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Proposition 4.6. Let \( R \) be a commutative ring, and let \( f\left( x\right) \in R\left\lbrack x\right\rbrack \) be a monic polynomial of degree \( d \) . Then the function\n\n\[ \n\varphi : R\left\lbrack x\right\rbrack \rightarrow {R}^{\oplus d} \]\n\n\ndefined by sending \( g\left( x\right) \in R\left\lbrack x\rig... | Proof. The given function \( \varphi \) is well-defined by Lemma 4.5, and it is surjective since it has a right inverse (that is, the function \( \psi : {R}^{\oplus d} \rightarrow R\left\lbrack x\right\rbrack \) defined above).\n\nI claim that \( \varphi \) is a homomorphism of abelian groups. Indeed, if\n\n\[ \n{g}_{1... | Yes |
Assume \( f\left( x\right) \) is monic of degree 1: \( f\left( x\right) = x - a \) for some \( a \in R \) . Then the remainder of \( g\left( x\right) \) after division by \( f\left( x\right) \) is simply the ’evaluation’ \( g\left( a\right) \) (cf. Example 2.3). | Indeed, \[ g\left( x\right) = \left( {x - a}\right) q\left( x\right) + r \] for some \( r \in R \) (the remainder must have degree \( < 1 \) ; hence it is a constant); evaluating at \( a \) gives \[ g\left( a\right) = \left( {a - a}\right) q\left( a\right) + r = 0 \cdot q\left( a\right) + r = r \] as claimed. In partic... | Yes |
For a concrete example, apply this procedure with \( f\left( x\right) = {x}^{2} + 1 \) : Proposition 4.6 gives an isomorphism of groups\n\n\[ R \oplus R \cong \frac{R\left\lbrack x\right\rbrack }{\left( {x}^{2} + 1\right) } \]\n\nwhat multiplication does this isomorphism induce on \( R \oplus R \) ? | Take two elements \( \left( {{a}_{0},{a}_{1}}\right) ,\left( {{b}_{0},{b}_{1}}\right) \) of \( R \oplus R \) . With the notation used in Proposition 4.6, we have\n\n\[ \left( {{a}_{0},{a}_{1}}\right) = \varphi \left( {{a}_{0} + {a}_{1}x}\right) ,\;\left( {{b}_{0},{b}_{1}}\right) = \varphi \left( {{b}_{0} + {b}_{1}x}\ri... | Yes |
For all \( a \in R \), the ideal \( \left( {x - a}\right) \) is prime in \( R\left\lbrack x\right\rbrack \) if and only if \( R \) is an integral domain; it is maximal if and only if \( R \) is a field. | Indeed, \( R\left\lbrack x\right\rbrack /\left( {x - a}\right) \cong R \), as we have seen in Example 4.7 | No |
Proposition 4.11. Let \( I \neq \left( 1\right) \) be an ideal of a commutative ring \( R \) . Then\n\n- I is prime if and only if for all \( a, b \in R \)\n\n\[{ab} \in I \Rightarrow \left( {a \in I\text{ or }b \in I}\right) ;\]\n\n- I is maximal if and only if for all ideals \( J \) of \( R \)\n\n\[I \subseteq J \Rig... | Proof. The ring \( R/I \) is an integral domain if and only if \( \forall \bar{a},\bar{b} \in R/I \)\n\n\[ \bar{a} \cdot \bar{b} = 0 \Rightarrow \left( {\bar{a} = 0\text{ or }\bar{b} = 0}\right) .\]\n\nThis condition translates immediately to the given condition in \( R \), with \( \bar{a} = a + I \) , \( \bar{b} = b +... | No |
Proposition 4.12. Let \( I \) be an ideal of a commutative ring \( R \) . If \( R/I \) is finite, then \( I \) is prime if and only if it is maximal. | Proof. This follows immediately from Proposition 1.15 | No |
Proposition 4.13. Let \( R \) be a PID, and let \( I \) be a nonzero ideal in \( R \). Then \( I \) is prime if and only if it is maximal. | Proof. Maximal ideals are prime in every ring, so we only need to verify that nonzero prime ideals are maximal in a PID; we will use the characterization of prime and maximal ideals obtained in Proposition 4.11 Let \( I = \left( a\right) \) be a prime ideal in \( R \), with \( a \neq 0 \), and assume \( I \subseteq J \... | Yes |
Proposition 5.3. Every abelian group is a \( \mathbb{Z} \) -module, in exactly one way. | Proof. Let \( G \) be an abelian group. A \( \mathbb{Z} \) -module structure on \( G \) is a ring homomorphism\n\n\[ \mathbb{Z} \rightarrow {\operatorname{End}}_{\mathrm{{Ab}}}\left( G\right) \]\n\nSince \( \mathbb{Z} \) is initial in Ring (§2.1), there exists exactly one such homomorphism, proving the statement.\n\nTh... | Yes |
The category \( \mathbb{Z} \) -Mod of \( \mathbb{Z} \) -modules is ’the same as’ the category Ab | indeed, every abelian group is a \( \mathbb{Z} \) -module in exactly one way (Proposition 5.3), and \( \mathbb{Z} \) -module homomorphisms are simply homomorphisms of abelian groups. | Yes |
Any homomorphism of rings \( \alpha : R \rightarrow S \) may be used to define an interesting \( R \) -module: define \( \rho : R \times S \rightarrow S \) by\n\n\[ \rho \left( {r, s}\right) \mathrel{\text{:=}} \alpha \left( r\right) s \] \n\nfor all \( r \in R \) and \( s \in S \). | The operation on the right is simply multiplication in \( S \) , and the axioms of Definition 5.2 are immediate consequence of the ring axioms and of the fact that \( \alpha \) is a homomorphism. For instance, taking \( S = R \) and \( \alpha = {\operatorname{id}}_{R} \) makes \( R \) a (left-) module over itself. | Yes |
Theorem 5.14. Let \( N \) be a submodule of an \( R \) -module \( M \) . Then for every homomorphism of \( R \) -modules \( \varphi : M \rightarrow P \) such that \( N \subseteq \ker \varphi \) there exists a unique homomorphism of \( R \) -modules \( \widetilde{\varphi } : M/N \rightarrow P \) so that the diagram\n\n!... | As in previous appearances of such statements, this is an immediate consequence of the set-theoretic version ( 115.3) and of easy notation matching and compatibility checks. For an even faster proof, one can just apply Theorem 117.12 and verify that \( \widetilde{\varphi } \) is an \( R \) -module homomorphism. | No |
Proposition 6.1. The direct sum \( M \oplus N \) satisfies the universal properties of both the product and the coproduct of \( M \) and \( N \) . | Proof. Product: Let \( P \) be an \( R \) -module, and let \( {\varphi }_{M} : P \rightarrow M,{\varphi }_{N} : P \rightarrow N \) be two \( R \) -module homomorphisms. The definition of an \( R \) -module homomorphism\n\n\[ \n{\varphi }_{M} \times {\varphi }_{N} : P \rightarrow M \oplus N \n\]\n\nis forced by the need... | No |
