . Then \( M ... | Yes |
Proposition 2. Let \( {L}_{1},{L}_{2} \) be two splitting fields of a family \( \mathfrak{F} \) of nonconstant polynomials in \( K\left\lbrack X\right\rbrack \), for a field \( K \), and let \( {\bar{L}}_{2} \) be an algebraic closure of \( {L}_{2} \) . Then every \( K \) -homomorphism \( \bar{\sigma } : {L}_{1} \right... | In particular, since the inclusion \( K \hookrightarrow {\bar{L}}_{2} \) extends to a \( K \) -homomorphism \( \bar{\sigma } : {L}_{1} \rightarrow {\bar{L}}_{2} \) by \( {3.4}/9 \), it follows that \( {L}_{1} \) is \( K \) -isomorphic to \( {L}_{2} \) . Therefore, we can state the following corollary: | Yes |
Corollary 3. Let \( {L}_{1},{L}_{2} \) be two splitting fields of a family of nonconstant polynomials in \( K\left\lbrack X\right\rbrack \) . Then there exists a \( K \) -isomorphism \( {L}_{1} \rightarrow {L}_{2} \) . | Proof of Proposition 2. First we consider the case in which \( \mathfrak{F} \) consists of a single polynomial \( f \), which we may assume to be monic. Let \( {a}_{1},\ldots ,{a}_{n} \) be the zeros of \( f \) in \( {L}_{1} \) and \( {b}_{1},\ldots ,{b}_{n} \) the zeros of \( f \) in \( {L}_{2} \subset {\bar{L}}_{2} \... | Yes |
Theorem 4. The following conditions are equivalent for a field \( K \) and an algebraic extension \( K \subset L \) :\n\n(i) Every \( K \) -homomorphism \( L \rightarrow \bar{L} \) into an algebraic closure \( \bar{L} \) of \( L \) restricts to an automorphism of \( L \) .\n\n(ii) \( L \) is a splitting field of a fami... | Proof of Theorem 4. We start with the implication from (i) to (iii) and consider an irreducible polynomial \( f \in K\left\lbrack X\right\rbrack \) admitting a zero \( a \in L \) . If \( b \in \bar{L} \) is another zero of \( f \), we can conclude from \( {3.4}/8 \) that there is a \( K \) -homomorphism \( \sigma : K\l... | Yes |
Proposition 7. Let \( L/K \) be an algebraic field extension.\n\n(i) \( L/K \) admits a normal closure \( {L}^{\prime }/K \), where \( {L}^{\prime } \) is unique up to (noncanonical) isomorphism over \( L \) . | Proof. Assume \( L = K\left( \mathfrak{A}\right) \), where \( \mathfrak{A} = {\left( {a}_{j}\right) }_{j \in J} \) is a family of elements in \( L \) . Let \( {f}_{j} \) be the minimal polynomial of \( {a}_{j} \) over \( K \) . If \( M \) is an algebraic extension field of \( L \) such that \( M/K \) is normal (for exa... | Yes |
Proposition 2. Let \( K \) be a field and \( f \in K\left\lbrack X\right\rbrack \) an irreducible polynomial.\n\n(i) If char \( K = 0 \), then \( f \) is separable.\n\n(ii) If \( \operatorname{char}K = p > 0 \), choose \( r \in \mathbb{N} \) maximal such that \( f \) is a polynomial in \( {X}^{{p}^{r}} \), i.e., such t... | Proof. The case char \( K = 0 \) was discussed before, so assume char \( K = p > 0 \) . Furthermore, write\n\n\[ f = \mathop{\sum }\limits_{{i = 0}}^{n}{c}_{i}{X}^{i},\;{f}^{\prime } = \mathop{\sum }\limits_{{i = 1}}^{n}i{c}_{i}{X}^{i - 1}. \]\n\nThen \( {f}^{\prime } = 0 \) is equivalent to \( i{c}_{i} = 0 \) for \( i... | Yes |
Lemma 6. Let \( K \subset L = K\left( \alpha \right) \) be a simple algebraic field extension with minimal polynomial \( f \in K\left\lbrack X\right\rbrack \) of \( \alpha \) over \( K \) .\n\n(i) The separable degree \( {\left\lbrack \bar{L} : K\right\rbrack }_{s} \) equals the number of different zeros of \( f \) in ... | Proof. Assertion (i) is a reformulation of \( {3.4}/8 \) . To justify (ii), let \( n = \deg f \) . Then \( \alpha \) is separable if and only if \( f \) does not admit multiple zeros and hence has \( n \) distinct zeros, or according to (i), if and only if \( n = {\left\lbrack L : K\right\rbrack }_{s} \) . However, due... | Yes |
