Monthly Archives: April 2011

As a ZZ-module, the p-primary components of a finite abelian group are its Sylow subgroups

Let M be a finite abelian group of order a = \prod p_i^{k_i} and consider M as a \mathbb{Z}-module in the natural way. Prove that M is annihilated by (a), that the p_i-primary component of M is the (unique) Sylow p_i-subgroup of M, and that M is isomorphic to the direct product of its Sylow subgroups.


Certainly for all m \in M, a \cdot m = 0. Thus (a) \subseteq \mathsf{Ann}_\mathbb{Z}(M). Since \mathbb{Z} is a principal ideal domain, by this previous exercise, M = \bigoplus (q_t)M, where q_t = \prod_{i \neq t} p_i^{k_i}.

We claim that each (q_t)M is a Sylow subgroup of M. To that end, note that p_t^{k_t} \cdot (q_t)M = 0; in particular, every element of (q_t)M has p_t-power order. Conversely, if m \in M has p_t-power order, then m \in \mathsf{Ann}_M(p_t^{k_t}) = (q_t)M. Hence the (q_t)M, that is, the p_t-primary components, are Sylow subgroups of M. Since M is abelian, there is a unique Sylow p_t-subgroup for each p_t; thus M = \bigoplus (q_t)M is the internal direct sum of its Sylow subgroups. Since the direct sum is finite, M is in fact the internal direct product of its Sylow subgroups.

Over a PID, every module with a nonzero annihilator is an internal direct sum of its p-primary components

Let R be a principal ideal domain. Let M be a left unital R-module, and suppose \mathsf{Ann}_R(M) \supseteq (a) where a is nonzero. Say a = \prod p_i^{k_i} is the irreducible factorization of a in R. Prove that M is the internal direct sum of its submodules of the form \mathsf{Ann}_M(p_i^{k_i}). (This \mathsf{Ann}_M(p_i^{k_i}) is called the p_i-primary component of M.)


For each t, let q_t = \prod_{i \neq t} p_i^{k_i}.

We claim that \mathsf{Ann}_M(p_t^{k_t}) = (q_t)M for all t. (\supseteq) Suppose q_t \cdot m \in (q_t)M. Note that p_t^{k_t} \cdot (q_t \cdot m) = a \cdot m = 0, so that q_t \cdot m \in \mathsf{Ann}_M(p_t^{k_t}). Thus (q_t)M \subseteq \mathsf{Ann}_M(p_t^{k_t}). (\subseteq) Suppose now that m \in \mathsf{Ann}_M(p_t^{k_t}). Since R is a principal ideal domain and p_t^{k_t} and q_t are relatively prime, we have 1 = q_tx + p_t^{k_t}y for some x,y \in R. Note then that m = 1 \cdot m = (q_tx + p_t^{k_t}y) \cdot m = q_tx \cdot m + p_t^{k_t}y \cdot m q_tx \cdot m \in (q_t)M. Thus \mathsf{Ann}_M(p_t^{k_t}) \subseteq (q_t)M, and we have \mathsf{Ann}_M(p_t^{k_t}) = (q_t)M.

Next, we claim that (q_t)M \cap \sum (q_s)M = 0 for all t. To that end, let m \in (q_t)M \cap \sum (q_s)M. Note that p_t^{k_t} and q_t are relatively prime (by definition). Now m = 1 \cdot m = xp_t^{k_t} \cdot m + yq_t \cdot m = 0, since p_t^{k_t} \cdot (q_t)M = 0 while q_t \cdot (\sum (q_s)M) = 0. Thus (q_t)M \cap (\sum (q_s)M) = 0 as desired.

Next, we claim that (q_i \ |\ 1 \leq i \leq n) = R. To see this, note that we have the irreducible factorization of each q_i, and that these have no irreducible factors in common.

Now let m \in M. Since (q_i \ |\ 1 \leq i \leq n) = R, we have 1 = \sum r_i q_i for some r_i \in R. Now m = 1 \cdot m = \sum r_iq_i \cdot m \in \sum (q_i)M.

Thus M = \sum_{i=1}^n (q_i)M, and in fact this sum is direct. So M = \oplus_{i=1}^n \mathsf{Ann}_M(p_i^{k_i}).