Proposition 6.2. The following hold in \( R \) -Mod:\n\n- kernels and cokernels exist;\n\n- \( \varphi \) is a monomorphism \( \Leftrightarrow \ker \varphi \) is trivial \( \Leftrightarrow \varphi \) is injective as a set-function;\n\n- \( \varphi \) is an epimorphism \( \Leftrightarrow \operatorname{coker}\varphi \) i... | This proposition of course simply generalizes to \( R \) -Mod facts we know already from our study of \( \mathrm{{Ab}} \), and a quick review should suffice for the careful reader. Kernels exist: indeed, the 'standard' definition of kernel satisfies the universal properties spelled out above (same argument as in Propos... | No |
Proposition 6.4. \( R\left\lbrack A\right\rbrack \) is a free commutative \( R \) -algebra on the set \( A \) . | Proof. The statement translates into the following: for every commutative \( R \) - algebra \( S \) and every set-function \( f : A \rightarrow S \), there exists a unique \( R \) -algebra homomorphism \( \varphi : R\left\lbrack A\right\rbrack \rightarrow S \) such that the diagram\n\n![23387543-548b-40c2-8595-20075621... | No |
Proposition 6.7. Let \( M \) be an \( R \) -module, and let \( N \) be a submodule of \( M \) . Then \( M \) is Noetherian if and only if both \( N \) and \( M/N \) are Noetherian. | Proof. If \( M \) is Noetherian, then so is \( M/N \) (same proof as for Exercise 4.2), and so is \( N \) (because every submodule of \( N \) is a submodule of \( M \), so it is finitely generated because \( M \) is Noetherian). This proves the ’only if’ part of the statement.\n\nFor the converse, assume \( N \) and \(... | No |
Corollary 6.8. Let \( R \) be a Noetherian ring, and let \( M \) be a finitely generated \( R \) -module. Then \( M \) is Noetherian (as an \( R \) -module). | Proof. Indeed, by hypothesis there is an onto homomorphism \( {R}^{\oplus n} \rightarrow M \) of \( R \) - modules; hence (by the first isomorphism theorem, Corollary 5.16) \( M \) is isomorphic to a quotient of \( {R}^{\oplus n} \) . By Proposition 6.7, it suffices to prove that \( {R}^{\oplus n} \) is Noetherian.\n\n... | No |
A complex \[ \cdots \rightarrow 0 \rightarrow L\xrightarrow[]{\alpha }M \rightarrow \cdots \] is exact at \( L \) if and only if \( \alpha \) is a monomorphism. | Indeed, exactness at \( L \) is equivalent to \( \ker \alpha = \) image of the trivial homomorphism \( 0 \rightarrow L \), that is, to \[ \ker \alpha = 0. \] This is equivalent to the injectivity of \( \alpha \) (Proposition 6.2). | Yes |
A complex \[ \cdots \rightarrow M\overset{\beta }{ \rightarrow }N \rightarrow 0 \rightarrow \cdots \] is exact at \( N \) if and only if \( \beta \) is an epimorphism. | Indeed, the complex is exact at \( N \) if and only if \( \operatorname{im}\beta = \) kernel of the trivial homomorphism \( N \rightarrow 0 \), that is, \( \operatorname{im}\beta = N \). | Yes |
Let \( \varphi : M \rightarrow N \) be an \( R \) -module homomorphism. Then\n\n- \( \varphi \) has a left-inverse if and only if the sequence\n\n\[ 0 \rightarrow M\overset{\varphi }{ \rightarrow }N \rightarrow \operatorname{coker}\varphi \rightarrow 0 \]\n\nsplits. | If the sequence splits, then \( \varphi \) may be identified with the embedding of \( M \) into a direct sum \( M \oplus {M}^{\prime } \), and the projection \( M \oplus {M}^{\prime } \rightarrow M \) gives a left-inverse of \( \varphi \) . Conversely, assume that \( \varphi \) has a left-inverse \( \psi \) :\n\nThen I... | Yes |
Lemma 7.8 (The snake lemma). With notation as above, there is an exact sequence\n\n\[ 0 \rightarrow \ker \lambda \rightarrow \ker \mu \rightarrow \ker \nu \overset{\delta }{ \rightarrow }\operatorname{coker}\lambda \rightarrow \operatorname{coker}\mu \rightarrow \operatorname{coker}\nu \rightarrow 0. \] | Proving the snake lemma is something that should not be done in public, and it is notoriously useless to write down the details of the verification for others to read: the details are all essentially obvious, but they lead quickly to a notational quagmire. Such proofs are collectively known as the sport of diagram chas... | No |
Corollary 7.12. In the same situation presented in the snake lemma (notation as in [7.3]), assume that \( \mu \) is surjective and \( \nu \) is injective. Then \( \lambda \) is surjective and \( \nu \) is an isomorphism. | Proof. Indeed, \( \mu \) surjective \( \Rightarrow \operatorname{coker}\mu = 0;\nu \) injective \( \Rightarrow \ker \nu = 0 \) (Proposition 6.2). Feeding this information into the sequence of the snake lemma gives an exact sequence\n\n\[ 0 \rightarrow \ker \lambda \rightarrow \ker \mu \rightarrow 0 \rightarrow \operato... | No |
Proposition 1.1. Let \( S \) be a finite set, and let \( G \) be a group acting on \( S \). With notation as above,\n\n\[ \left| S\right| = \left| Z\right| + \mathop{\sum }\limits_{{a \in A}}\left\lbrack {G : {G}_{a}}\right\rbrack \]\n\nwhere \( A \subseteq S \) has exactly one element for each nontrivial orbit of the ... | Proof. The orbits form a partition of \( S \), and \( Z \) collects the trivial orbits; hence\n\n\[ \left| S\right| = \left| Z\right| + \mathop{\sum }\limits_{{a \in A}}\left| {O}_{a}\right| \]\n\nwhere \( {O}_{a} \) denotes the orbit of \( a \). By Proposition 1119.9, the order \( \left| {O}_{a}\right| \) equals the i... | Yes |
Corollary 1.3. Let \( G \) be a p-group acting on a finite set \( S \), and let \( Z \) be the fixed point set of the action. Then \[ \left| Z\right| \equiv \left| S\right| \;{\;\operatorname{mod}\;p}. \] | Proof. Indeed, each summand \( \left\lbrack {G : {G}_{a}}\right\rbrack \) in Proposition 1.1 is a number larger than 1, and a power of \( p \) ; hence it is 0 mod \( p \) . | Yes |
Lemma 1.5. Let \( G \) be a finite group, and assume \( G/Z\left( G\right) \) is cyclic. Then \( G \) is commutative (and hence \( G/Z\left( G\right) \) is in fact trivial). | Proof. (Cf. Exercise 1.5) As \( G/Z\left( G\right) \) is cyclic, there exists an element \( g \in G \) such that the class \( {gZ}\left( G\right) \) generates \( G/Z\left( G\right) \) . Then \( \forall a \in G \)\n\n\[ \n{aZ}\left( G\right) = {\left( gZ\left( G\right) \right) }^{r} \n\]\n\nfor some \( r \in \mathbb{Z} ... | No |