Proposition 7 (Multiplicativity formula). Let \( K \subset L \subset M \) be algebraic field extensions. Then\n\n\[{\left\lbrack M : K\right\rbrack }_{s} = {\left\lbrack M : L\right\rbrack }_{s} \cdot {\left\lbrack L : K\right\rbrack }_{s}.\] | Proof. Fix an algebraic closure \( \bar{K} \) of \( M \) . Then \( K \subset L \subset M \subset \bar{K} \), and we may view \( \bar{K} \) also as an algebraic closure of \( K \) and of \( L \) . Furthermore, let\n\n\[{\operatorname{Hom}}_{K}\left( {L,\bar{K}}\right) = \left\{ {{\sigma }_{i};i \in I}\right\} ,\;{\opera... | Yes |
Theorem 9. For a finite field extension \( K \subset L \) the following conditions are equivalent:\n\n(i) \( L/K \) is separable.\n\n(ii) There exist elements \( {a}_{1},\ldots ,{a}_{n} \in L \) that are separable over \( K \) and satisfy \( L = K\left( {{a}_{1},\ldots ,{a}_{n}}\right) .\n\n(iii) \( {\left\lbrack L : K... | Proof. The implication from (i) to (ii) is trivial. If \( a \in L \) is separable over \( K \) , then the same is true over every intermediate field of \( L/K \) . Therefore, using the multiplicativity formulas in \( {3.2}/2 \) and in Proposition 7, the implication from (ii) to (iii) can be reduced to the case of a sim... | Yes |
Corollary 10. Let \( K \subset L \) be an algebraic field extension and \( \mathfrak{A} \) a family of elements in \( L \) such that \( L/K \) is generated by \( \mathfrak{A} \). Then the following conditions are equivalent:\n\n(i) \( L/K \) is separable.\n\n(ii) Every \( a \in \mathfrak{A} \) is separable over \( K \)... | Proof. Every \( a \in L \) is contained in a subfield of type \( K\left( {{a}_{1},\ldots ,{a}_{n}}\right) \), where \( {a}_{1},\ldots ,{a}_{n} \in \mathfrak{A} \). In this way, the equivalence between (i) and (ii) is a direct consequence of Theorem 9. Furthermore, for \( L/K \) finite separable, we conclude that \( \le... | Yes |
Corollary 11. Let \( K \subset L \subset M \) be algebraic field extensions. Then \( M/K \) is separable if and only if \( M/L \) and \( L/K \) are separable. | Proof. We have only to show that the separability of \( M/L \) and \( L/K \) implies the separability of \( M/K \) . Fix an element \( a \in M \) with minimal polynomial \( f \in L\left\lbrack X\right\rbrack \) over \( L \) . Furthermore, let \( {L}^{\prime } \) be the intermediate field of \( L/K \) that is generated ... | Yes |
Lemma 13. Let \( a \) and \( b \) be two elements of finite order in an abelian group \( G \) , say \( \operatorname{ord}a = m \) and \( \operatorname{ord}b = n \) . Then there exists an element of order \( \operatorname{lcm}\left( {m, n}\right) \) in \( G \) . | More precisely, choose integer decompositions \( m = {m}_{0}{m}^{\prime }, n = {n}_{0}{n}^{\prime } \), where \( \operatorname{lcm}\left( {m, n}\right) = {m}_{0}{n}_{0} \) and \( \gcd \left( {{m}_{0},{n}_{0}}\right) = 1 \) . Then \( {a}^{{m}^{\prime }}{b}^{{n}^{\prime }} \) is an element of order \( \operatorname{lcm}\... | Yes |
Proposition 14. Let \( K \) be a field and \( H \) a finite subgroup of the multiplicative group \( {K}^{ * } \) . Then \( H \) is cyclic. | Proof. Fix an element \( a \in H \) of maximal order \( m \) and let \( {H}_{m} \) be the subgroup of all elements in \( H \) whose order divides \( m \) . Then all elements of \( {H}_{m} \) are zeros of the polynomial \( {X}^{m} - 1 \), so that \( {H}_{m} \) can contain at most \( m \) elements. On the other hand, \( ... | Yes |