Chinese Remainder Theorem for Modules, Part 2

Let R be a commutative ring with 1 and let \{A_i\}_{i =1}^n be a finite family of ideals of R. Suppose further that A_i + A_j = R if i \neq j. (I.e. the ideals A_i are pairwise comaximal.) Let M be a unital left R-module. Prove that the R-module homomorphism \Phi : M \rightarrow \prod_{i=1}^n M/A_iM discussed here is surjective and has kernel (\prod_{i=1}^n A_i)M, where \prod_{i=1}^n A_i denotes the ideal product in R rather than the direct product of sets. Deduce that M/(\prod_{i=1}^n A_i)M \cong_R \prod_{i=1}^n M/A_iM.


We will prove this result by induction on n. For the base case n = 2, we have \Phi(m) = (m+A_1M, m+A_2M). First we show that \Phi is surjective. Let (m_1 + A_1M, m_2 + A_2M) be in M/A_1M \times M/A_2M. Since A_1 and A_2 are comaximal and R has a 1, there exist a_1 \in A_1 and a_2 \in A_2 such that a_1+a_2 = 1. Note then that \Phi(a_1m_2 + a_2m_1) = (a_1m_2 + a_2m_1 A_1M, a_1m_2 + a_2m_1 + A_2M) = (a_2m_1 + A_1M, a_1m_2 + A_2M) = ((1-a_1)m_1 + A_1M, (1-a_2)m_2 + A_2M) = (m_1 - a_1m_1 + A_1M, m_2 - a_2m_2 + A_2M) = (m_1 + A_1M, m_2 + A_2M). Thus \Phi is surjective. Next we show that \mathsf{ker}\ \Phi = (A_1A_2)M. We showed in Part 1 that \mathsf{ker}\ \Phi = A_1M \cap A_2M. Clearly (A_1A_2)M \subseteq A_1M \cap A_2M. Now suppose m \in A_1M \cap A_2M. Recall that a_1+a_2 = 1 for some a_1 \in A_1 and a_2 \in A_2; now m = 1 \cdot m = (a_1+a_2) \cdot m = a_1 \cdot m + a_2 \cdot m. Since m \in A_2, a_1 \cdot m \in (A_1A_2)M. Since m \in A_1, a_2 \cdot m \in (A_2A_1)M = (A_1A_2)M. Thus m \in (A_1A_2)M as desired. By the First Isomorphism Theorem for modules, the induced mapping \Psi : M/(A_1A_2)M \rightarrow M/A_1M \times M/A_2M given by m+(A_1A_2)M \mapsto (m+A_1M, m+A_2M) is an R-module isomorphism.

Now for the inductive step, suppose the result holds for 2 and n. Let \{A_i\}_{i=1}^{n+1} be a family of pairwise comaximal ideals in R and let M be a unital left R-module. We claim that A_1 and \prod_{i=2}^{n+1} are comaximal; to see this, note that 1 = a_{1,i} + a_i for some a_{1,i} \in A_i and a_i \in A_i for all 2 \leq i \leq n+1. Then 1 = \prod_{i=2}^{n+1}(a_{1,i} + a_i) \in A_1 + \prod_{i=2}^{n+1} A_i, so that A_1 + \prod_{i=2}^{n+1} A_i = R. Now M/(\prod_{i=1}^{n+1} A_i)M \cong_R M/A_1M \times M/(\prod_{i=2}^{n+1} A_i)M \cong_R M/A_1M \times \prod_{i=2}^{n+1} M/A_iM \cong_R \prod_{i=1}^{n+1} M/A_iM.

Chinese Remainder Theorem for Modules, Part 1

Let R be a ring with 1 and let \{A_i\}_{i \in I} be a family of ideals in R indexed by a set I (I need not be finite for the moment.) Let M be a unital left R-module. Prove that the mapping \Phi : M \rightarrow \prod_{i \in I} M/A_iM given by \Phi(m)_i = m + A_iM is an R-module homomorphism with kernel \bigcap A_iM.


[Note: If you’re following along with Dummit & Foote, you may notice that we jumped ahead a bit by assuming that \prod M/A_iM is naturally an R-module. It is, and this result could have been proved shortly after the definition of module, so we will let slide the fact that we haven’t yet formally proven that the product of modules is a module.]

First, note that for all x,y \in M and r \in R, we have \Phi(x + r \cdot y)_i = (x+r \cdot y) + A_iM = (x + A_iM) + r \cdot (y + A_iM) = \Phi(x) + r \cdot \Phi(y). Thus \Phi is an R-module homomorphism.