Proposition 1.8 (Class formula). Let \( G \) be a finite group. Then\n\n\[ \left| G\right| = \left| {Z\left( G\right) }\right| + \mathop{\sum }\limits_{{a \in A}}\left\lbrack {G : Z\left( a\right) }\right\rbrack \]\n\nwhere \( A \subseteq G \) is a set containing one representative for each nontrivial conjugacy class i... | Proof. The set of fixed points is \( Z\left( G\right) \), and the stabilizer of \( a \) is the centralizer \( Z\left( a\right) \) ; apply Proposition 1.1 | No |
Corollary 1.9. Let \( G \) be a nontrivial p-group. Then \( G \) has a nontrivial center. | Proof. Since \( \left| {Z\left( G\right) }\right| \equiv \left| G\right| {\;\operatorname{mod}\;p} \) and \( \left| G\right| > 1 \) is a power of \( p \), necessarily \( \left| {Z\left( G\right) }\right| \) is a multiple of \( p \) . As \( Z\left( G\right) \neq \varnothing \) (since \( {e}_{G} \in Z\left( G\right) \) )... | Yes |
Example 1.10. Consider a group \( G \) of order 6 ; what are the possibilities for its class formula? | If \( G \) is commutative, then the class formula will tell us very little:\n\n\[6 = 6\text{.}\]\n\nIf \( G \) is not commutative, then its center must be trivial (as a consequence of Lagrange’s theorem and Lemma 1.5); so the class formula is \( 6 = 1 + \cdots \), where \( \cdots \) collects the sizes of the nontrivial... | No |
Lemma 1.13. Let \( H \subseteq G \) be a subgroup. Then (if finite) the number of subgroups conjugate to \( H \) equals the index \( \left\lbrack {G : {N}_{G}\left( H\right) }\right\rbrack \) of the normalizer of \( H \) in \( G \) . | Proof. This is again an immediate consequence of Proposition 119.9 | No |
Corollary 1.14. If \( \left\lbrack {G : H}\right\rbrack \) is finite, then the number of subgroups conjugate to \( H \) is finite and divides \( \left\lbrack {G : H}\right\rbrack \) . | Proof.\n\n\[ \left\lbrack {G : H}\right\rbrack = \left\lbrack {G : {N}_{G}\left( H\right) }\right\rbrack \cdot \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack \]\n\n(cf. \( §\overline{118.5} \) ). | Yes |
Theorem 2.1 (Cauchy’s theorem). Let \( G \) be a finite group, and let \( p \) be a prime divisor of \( \left| G\right| \) . Then \( G \) contains an element of order \( p \) . | Proof of Theorem 2.1. Consider the set \( S \) of ordered \( p \) -tuples of elements of \( G \) :\n\n\[ \n\left( {{a}_{1},\ldots ,{a}_{p}}\right) \n\]\n\nsuch that \( {a}_{1}\cdots {a}_{p} = e \) . I claim that \( \left| S\right| = {\left| G\right| }^{p - 1} \) : indeed, once \( {a}_{1},\ldots ,{a}_{p - 1} \) are chos... | Yes |
Example 2.4. Let \( p \) be a positive prime integer. If \( \left| G\right| = {mp} \), with \( 1 < m < p \) , then \( G \) is not simple. | Indeed, consider the subgroups of \( G \) with \( p \) elements. By Claim 2.2, the number of such subgroups is \( \equiv 1{\;\operatorname{mod}\;p} \) . Thus, if there is more than one such subgroup, then there must be at least \( p + 1 \) . Any two distinct subgroups of prime order can only meet at the identity (why?)... | Yes |
Proposition 2.6. If \( {p}^{k} \) divides the order of \( G \), then \( G \) has a subgroup of order \( {p}^{k} \) . | Proof of Proposition 2.6. If \( k = 0 \), there is nothing to prove, so we may assume \( k \geq 1 \) and in particular that \( \left| G\right| \) is a multiple of \( p \) .\n\nArgue by induction on \( \left| G\right| \) : if \( \left| G\right| = p \), again there is nothing to prove; if \( \left| G\right| > p \) and \(... | Yes |
Theorem 2.8 (Second Sylow theorem). Let \( G \) be a finite group, let \( P \) be a p-Sylow subgroup, and let \( H \subseteq G \) be a p-group. Then \( H \) is contained in a conjugate of \( P \) : there exists \( g \in G \) such that \( H \subseteq {gP}{g}^{-1} \) . | Proof. Act with \( H \) on the set of left-cosets of \( P \), by left-multiplication. Since there are \( \left\lbrack {G : P}\right\rbrack \) cosets and \( p \) does not divide \( \left\lbrack {G : P}\right\rbrack \), we know this action must have fixed points (Exercise 1.1): let \( {gP} \) be one of them. This means t... | No |
Lemma 2.9. Let \( H \) be a p-group contained in a finite group \( G \) . Then\n\n\[ \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack \equiv \left\lbrack {G : H}\right\rbrack \;{\;\operatorname{mod}\;p}. \] | Proof. If \( H \) is trivial, then \( {N}_{G}\left( H\right) = G \) and the two numbers are equal.\n\nAssume then that \( H \) is nontrivial, and act with \( H \) on the set of left-cosets of \( H \) in \( G \), by left-multiplication. The fixed points of this action are the cosets \( {gH} \) such that \( \forall h \in... | Yes |
Proposition 2.10. Let \( H \) be a p-subgroup of a finite group \( G \), and assume that \( H \) is not a p-Sylow subgroup. Then there exists a p-subgroup \( {H}^{\prime } \) of \( G \) containing \( H \) , such that \( \left\lbrack {{H}^{\prime } : H}\right\rbrack = p \) and \( H \) is normal in \( {H}^{\prime } \) . | Proof. Since \( H \) is not a \( p \) -Sylow subgroup of \( G, p \) divides \( \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack \), by Lemma 2.9. Since \( H \) is normal in \( {N}_{G}\left( H\right) \), we may consider the quotient group \( {N}_{G}\left( H\right) /H \), and \( p \) divides the order of this group... | Yes |
Theorem 2.11 (Third Sylow theorem). Let \( p \) be a prime integer, and let \( G \) be a finite group of order \( \left| G\right| = {p}^{r}m \) . Assume that \( p \) does not divide \( m \) . Then the number of p-Sylow subgroups of \( G \) divides \( m \) and is congruent to 1 modulo \( p \) . | Proof. Let \( {N}_{p} \) denote the number of \( p \) -Sylow subgroups of \( G \) .\n\nBy Theorem 2.8, the \( p \) -Sylow subgroups of \( G \) are the conjugates of any given \( p \) -Sylow subgroup \( P \) . By Lemma 1.13, \( {N}_{p} \) is the index of the normalizer \( {N}_{G}\left( P\right) \) of \( P \) ; thus (Cor... | Yes |