Corollary 3. Let \( K \subset L \subset M \) be algebraic field extensions. Then \( M/K \) is purely inseparable if and only if \( M/L \) and \( L/K \) are purely inseparable. | \[ \text{Proof.}{\left\lbrack M : K\right\rbrack }_{s} = {\left\lbrack M : L\right\rbrack }_{s} \cdot {\left\lbrack L : K\right\rbrack }_{s}\text{; cf.}{3.6}/7\text{.} \] | No |
Proposition 4. Let \( L/K \) be an algebraic field extension. Then there exists a unique intermediate field \( {K}_{s} \) of \( L/K \) such that \( L/{K}_{s} \) is purely inseparable and \( {K}_{s}/K \) is separable. The field \( {K}_{s} \) is called the separable closure of \( K \) in \( L \), i.e., | \[ {K}_{s} = \{ a \in L;a\text{ separable over }K\} ,\] and we have \( {\left\lbrack L : K\right\rbrack }_{s} = \left\lbrack {{K}_{s} : K}\right\rbrack \) . If \( L/K \) is normal, the extension \( {K}_{s}/K \) is normal, too. | Yes |
Proposition 5. Let \( L/K \) be a normal algebraic field extension. Then there exists a unique intermediate field \( {K}_{i} \) of \( L/K \) such that \( L/{K}_{i} \) is separable and \( {K}_{i}/K \) is purely inseparable. | Proof of Proposition 5. Since the extension \( L/K \) is assumed to be normal, we can identify the set of \( K \) -homomorphisms of \( L \) into an algebraic closure \( \bar{L} \) of \( L \) with the set of \( K \) -automorphisms of \( L \), the latter forming a group \( G \) . Let\n\n\[ {K}_{i} = \{ a \in L;\sigma \le... | Yes |
Lemma 1. Let \( \mathbb{F} \) be a finite field. Then \( p = \operatorname{char}\mathbb{F} > 0 \), and \( \mathbb{F} \) contains \( {\mathbb{F}}_{p} \) as its prime subfield. Moreover, \( \mathbb{F} \) consists of precisely \( q = {p}^{n} \) elements, where \( n = \left\lbrack {\mathbb{F} : {\mathbb{F}}_{p}}\right\rbra... | Proof. Since \( \mathbb{F} \) is finite, the same is true for its prime subfield. Hence, the latter is of type \( {\mathbb{F}}_{p} \), where \( p = \operatorname{char}\mathbb{F} > 0 \) . Furthermore, the finiteness of \( \mathbb{F} \) shows that the degree \( n = \left\lbrack {\mathbb{F} : {\mathbb{F}}_{p}}\right\rbrac... | Yes |
Theorem 2. Let \( p \) be a prime number. For every integer \( n \in \mathbb{N} - \{ 0\} \) there exists an extension field \( {\mathbb{F}}_{q}/{\mathbb{F}}_{p} \) consisting of \( q = {p}^{n} \) elements. Furthermore, up to isomorphism, \( {\mathbb{F}}_{q} \) is uniquely characterized as a splitting field of the polyn... | Proof. Write \( f = {X}^{q} - X \) . Since \( {f}^{\prime } = - 1 \), the polynomial \( f \) does not admit multiple zeros and therefore has \( q \) simple zeros in an algebraic closure \( {\overline{\mathbb{F}}}_{p} \) of \( {\mathbb{F}}_{p} \) . If \( a, b \in {\overline{\mathbb{F}}}_{p} \) are two zeros of \( f \), ... | Yes |
Embed the fields \( {\mathbb{F}}_{q} \) for \( q = {p}^{n}, n \in \mathbb{N} - \{ 0\} \), into an algebraic closure \( {\overline{\mathbb{F}}}_{p} \) of \( {\mathbb{F}}_{p} \) . Then an inclusion \( {\mathbb{F}}_{q} \subset {\mathbb{F}}_{{q}^{\prime }} \) holds for \( q = {p}^{n} \) and \( {q}^{\prime } = {p}^{{n}^{\pr... | Proof. Assume \( {\mathbb{F}}_{q} \subset {\mathbb{F}}_{{q}^{\prime }} \) and let \( m = \left\lbrack {{\mathbb{F}}_{{q}^{\prime }} : {\mathbb{F}}_{q}}\right\rbrack \) . Then\n\n\[ \n{p}^{{n}^{\prime }} = \# {\mathbb{F}}_{{q}^{\prime }} = {\left( \# {\mathbb{F}}_{q}\right) }^{m} = {p}^{mn}, \n\] \nand we see that \( n ... | Yes |