Now suppose m \in \mathsf{ker}\ \Phi. Then for all i, we have 0 = \Phi(m)_i = m + A_iM, and thus m \in A_iM. So m \in \bigcap A_iM. Conversely, if m \in \bigcap A_iM, then \Phi(m)_i = m + A_iM = 0 for all i. Thus m \in \mathsf{ker}\ \Phi.

Over a ring with a central idempotent, every module is an internal direct sum

Let R be a ring with 1 and let M be a left unital R-module. Suppose there exists e \in R such that e^2 = e and e \in Z(R). Prove that M = eM \oplus (1-e)M.


First, recall by this previous exercise that eM and (1-e)M are both submodules of M. Suppose m \in M. Then m = e \cdot m + 1 \cdot m - e \cdot m \in eM + (1-e)M, so that M = eM + (1-e)M. Finally, suppose m \in eM \cap (1-e)M; say m = e \cdot a = (1-e) \cdot b. Now e \cdot a = 1 \cdot b - e \cdot b, so that e \cdot (a+b) = b. Now e \cdot b = e^2 \cdot (a+b) = e \cdot (a + b) = e \cdot a + e \cdot b, so that e \cdot a = 0. Thus m = 0, and we have eM \cap (1-e)M = 0. So M = eM \oplus (1-e)M.

Compute Hom(F,M) over a commutative ring, where F is free of finite rank

Let R be a commutative ring with 1, let F be the free left unital R-module of finite rank n, and let M be a left unital R-module. Prove that \mathsf{Hom}_R(F,M) \cong_R M^n.


Recall by Theorem 6 in D&F that F \cong R^n, since F has finite rank. Now \mathsf{Hom}_R(F,M) \cong_R \mathsf{Hom}_R(R^n,M) \cong_R \mathsf{Hom}_R(R,M)^n (by this previous exercise) \cong_R M^n (by this previous exercise).

Thus \mathsf{Hom}_R(F,M) \cong_R M^n.

If R is a commutative unital ring and F a free unital left R-module of finite rank then Hom(F,R) and F are isomorphic as R-modules

Let R be a commutative ring with 1 and let F be a free left unital R-module of finite rank. Prove that \mathsf{Hom}_R(F,R) \cong_R F, where \cong_R means “isomorphic as R-modules”.


Suppose F is free on the set \{a_i\}_{i=1}^n; then every element of F can be written uniquely as \sum r_i \cdot a_i for some r_i \in R. Now define \Phi : \mathsf{Hom}_R(F,R) \rightarrow F by \Phi(\varphi) = \sum \varphi(a_i) \cdot a_i. We claim that \Phi is an R-module isomorphism.

Note that if \varphi, \psi \in \mathsf{Hom}_R(F,R) and r \in R, then \Phi(\varphi + r \cdot \psi) = \sum (\varphi + r \cdot \psi)(a_i) \cdot a_i = (\sum \varphi(a_i) \cdot a_i) + r \cdot (\sum \psi(a_i) \cdot a_i) = \Phi(\varphi) + r \cdot \Phi(\psi). Thus \Phi is an R-module homomorphism.

Now suppose \varphi \in \mathsf{ker}\ \Phi. Then 0 = \Phi(\varphi) = \sum \varphi(a_i) \cdot a_i. Since F is free on the a_i, we have \varphi(a_i) = 0 for all a_i, and thus \varphi = 0. So \mathsf{ker}\ \Phi = 0, and thus \Phi is injective.

Finally, let \sum t_i \cdot a_i \in F. Define \theta : F \rightarrow R by \theta(\sum r_i \cdot a_i) = \sum r_it_i. Evidently, \theta is an R-module homomorphism. Moreover, \theta(a_i) = t_i, so that \Phi(\theta) = \sum t_i \cdot a_i. Hence \Phi is surjective.

Thus \Phi is an R-module isomorphism, so that we have \mathsf{Hom}_R(F,R) \cong_R F.

Irrelevant ads are irrelevant.

Hello.

I was testing my backup script today and noticed that if you browse this blog while not logged in to WordPress, every post is accompanied by hideous and irrelevant advertisements. These are automatically added by WordPress. I have nothing to do with them and find them to be incredibly obnoxious.

Now I can’t complain too much about this, since after all WordPress.com is providing this platform to broadcast my garbled rantings for free. I do think these particular ads are tacky and detract from the usefulness of the site, though. If I were going to put ads up here on purpose (which I have no intention of doing for the foreseeable future) they would be unobtrusive and only for relevant things I actually use and enjoy myself- like good books and tools, for instance. Certainly not this PopPressed crap.