Example 2.13. There are no simple groups of order 2002. | Indeed9,\n\n\[ \n{2002} = 2 \cdot 7 \cdot {11} \cdot {13} \n\]\n\nthe divisors of \( 2 \cdot 7 \cdot {13} \) are\n\n\[ \n1,2,7,{13},{14},{26},{91},{182} :\n\]\n\nof these, only 1 is congruent to \( 1{\;\operatorname{mod}\;{11}} \) . Thus there is a normal subgroup of order 11 in every group of order 2002. | Yes |
Example 2.14. There are no simple groups of order 12. | Note that \( 3 \equiv 1{\;\operatorname{mod}\;2} \) and \( 4 \equiv 1{\;\operatorname{mod}\;3} \) : thus the argument used above does not guarantee the existence of either a normal 2-Sylow subgroup or a normal 3-Sylow subgroup.\n\nHowever, suppose that there is more than one 3-Sylow subgroup. Then there must be 4 , by ... | Yes |
Example 2.15. There are no simple groups of order 24. | Indeed, let \( G \) be a group of order 24, and consider its 2-Sylow subgroups; by the third Sylow theorem, there are either 1 or 3 such subgroups. If there is 1 , the 2-Sylow subgroup is normal and \( G \) is not simple. Otherwise, \( G \) acts (nontrivially) by conjugation on this set of three 2-Sylow subgroups; this... | Yes |
Theorem 3.2 (Jordan-Hölder). Let \( G \) be a group, and let\n\n\[ G = {G}_{0} \supsetneq {G}_{1} \supsetneq {G}_{2} \supsetneq \cdots \supsetneq {G}_{n} = \{ e\} ,\]\n\n\[ G = {G}_{0}^{\prime } \supsetneq {G}_{1}^{\prime } \supsetneq {G}_{2}^{\prime } \supsetneq \cdots \supsetneq {G}_{m}^{\prime } = \{ e\}\]\n\nbe two... | Proof. Let\n\n\( \left( *\right) \)\n\n\[ G = {G}_{0} \supsetneq {G}_{1} \supsetneq {G}_{2} \supsetneq \cdots \supsetneq {G}_{n} = \{ e\} \]\n\nbe a composition series. Argue by induction on \( n \) : if \( n = 0 \), then \( G \) is trivial, and there is nothing to prove. Assume \( n > 0 \), and let\n\n\( \left( {* * }... | Yes |
Example 3.3. Let \( G = \mathbb{Z}/6\mathbb{Z} = \{ \left\lbrack 0\right\rbrack ,\left\lbrack 1\right\rbrack ,\left\lbrack 2\right\rbrack ,\left\lbrack 3\right\rbrack ,\left\lbrack 4\right\rbrack ,\left\lbrack 5\right\rbrack \} \) . Then \[ \{ \left\lbrack 0\right\rbrack ,\left\lbrack 1\right\rbrack ,\left\lbrack 2\rig... | The (normal) subgroup \( N = \{ \left\lbrack 0\right\rbrack ,\left\lbrack 2\right\rbrack ,\left\lbrack 4\right\rbrack \} \) ’turns off’ the second factor: indeed, intersecting the series with \( N \) gives \[ \{ \left\lbrack 0\right\rbrack ,\left\lbrack 2\right\rbrack ,\left\lbrack 4\right\rbrack \} \supsetneq \{ \left... | Yes |
Proposition 3.4. Let \( G \) be a group, and let \( N \) be a normal subgroup of \( G \) . Then \( G \) has a composition series if and only if both \( N \) and \( G/N \) have composition series. Further, if this is the case, then\n\n\[ \ell \left( G\right) = \ell \left( N\right) + \ell \left( {G/N}\right) \]\n\nand th... | Proof. If \( G/N \) has a composition series, the subgroups appearing in it correspond to subgroups of \( G \) containing \( N \), with isomorphic quotients, by Proposition 118.10 (the \ | No |
Proposition 3.5. Any two normal series of a finite group ending with \( \{ e\} \) admit equivalent refinements. | Proof. Refine the series to a composition series; then apply the Jordan-Hölder theorem. | No |
Proposition 3.8. Let \( {G}^{\prime } \) be the commutator subgroup of \( G \) . Then\n\n- \( {G}^{\prime } \) is normal in \( G \) ;\n\n- the quotient \( G/{G}^{\prime } \) is commutative;\n\n- if \( \alpha : G \rightarrow A \) is a homomorphism of \( G \) to a commutative group, then \( {G}^{\prime } \subseteq \) \( ... | Proof. These are all easy consequences of Lemma 3.7\n\n-By Lemma 3.7, the commutator subgroup is characteristic, hence normal (cf. Exercise 2.2).\n\n-By Lemma 3.7, the commutator of any two cosets \( g{G}^{\prime }, h{G}^{\prime } \) is the coset of the commutator \( \left\lbrack {g, h}\right\rbrack \) ; hence it is th... | Yes |
Proposition 3.11. For a finite group \( G \), the following are equivalent:\n\n(i) All composition factors of \( G \) are cyclic.\n\n(ii) \( G \) admits a cyclic series ending in \( \{ e\} \) .\n\n(iii) \( G \) admits an abelian series ending in \( \{ e\} \) .\n\n(iv) \( G \) is solvable. | Proof. (i) \( \Rightarrow \) (ii) \( \Rightarrow \) (iii) are trivial. (iii) \( \Rightarrow \) (i) is obtained by refining an abelian series to a composition series (keeping in mind that the simple abelian groups are cyclic \( p \) -groups).\n\n(iv) \( \Rightarrow \) (iii) is also trivial, since the derived series is a... | Yes |
All \( p \) -groups are solvable. | Indeed, the composition factors of a \( p \) -group are simple \( p \) -groups (what else could they be?), hence cyclic. | No |
Corollary 3.13. Let \( N \) be a normal subgroup of a group \( G \) . Then \( G \) is solvable if and only if both \( N \) and \( G/N \) are solvable. | Proof. This follows immediately from Proposition 3.4 and the formulation of solvability in terms of composition factors given in Proposition 3.11 | Yes |
Lemma 4.3. Every \( \sigma \in {S}_{n},\sigma \neq e \), can be written as a product of disjoint nontrivial cycles, in a unique way up to permutations of the factors. | Proof. As we have seen, every \( \sigma \in {S}_{n} \) determines a partition of \( \{ \mathbf{1},\ldots ,\mathbf{n}\} \) into orbits under the action of \( \langle \sigma \rangle \) . If \( \sigma \neq e \), then \( \langle \sigma \rangle \) has nontrivial orbits. As \( \sigma \) acts as a cycle on each orbit, it foll... | No |
Lemma 4.5. Let \( \tau \in {S}_{n} \), and let \( \left( {{a}_{1}\ldots {a}_{r}}\right) \) be a cycle. Then\n\n\[ \tau \left( {{a}_{1}\ldots {a}_{r}}\right) {\tau }^{-1} = \left( {{a}_{1}{\tau }^{-1}\ldots {a}_{r}{\tau }^{-1}}\right) . \] | Proof. This is verified by checking that both sides act in the same way on \( \{ \mathbf{1},\ldots ,\mathbf{n}\} \) . For example, for \( 1 \leq i < r \)\n\n\[ \left( {{a}_{i}{\tau }^{-1}}\right) \left( {\tau \left( {{a}_{1}\ldots {a}_{r}}\right) {\tau }^{-1}}\right) = {a}_{i}\left( {{a}_{1}\ldots {a}_{r}}\right) {\tau... | No |