Corollary 4. Every algebraic extension of a finite field is normal and separable. In particular, finite fields are perfect. | Proof. Let \( \mathbb{F} \subset K \) be an algebraic field extension, where \( \mathbb{F} \) is finite. If \( K \) is finite as well, say \( K = {\mathbb{F}}_{q} \) for \( q = {p}^{n} \), then \( K \) is a splitting field of the separable polynomial \( {X}^{q} - X \) and therefore is normal and separable over \( {\mat... | No |
Lemma 1. For ideals \( {\mathfrak{a}}_{1},{\mathfrak{a}}_{2} \), resp. a family \( {\left( {\mathfrak{a}}_{i}\right) }_{i \in I} \) of ideals in \( K\left\lbrack X\right\rbrack \), as well as for subsets \( {U}_{1},{U}_{2} \subset {\bar{K}}^{n} \), we have:\n\n(i) \( {\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \Right... | Proof. The assertions (i), (ii), and (iii) are easy to verify; we show only how to obtain (iv). Since\n\n\[ \n{\mathfrak{a}}_{1} \cdot {\mathfrak{a}}_{2} \subset {\mathfrak{a}}_{1} \cap {\mathfrak{a}}_{2} \subset {\mathfrak{a}}_{i},\;i = 1,2, \n\]\n\nwe conclude from (i) that\n\n\[ \nV\left( {{\mathfrak{a}}_{1} \cdot {... | Yes |
Theorem 2 (Hilbert’s basis theorem). Let \( R \) be a Noetherian ring. Then the polynomial ring \( R\left\lbrack Y\right\rbrack \) in a variable \( Y \) is Noetherian as well. In particular, the polynomial ring \( K\left\lbrack X\right\rbrack = K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) in finitely many va... | Proof of Theorem 2. Let \( R \) be a Noetherian ring and \( \mathfrak{a} \subset R\left\lbrack Y\right\rbrack \) an ideal. For \( i \in \mathbb{N} \) define \( {\mathfrak{a}}_{i} \subset R \) as the set of all elements \( a \in R \) such that there exists a polynomial of type\n\n\[ a{Y}^{i} + \text{ terms of lower degr... | Yes |
Lemma 5. Let \( A = K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \neq 0 \) be a ring of finite type over a field \( K \) . Then the inclusion \( K \hookrightarrow \bar{K} \) can be extended to a \( K \) -homomorphism \( A \rightarrow \bar{K} \) . | Proof. For a maximal ideal \( \mathfrak{m} \subset A \), consider the canonical map \( K \rightarrow A/\mathfrak{m} \) . Since \( A/\mathfrak{m} \) is a field that is of finite type over \( K \) in the ring-theoretic sense, we conclude from \( {3.3}/8 \) that \( A/\mathfrak{m} \) is finite over \( K \) . Then \( {3.4}/... | Yes |
Corollary 6. Let \( K \) be an algebraically closed field. An ideal \( \mathfrak{m} \) of the polynomial ring \( K\left\lbrack X\right\rbrack = K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is maximal if and only if there exists a point \( x = \left( {{x}_{1},\ldots ,{X}_{n}}\right) \in {K}^{n} \) such that \... | Proof. First note that \( \left( {{X}_{1},\ldots ,{X}_{n}}\right) \subset K\left\lbrack X\right\rbrack \) is a maximal ideal, since the residue class ring \( K\left\lbrack X\right\rbrack /\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) is isomorphic to \( K \) . In the same way, we see that ideals of type \( \left( {{X}_{1}... | Yes |
Proposition 4. Let \( L \) be a field and \( G \) a subgroup of \( \operatorname{Aut}\left( L\right) \), the group of automorphisms of L. Furthermore, consider\n\n\[ K = {L}^{G} = \{ a \in L;\sigma \left( a\right) = a\text{ for all }\sigma \in G\} ,\]\n\n the fixed field attached to \( G \) .\n\n(i) If \( G \) is finit... | Proof. First, it is easily checked that \( K = {L}^{G} \) is indeed a subfield of \( L \) . Now assume that \( G \) is finite, or if not, that \( L/K \) is algebraic. To see that \( L/K \) is separable algebraic, consider an element \( a \in L \), as well as a maximal system of elements \( {\sigma }_{1},\ldots ,{\sigma... | Yes |
Corollary 5. Let \( L/K \) be a normal algebraic field extension with automorphism group \( G = {\operatorname{Aut}}_{K}\left( L\right) \) . Then:\n\n(i) \( L/{L}^{G} \) is a Galois extension with Galois group \( G \) .\n\n(ii) If \( L/K \) is separable and therefore Galois, then \( {L}^{G} = K \) .\n\n(iii) Assume cha... | Proof. We know by Proposition 4 that \( L/{L}^{G} \) is a Galois extension. The corresponding Galois group coincides with \( G \) in this case, since \( {\operatorname{Aut}}_{{L}^{G}}\left( L\right) = {\operatorname{Aut}}_{K}\left( L\right) \) . Furthermore, the definition of \( {L}^{G} \) shows that \( {\left\lbrack {... | Yes |
Theorem 6 (Fundamental theorem of Galois theory). Let \( L/K \) be a finite Galois extension with Galois group \( G = \operatorname{Gal}\left( {L/K}\right) \) . Then the maps\n\n\[ \n\{ \text{ subgroups of }G\} \xrightarrow[\Psi ]{\Phi }\{ \text{ intermediate fields of }L/K\}\n\]\n\n\[ \nH \mapsto {L}^{H}\n\]\n\n\[ \n\... | Proof of Theorem 6 and Remark 7. Let \( L/K \) be a Galois extension that is not necessarily finite. If \( E \) is an intermediate field of \( L/K \), then \( L/E \) is Galois and the Galois group \( H = \operatorname{Gal}\left( {L/E}\right) \) is a subgroup of \( G = \operatorname{Gal}\left( {L/K}\right) \) ; cf. Rema... | Yes |
Corollary 8. Every finite separable field extension \( L/K \) admits only finitely many intermediate fields. | Proof. Passing to a normal closure of \( L/K \), see \( {3.5}/7 \), we may assume that \( L/K \) is finite and Galois. Then the intermediate fields of \( L/K \) correspond bijectively to the subgroups of the finite group \( \operatorname{Gal}\left( {L/K}\right) \) . | No |
Corollary 9. Let \( L/K \) be a finite Galois extension. For intermediate fields \( E \) and \( {E}^{\prime } \) of \( L/K \), consider \( H = \operatorname{Gal}\left( {L/E}\right) \) and \( {H}^{\prime } = \operatorname{Gal}\left( {L/{E}^{\prime }}\right) \) as subgroups of \( G = \operatorname{Gal}\left( {L/K}\right)... | Proof. (i) If \( E \subset {E}^{\prime } \), then every \( {E}^{\prime } \) -automorphism of \( L \) is an \( E \) -automorphism as well, i.e., we have \( H = \operatorname{Gal}\left( {L/E}\right) \supset \operatorname{Gal}\left( {L/{E}^{\prime }}\right) = {H}^{\prime } \) . On the other hand, we see that \( H \supset ... | Yes |
Corollary 11. Let \( L/K \) be a finite abelian (resp. cyclic) Galois extension. Then, for every intermediate field \( E \) of \( L/K \), the extension \( E/K \) is a finite abelian (resp. cyclic) Galois extension. | Proof. In each case, \( \operatorname{Gal}\left( {L/E}\right) \) is a normal subgroup in \( \operatorname{Gal}\left( {L/K}\right) \), since cyclic groups are abelian. It follows that the extension \( E/K \) is Galois. Furthermore, the Galois group \( \operatorname{Gal}\left( {E/K}\right) = \operatorname{Gal}\left( {L/K... | Yes |
Proposition 12. Let \( L/K \) be a field extension together with intermediate fields \( E \) and \( {E}^{\prime } \) such that \( E/K \) and \( {E}^{\prime }/K \) are finite Galois extensions. Then:\n\n(i) \( E \cdot {E}^{\prime } \) is finite and Galois over \( K \), and the homomorphism\n\n\[ \varphi : \operatorname{... | Proof. We start with assertion (i). First, using the fact that \( E \cdot {E}^{\prime } = K\left( {E,{E}^{\prime }}\right) \) , we see that \( E \cdot {E}^{\prime } \) is normal, separable, and finite over \( K \), since \( E/K \) and \( {E}^{\prime }/K \) admit these properties. To show that \( \varphi \) is injective... | Yes |
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