Proposition 4.6. Two elements of \( {S}_{n} \) are conjugate in \( {S}_{n} \) if and only if they have the same type. | Proof. The 'only if' part of this statement follows immediately from the preceding considerations: conjugating a permutation yields a permutation of the same type. As for the 'if' part, suppose\n\n\[ \n{\sigma }_{1} = \left( {{a}_{1}\ldots {a}_{r}}\right) \left( {{b}_{1}\ldots {b}_{s}}\right) \cdots \left( {{c}_{1}\ldo... | Yes |
In \( {S}_{8} \), \(\left( {18632}\right) \left( {47}\right) \text{ and }\left( {12345}\right) \left( {67}\right)\) must be conjugate, since they have the same type. | The proof of Proposition 4.6 tells us that \(\tau \left( {18632}\right) \left( {47}\right) {\tau }^{-1} = \left( {12345}\right) \left( {67}\right)\) for \(\tau = \left( \begin{array}{llllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 1 & 8 & 6 & 3 & 2 & 4 & 7 & 5 \end{array}\right)\) and of course this may be checked by hand i... | Yes |
There are no normal subgroups of size 30 in \( {S}_{5} \) . | Indeed, normal subgroups are unions of conjugacy classes (4.3); since the identity is in every subgroup and \( {30} - 1 = {29} \) cannot be written as a sum of the numbers appearing in the class formula for \( {S}_{5} \), there is no such subgroup. | Yes |
Lemma 4.11. Transpositions generate \( {S}_{n} \) . | Proof. Indeed, by Lemma 4.3 it suffices to show that every cycle is a product of transpositions, and indeed\n\n\[ \left( {{a}_{1}\ldots {a}_{r}}\right) = \left( {{a}_{1}{a}_{2}}\right) \left( {{a}_{1}{a}_{3}}\right) \cdots \left( {{a}_{1}{a}_{r}}\right) ,\]\n\nas may be checked by applying \( {}^{18} \) both sides to e... | No |
Lemma 4.12. Let \( \sigma = {\tau }_{1}\cdots {\tau }_{r} \) be a product of transpositions. Then \( \sigma \) is even, resp., odd, according to whether \( r \) is even, resp., odd. | Proof. This follows immediately from the facts that \( \epsilon \) is a homomorphism and the sign of a transposition is -1 : indeed, \( \left( {ij}\right) \) acts on \( {\Delta }_{n} \) by permuting its factors and changing the sign of an odd number of factors (for \( i < j \), the factor \( \left( {{x}_{i} - {x}_{j}}\... | Yes |
Lemma 4.14. Let \( n \geq 2 \), and let \( \sigma \in {A}_{n} \) . Then \( {\left\lbrack \sigma \right\rbrack }_{{A}_{n}} = {\left\lbrack \sigma \right\rbrack }_{{S}_{n}} \) or the size of \( {\left\lbrack \sigma \right\rbrack }_{{A}_{n}} \) is half the size of \( {\left\lbrack \sigma \right\rbrack }_{{S}_{n}} \), acco... | Proof. (Cf. Exercise 1.16.) Note that\n\n\[ \n{Z}_{{A}_{n}}\left( \sigma \right) = {A}_{n} \cap {Z}_{{S}_{n}}\left( \sigma \right) :\n\]\n\nthis follows immediately from the definition of centralizer (Definition 1.6). Now recall that the centralizer of \( \sigma \) is its stabilizer under conjugation, and therefore the... | Yes |
Looking again at \( {A}_{5} \), we have noted in \( \$ \underline{4.3} \) that the types of the even permutations in \( {S}_{5} \) are \( \left\lbrack {1,1,1,1,1}\right\rbrack ,\left\lbrack {2,2,1}\right\rbrack ,\left\lbrack {3,1,1}\right\rbrack \), and \( \left\lbrack 5\right\rbrack \) . By Proposition 4.15 the conjug... | Therefore there are exactly 5 conjugacy classes in \( {A}_{5} \), and the class formula for \( {A}_{5} \) is\n\n\[ \n{60} = 1 + {15} + {20} + {12} + {12}.\n\] | Yes |
Corollary 4.17. The alternating group \( {A}_{5} \) is a simple noncommutative group of order 60. | Proof. A normal subgroup of \( {A}_{5} \) is necessarily the union of conjugacy classes, contains the identity, and has order equal to a divisor of 60 (by Lagrange's theorem). The divisors of 60 other than 1 and 60 are\n\n\[ 2,3,4,5,6,{10},{12},{15},{20},{30} \]\n\ncounting the elements other than the identity would gi... | Yes |
Lemma 4.18. The alternating group \( {A}_{n} \) is generated by 3-cycles. | Proof. Since every even permutation is a product of an even number of 2-cycles, it suffices to show that every product of two 2-cycles may be written as product of 3-cycles. Therefore, consider a product\n\n\[ \left( {ab}\right) \left( {cd}\right) \]\n\nwith \( a \neq b, c \neq d \) . If \( \left( {ab}\right) = \left( ... | Yes |
Theorem 4.20. The alternating group \( {A}_{n} \) is simple for \( n \geq 5 \) . | Proof. We have already checked this for \( n = 5 \), and the reader has checked it for \( n = 6 \). For \( n > 6 \), let \( N \) be a nontrivial normal subgroup of \( {A}_{n} \); we will show that necessarily \( N = {A}_{n} \), by proving that \( N \) contains 3-cycles.\n\nLet \( \tau \in N,\tau \neq \left( 1\right) \)... | No |
Corollary 4.21. For \( n \geq 5 \), the group \( {S}_{n} \) is not solvable. | Proof. Since \( {A}_{n} \) is simple, the sequence\n\n\[ \n{S}_{n} \supsetneq {A}_{n} \supsetneq \{ \left( 1\right) \} \n\] \n\nis a composition series for \( {S}_{n} \) . It follows that the composition factors of \( {S}_{n} \) are \( \mathbb{Z}/2\mathbb{Z} \) and \( {A}_{n} \) . By Proposition 3.11, \( {S}_{n} \) is ... | Yes |
Lemma 5.1. Let \( N, H \) be normal subgroups of a group \( G \) . Then\n\n\[ \left\lbrack {N, H}\right\rbrack \subseteq N \cap H \] | Proof. It suffices to verify this on generators; that is, it suffices to check that\n\n\[ \left\lbrack {n, h}\right\rbrack = n\left( {h{n}^{-1}{h}^{-1}}\right) = \left( {{nh}{n}^{-1}}\right) {h}^{-1} \in N \cap H \]\n\nfor all \( n \in N, h \in H \) . But the first expression and the normality of \( N \) show that \( \... | Yes |
Corollary 5.2. Let \( N, H \) be normal subgroups of a group \( G \) . Assume \( N \cap H = \{ e\} \) . Then \( N, H \) commute with each other:\n\n\[ \left( {\forall n \in N}\right) \left( {\forall h \in H}\right) \;{nh} = {hn}. \] | Proof. By Lemma 5.1, \( \left\lbrack {N, H}\right\rbrack = \{ e\} \) if \( N \cap H = \{ e\} \) ; the result follows immediately. | Yes |
Proposition 5.3. Let \( N, H \) be normal subgroups of a group \( G \), such that \( N \cap H = \) \( \{ e\} \) . Then \( {NH} \cong N \times H \) . | Proof. Consider the function\n\n\[ \varphi : N \times H \rightarrow {NH} \]\n\ndefined by \( \varphi \left( {n, h}\right) = {nh} \) . Under the stated hypothesis, \( \varphi \) is a group homomorphism: indeed\n\n\[ \varphi \left( {\left( {{n}_{1},{h}_{1}}\right) \cdot \left( {{n}_{2},{h}_{2}}\right) }\right) = \varphi ... | Yes |
Lemma 5.8. The resulting structure \( \left( {N \times H,{ \bullet }_{\theta }}\right) \) is a group, with identity element \( \left( {{e}_{N},{e}_{H}}\right) \) . | Proof. The reader should carefully verify this. For example, inverses exist because\n\n\[ \left( {{n}_{1},{h}_{1}}\right) { \bullet }_{\theta }\left( {{\theta }_{{h}_{1}^{-1}}\left( {n}_{1}^{-1}\right) ,{h}_{1}^{-1}}\right) = \left( {{n}_{1}{\theta }_{{h}_{1}}\left( {{\theta }_{{h}_{1}^{-1}}\left( {n}_{1}^{-1}\right) }... | No |
Proposition 5.10. Let \( N, H \) be groups, and let \( \theta : H \rightarrow {\operatorname{Aut}}_{\mathrm{{Grp}}}\left( N\right) \) be a homomorphism; let \( G = N{ \rtimes }_{\theta }H \) be the corresponding semidirect product. Then\n\n- \( G \) contains isomorphic copies of \( N \) and \( H \) ;\n\n- the natural p... | Proof. The functions \( N \rightarrow G, H \rightarrow G \) defined for \( n \in N, h \in H \) by\n\n\[ n \mapsto \left( {n,{e}_{H}}\right) ,\;h \mapsto \left( {{e}_{N}, h}\right) \]\n\nare manifestly injective homomorphisms, allowing us to identify \( N, H \) with the corresponding subgroups of \( G \) . It is clear t... | Yes |
Proposition 5.11. Let \( N, H \) be subgroups of a group \( G \), with \( N \) normal in \( G \) . Assume that \( N \cap H = \{ e\} \), and \( G = {NH} \) . Let \( \gamma : H \rightarrow {\operatorname{Aut}}_{\mathrm{{Grp}}}\left( N\right) \) be defined by conjugation: for \( h \in H, n \in N \), \[ {\gamma }_{h}\left(... | Proof. Define a function \[ \varphi : N{ \rtimes }_{\gamma }H \rightarrow G \] by \( \varphi \left( {n, h}\right) = {nh} \) ; this is clearly a bijection. We need to verify that \( \varphi \) is a homomorphism, and indeed \( \left( {\forall {n}_{1},{n}_{2} \in N}\right) ,\left( {\forall {h}_{1},{h}_{2} \in H}\right) \)... | Yes |
The automorphism group of \( {C}_{3} \) is isomorphic to the cyclic group \( {C}_{2} \) : if \( {C}_{3} = \left\{ {e, y,{y}^{2}}\right\} \), then the two automorphisms of \( {C}_{3} \) are | \[ \text{ id : }\left\{ {\begin{aligned} e & \mapsto e, \\ y & \mapsto y, \\ {y}^{2} & \mapsto {y}^{2}, \end{aligned}\;\sigma : \begin{cases} e & \mapsto e, \\ y & \mapsto {y}^{2}, \\ {y}^{2} & \mapsto y. \end{cases}}\right. \] | Yes |
Lemma 6.1. Let \( G \) be an abelian group, and let \( H, K \) be subgroups such that \( \left| H\right| \) , \( \left| K\right| \) are relatively prime. Then \( H + K \cong H \oplus K \) . | Proof. By Lagrange’s theorem (Corollary 118.14), \( H \cap K = \{ 0\} \) . Since subgroups of abelian groups are automatically normal, the statement follows from Proposition 5.3 | No |
Corollary 6.2. Every finite abelian group is the direct sum of its nontrivial Sylow subgroups. | (The diligent reader knew already that this had to be the case, since abelian groups are nilpotent; cf. Exercise 5.1.) Thus, we already know that every finite abelian group is a direct sum of \( p \) -groups, and our main task amounts to classifying abelian \( p \) -groups for a fixed prime \( p \) . This is somewhat t... | No |
Lemma 6.3. Let \( G \) be an abelian p-group, and let \( g \in G \) be an element of maximal order. Then the exact sequence\n\n\[ 0 \rightarrow \langle g\rangle \rightarrow G \rightarrow G/\langle g\rangle \rightarrow 0 \]\n\nsplits. | Put otherwise, there is a subgroup \( L \) of \( G \) such that \( L \) maps isomorphically to \( G/\langle g\rangle \) via the canonical projection, that is, such that \( \langle g\rangle \cap L = \{ 0\} \) and \( \langle g\rangle + L = G \) . Note that it will follow that \( G \cong \langle g\rangle \oplus L \), by P... | Yes |
Lemma 6.4. Let \( p \) be a prime integer and \( r \geq 1 \) . Let \( G \) be a noncyclic abelian group of order \( {p}^{r + 1} \), and let \( g \in G \) be an element of order \( {p}^{r} \) . Then there exists an element \( h \in G, h \notin \langle g\rangle \), such that \( \left| h\right| = p \) . | Proof of Lemma 6.4. Denote \( \langle g\rangle \) by \( K \), and let \( {h}^{\prime } be any element of \( G,{h}^{\prime } \notin K \) . The subgroup \( K \) is normal in \( G \) since \( G \) is abelian; the quotient group \( G/K \) has order \( p \) . Since \( {h}^{\prime } \notin K \), the coset \( {h}^{\prime } + ... | Yes |
Corollary 6.5. Let \( G \) be a finite abelian group. Then \( G \) is a direct sum of cyclic groups, which may be assumed to be cyclic p-groups. | Proof. As noted in Corollary 6.2, \( G \) is a direct sum of \( p \) -groups (as a consequence of the Sylow theorems). I claim that every abelian \( p \) -group \( P \) is a direct sum of cyclic \( p \) -groups.\n\nTo establish this, argue by induction on \( \left| P\right| \) . There is nothing to prove if \( P \) is ... | Yes |
Theorem 6.6. Let \( G \) be a finite nontrivial abelian group. Then\n\n- there exist prime integers \( {p}_{1},\ldots ,{p}_{r} \) and positive integers \( {n}_{ij} \) such that \( \left| G\right| = \) \( \mathop{\prod }\limits_{{i, j}}{p}_{i}^{{n}_{i, j}} \) and\n\n\[ G \cong {\bigoplus }_{i, j}\frac{\mathbb{Z}}{{p}_{i... | The first form is nothing but a more explicit version of the statement of Corollary 6.5, so it has already been proven. I will explain how to obtain the second form from the first. The uniqueness statement \( {}^{29} \) is left to the reader (Exercise 6.1).\n\nThe prime powers appearing in the first form of Theorem 6.6... | No |
There are exactly 6 isomorphism classes of abelian groups of order 360. | Indeed, \( 360 = 2^3 \cdot 3^2 \cdot 5 \) ; the six possible tables of elementary divisors are shown below. In terms of invariant factors, the six distinct abelian groups of order 360 (up to isomorphism, by the uniqueness part of Theorem 6.6) are therefore\n\n\[ \frac{\mathbb{Z}}{3\mathbb{Z}} \oplus \frac{\mathbb{Z}}{{... | Yes |
Lemma 6.9. Let \( G \) be a finite abelian group, and assume that for every integer \( n > 0 \) the number of elements \( g \in G \) such that \( {ng} = 0 \) is at most \( n \) . Then \( G \) is cyclic. | Indeed, by Theorem 6.6\n\n\[ \nG \cong \frac{\mathbb{Z}}{{d}_{1}\mathbb{Z}} \oplus \cdots \oplus \frac{\mathbb{Z}}{{d}_{s}\mathbb{Z}} \n\]\n\nfor some positive integers \( 1 < {d}_{1}\left| \cdots \right| {d}_{s} \) . But if \( s > 1 \), then \( \left| G\right| > {d}_{s} \) and \( {d}_{s}g = 0 \) for all \( g \in G \) ... | Yes |
Theorem 6.10. Let \( F \) be a field, and let \( G \) be a finite subgroup of the multiplicative group \( \left( {{F}^{ * }, \cdot }\right) \) . Then \( G \) is cyclic. | Proof. By the considerations preceding the statement, for every \( n \) there are at most \( n \) elements \( a \in F \) such that \( {a}^{n} - 1 = 0 \), that is, at most \( n \) elements \( a \in G \) such that \( {a}^{n} = 1 \) . Lemma 6.9 implies then that \( G \) is cyclic. | No |
Proposition 1.1. Let \( R \) be a commutative ring, and let \( M \) be an \( R \)-module. Then the following are equivalent:\n\n(1) \( M \) is Noetherian; that is, every submodule of \( M \) is finitely generated.\n\n(2) Every ascending chain of submodules of \( M \) stabilizes; that is, if\n\n\[ \n{N}_{1} \subseteq {N... | Proof. (1) \( \Rightarrow \) (2): Assume that \( M \) is Noetherian, and let\n\n\[ \n{N}_{1} \subseteq {N}_{2} \subseteq {N}_{3} \subseteq \cdots \n\]\n\nbe a chain of submodules of \( M \) . Consider the union\n\n\[ \nN = \mathop{\bigcup }\limits_{i}{N}_{i} \n\]\n\nthe reader will verify that \( N \) is a submodule of... | No |
Theorem 1.2. Let \( R \) be a Noetherian ring, and let \( J \) be an ideal of the polynomial ring \( R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then the ring \( R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /J \) is Noetherian. | The proof of this deep fact is surprisingly easy. By Exercise 1.1 it suffices to prove that\n\n\[ R\text{Noetherian} \Rightarrow R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \text{Noetherian;} \] | No |
Lemma 1.5. Let \( a, b \) be nonzero elements of an integral domain \( R \) . Then \( a \) and \( b \) are associates if and only if \( a = {ub} \), for \( u \) a unit in \( R \) . | Proof. Assume \( a \) and \( b \) are associates. Then \( \exists c, d \in R \) such that\n\n\[ b = {ac},\;a = {bd}; \]\n\ntherefore \( a = {bd} = {acd} \), i.e.,\n\n\[ a\left( {1 - {cd}}\right) = 0. \]\n\nSince cancellation by nonzero elements hold in integral domains, this implies \( {cd} = 1 \) . Thus \( c \) is a u... | No |
Lemma 1.7. Let \( R \) be an integral domain, and let \( a \in R \) be a nonzero prime element. Then a is irreducible. | Proof. Since \( \left( a\right) \) is prime, \( \left( a\right) \neq \left( 1\right) \) ; hence \( a \) is not a unit. If \( a = {bc} \), then \( {bc} = a \in \left( a\right) \) ; therefore \( b \in \left( a\right) \) or \( c \in \left( a\right) \) since \( \left( a\right) \) is prime. Assuming without loss of generali... | Yes |
Proposition 1.11. Let \( R \) be an integral domain, and let \( r \) be a nonzero, nonunit element of \( R \) . Assume that every ascending chain of principal ideals\n\n\[ \left( r\right) \subseteq \left( {r}_{1}\right) \subseteq \left( {r}_{2}\right) \subseteq \left( {r}_{3}\right) \subseteq \cdots \]\n\nstabilizes. T... | Proof. Assume that \( r \) does not have a factorization into irreducible elements. In particular, \( r \) is itself not irreducible; thus \( \exists {r}_{1},{s}_{1} \in R \) such that \( r = {r}_{1}{s}_{1} \) and \( \left( r\right) \varsubsetneq \left( {r}_{1}\right) ,\left( r\right) \varsubsetneq \left( {s}_{1}\right... | Yes |
Corollary 1.12. Let \( R \) be a Noetherian domain. Then factorizations exist in \( R \) . | Proof. By Proposition 1.1, Noetherian domains satisfy the ascending chain condition for all ideals. | No |
Lemma 2.1. Let \( R \) be a UFD, and let \( a, b, c \) be nonzero elements of \( R \) . Then\n\n- \( \left( a\right) \subseteq \left( b\right) \Leftrightarrow \) the multiset of irreducible factors of \( b \) is contained in the multiset of irreducible factors of \( a \) ;\n\n- \( a \) and \( b \) are associates (that ... | The proof is left to the reader (Exercise 2.1). | No |
Lemma 2.3. Let \( R \) be a UFD, and let \( a, b \) be nonzero elements of \( R \). Then \( a, b \) have a greatest common divisor. | Proof. We can write\n\n\[ a = u{q}_{1}^{{\alpha }_{1}}\cdots {q}_{r}^{{\alpha }_{r}},\;b = v{q}_{1}^{{\beta }_{1}}\cdots {q}_{r}^{{\beta }_{r}} \]\n\nwhere \( u \) and \( v \) are units, the elements \( {q}_{i} \) are irreducible, \( {q}_{i} \) is not an associate of \( {q}_{j} \) for \( i \neq j \), and \( {\alpha }_{... | Yes |
Lemma 2.4. Let \( R \) be a UFD, and let a be an irreducible element of \( R \) . Then a is prime. | Proof. The element \( a \) is not a unit, by definition of irreducible. Assume \( {bc} \in \left( a\right) \) : thus \( \left( {bc}\right) \subseteq \left( a\right) \), and by Lemma 2.1 the irreducible factors of \( a \), that is, \( a \) itself, must be among the factors of \( b \) or of \( c \) . We have \( b \in \le... | Yes |
Theorem 2.5. An integral domain \( R \) is a UFD if and only if\n\n- the a.c.c. for principal ideals holds in \( R \) and\n\n- every irreducible element of \( R \) is prime. | Proof. ( \( \Rightarrow \) ) Assume that \( R \) is a UFD. Lemma 2.4 shows that irreducible elements of \( R \) are prime. To prove that the a.c.c. for principal ideals holds, consider an ascending chain\n\n\[ \left( {r}_{1}\right) \varsubsetneq \left( {r}_{2}\right) \varsubsetneq \left( {r}_{3}\right) \varsubsetneq \c... | Yes |
Proposition 2.6. If \( R \) is a PID, then it is a UFD. | Proof. Let \( R \) be a PID. The a.c.c. (for principal ideals, as all ideals in \( R \) are principal!) holds in \( R \) since PIDs are Noetherian. We verify that irreducible elements are prime in \( R \), which implies that \( R \) is a UFD by Theorem 2.5\n\nLet \( a \in R \) be an irreducible element. Ideals generate... | No |
Proposition 2.8. Let \( R \) be a Euclidean domain. Then \( R \) is a PID. | The proof is modeled after the instances encountered for \( \mathbb{Z} \) (Proposition 1114.4) and \( k\left\lbrack x\right\rbrack \) (which the reader has hopefully worked out in Exercise 11114.4).\n\nProof. Let \( I \) be an ideal of \( R \) ; we have to prove that \( I \) is principal. If \( I = \{ 0\} \) , there is... | Yes |
Lemma 2.9. Let \( a = {bq} + r \) in \( a \) ring \( R \) . Then \( \left( {a, b}\right) = \left( {b, r}\right) \) . | Proof. Indeed, \( r = a - {bq} \in \left( {a, b}\right) \), proving \( \left( {b, r}\right) \subseteq \left( {a, b}\right) \) ; and \( a = {bq} + r \in \left( {b, r}\right) \) , proving \( \left( {a, b}\right) \subseteq \left( {b, r}\right) \) . | Yes |
Proposition 2.12. With notation as above, \( {r}_{N - 1} \) is a gcd of \( a, b \) . | Proof. By Corollary 2.10,\n\n\[ \gcd \left( {a, b}\right) = \gcd \left( {b,{r}_{1}}\right) = \gcd \left( {{r}_{1},{r}_{2}}\right) = \cdots = \gcd \left( {{r}_{N - 2},{r}_{N - 1}}\right) .\n\]\n\nBut \( {r}_{N - 2} = {r}_{N - 1}{q}_{N - 1} \) gives \( {r}_{N - 2} \in \left( {r}_{N - 1}\right) \) ; hence \( \left( {{r}_{... | Yes |
Theorem 3.3 (Well-ordering theorem). Every set admits a well-ordering. | Well-ordering theorem \( \Rightarrow \) Zorn’s lemma. Let \( \left( {Z, \leq }\right) \) be a nonempty poset such that every chain in \( Z \) has an upper bound in \( Z \) . By the well-ordering theorem, there is a well-ordering \( {}^{13} \preccurlyeq \) on \( Z \) . Define a function \( f \) from \( Z \) to the power... | Yes |
Proposition 3.5. Let \( I \neq \left( 1\right) \) be a proper ideal of a commutative ring \( R \) . Then there exists a maximal ideal \( \mathfrak{m} \) of \( R \) containing \( I \) . | Proof. The set \( \mathcal{I} \) of proper ideals of \( R \) containing \( I \) is ordered by inclusion. Then let \( \mathcal{C} \) be a chain of proper ideals, and consider\n\n\[ U \mathrel{\text{:=}} \mathop{\bigcup }\limits_{{J \in \mathcal{C}}}J \]\n\nI claim that \( U \) is a proper ideal containing \( I \) ; henc... | Yes |
Lemma 4.1. Let \( R \) be a ring, and let \( I \) be an ideal of \( R \) . Then\n\n\[ \frac{R\left\lbrack x\right\rbrack }{{IR}\left\lbrack x\right\rbrack } \cong \frac{R}{I}\left\lbrack x\right\rbrack \] | The proof of this lemma is a standard application of the first isomorphism theorem and is left to the reader (Exercise 4.1). | No |
Corollary 4.2. If \( I \) is a prime ideal of \( R \), then \( {IR}\left\lbrack x\right\rbrack \) is prime in \( R\left\lbrack x\right\rbrack \) . | Proof. If \( I \) is prime in \( R \), then \( R/I \) is an integral domain; hence so is \( R\left\lbrack x\right\rbrack /{IR}\left\lbrack x\right\rbrack \cong \) \( \left( {R/I}\right) \left\lbrack x\right\rbrack \), and therefore \( {IR}\left\lbrack x\right\rbrack \) is prime in \( R\left\lbrack x\right\rbrack \) . | Yes |
Lemma 4.4. Let \( R \) be a commutative ring. Then for \( f, g \in R\left\lbrack x\right\rbrack \)\n\n\( {fg} \) is primitive \( \Leftrightarrow \) both \( f \) and \( g \) are primitive. | Proof. This is an easy consequence of Corollary 4.2\n\n\( {fg} \) primitive \( \Leftrightarrow \forall \mathfrak{p} \) prime and principal in \( R,{fg} \notin \mathfrak{p}R\left\lbrack x\right\rbrack \)\n\n\( \Leftrightarrow \forall \mathfrak{p} \) prime and principal in \( R, f \notin \mathfrak{p}R\left\lbrack x\right... | Yes |
Lemma 4.5. Let \( R \) be a commutative ring and \( f = {a}_{0} + {a}_{1}x + \cdots + {a}_{d}{x}^{d} \in R\left\lbrack x\right\rbrack \) as above.\n\n- \( f \) is very primitive if and only if \( \left( {{a}_{0},\ldots ,{a}_{d}}\right) = \left( 1\right) \).\n\n- If \( R \) is a UFD, then \( f \) is primitive if and onl... | Proof. If \( \left( {{a}_{0},\ldots ,{a}_{d}}\right) = \left( 1\right) \), then no prime ideal can contain all coefficients \( {a}_{i} \) , and it follows that \( f \) is very primitive. Conversely, if \( f \) is very primitive, then the coefficients of \( f \) are not all contained in any one prime ideal, and in parti... | Yes |
Proposition 4.8 (Gauss’s lemma). Let \( R \) be a UFD, and let \( f, g \in R\left\lbrack x\right\rbrack \) . Then\n\n\[ \left( {\operatorname{cont}}_{fg}\right) = \left( {\operatorname{cont}}_{f}\right) \left( {\operatorname{cont}}_{g}\right) \] | Proof. This follows easily from our preparatory work. Write\n\n\[ \left( {fg}\right) = \left( {\left( {\operatorname{cont}}_{f}\right) \left( \underline{f}\right) }\right) \left( {\left( {\operatorname{cont}}_{g}\right) \left( \underline{g}\right) }\right) = \left( {\operatorname{cont}}_{f}\right) \left( {\operatorname... | Yes |
Example 4.12. With the notation introduced above, \( K\left( \mathbb{Z}\right) = \mathbb{Q} \) . | The universal property implies immediately that \( F \hookrightarrow K\left( F\right) \) is an isomorphism if \( F \) is itself a field. Thus, the construction adds nothing to \( \mathbb{Q},\mathbb{R},\mathbb{C},\mathbb{Z}/p\mathbb{Z} \), etc. | No |
Theorem 4.14. Let \( R \) be a UFD; then \( R\left\lbrack x\right\rbrack \) is a UFD. | By Theorem 2.5, in order to prove Theorem 4.14, we have to verify that \( R\left\lbrack x\right\rbrack \) satisfies the a.c.c. for principal ideals and that every irreducible element in \( R\left\lbrack x\right\rbrack \) is prime, provided that \( R \) is itself a UFD. The general idea is to reduce these questions to m... | Yes |
Lemma 4.15. Let \( R \) be a UFD, and let \( K = K\left( R\right) \) be its field of fractions. For nonzero \( f, g \in R\left\lbrack x\right\rbrack \), denote by \( \left( f\right) ,\left( g\right) \) the principal ideals \( {fR}\left\lbrack x\right\rbrack ,{gR}\left\lbrack x\right\rbrack \) in \( R\left\lbrack x\righ... | Proof. Since \( {\left( g\right) }_{K} \subseteq {\left( f\right) }_{K} \), we have \( g = {fh} \), where \( h \in K\left\lbrack x\right\rbrack \) . Write \( h = \frac{a}{b}\underline{h} \), where \( a, b \in R \) and \( \underline{h} \in R\left\lbrack x\right\rbrack \) is a primitive polynomial: this can be done by co... | Yes |
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