There are a couple of ways to remove these, most of which are inaccessible because they cost money. At the moment it appears that the easiest (and cheapest) way to view this site without the tacky ads is to get a WordPress account (which is free) and log in.

Sorry for the inconvenience.

Hom commutes with finite direct products in both slots

Let R be a ring with 1 and let A, B, and M be unital left R-modules. Prove that \mathsf{Hom}_R(A \times B, M) \cong_R \mathsf{Hom}_R(A,M) \times \mathsf{Hom}_R(B,M) and \mathsf{Hom}_R(M,A \times B) \cong_R \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B). (Where \cong_R means “isomorphic as R-modules”.)


Given R-module homomorphisms \alpha : A \rightarrow M and \beta : B \rightarrow M, define \Phi(\alpha,\beta) : A \times B \rightarrow M by \Phi(\alpha,\beta)(a,b) = \alpha(a) + \beta(b). Note that, for all a_1,a_2 \in A, b_1,b_2 \in B, and r \in R, we have

\Phi(\alpha,\beta)((a_1,b_1) + r \cdot (a_2,b_2))  =  \Phi(\alpha,\beta)((a_1 + r \cdot a_2, b_1 + r \cdot b_2))
 =  \alpha(a_1 + r \cdot a_2) + \beta(b_1 + r \cdot b_2)
 =  \alpha(a_1) + r \cdot \alpha(a_2) + \beta(b_1) + r \cdot \beta(b_2)
 =  \Phi(\alpha,\beta)(a_1,b_1) + r \cdot \Phi(\alpha,\beta)(a_2,b_2).

So \Phi(\alpha,\beta) : A \times B \rightarrow M is an R-module homomorphism. Consider the map \Phi : \mathsf{Hom}_R(A,M) \times \mathsf{Hom}_R(B,M) \rightarrow \mathsf{Hom}_R(A \times B, M). We claim that \Phi is an R-module homomorphism. To see this, let \alpha_1,\alpha_2 : A \rightarrow M and \beta_1,\beta_2 : B \rightarrow M be R-module homomorphisms and let r \in R. Now

\Phi((a_1,b_1) + r \cdot (a_2,b_2))(a,b)  =  \Phi(\alpha_1 + r \cdot \alpha_2, \beta_1 + r \cdot \beta_2)(a,b)
 =  (\alpha_1 + r \cdot \alpha_2)(a) + (\beta_1 + r \cdot \beta_2)(b)
 =  \alpha_1(a) + r \cdot \alpha_2(a) + \beta_1(b) + r \cdot \beta_2(b)
 =  \Phi(\alpha_1,\beta_1)(a,b) + r \cdot \Phi(\alpha_2,\beta_2)(a,b)
 =  (\Phi(\alpha_1,\beta_1) + r \cdot \Phi(\alpha_2,\beta_2))(a,b).

So \Phi is an R-module homomorphism. Now suppose (\alpha,\beta) \in \mathsf{ker}\ \Phi. In particular, 0 = \Phi(\alpha,\beta)(a,0) = \alpha(a) for all a \in A, so that \alpha = 0. Similarly, \beta = 0. Thus (\alpha,\beta) = 0, and in fact \mathsf{ker}\ \Phi = 0. Hence \Phi is injective. Finally, let \theta : A \times B \rightarrow M be an R-module homomorphism, and define \alpha_\theta : A \rightarrow M and \beta_\theta : B \rightarrow M by \alpha_\theta(a) = \theta(a,0) and \beta_\theta(b) = \theta(0,b). Note that \alpha_\theta(a_1 + r \cdot a_2) = \theta(a_1 + r \cdot a_2,0 = \theta(a_1,0) + r \cdot \theta(a_2,0) = \alpha_\theta(a_1) + r \cdot \alpha_\theta(a_2), so that \alpha_\theta is an R-module homomorphism. Similarly, \beta_\theta is an R-module homomorphism. Since \Phi(\alpha_\theta, \beta_\theta)(a,b) = \alpha_\theta(a) + \beta_\theta(b) = \theta(a,0) + \theta(0,b) = \theta(a,b), we have \Phi(\alpha_\theta,\beta_\theta) = \theta. Hence \Phi is surjective. Thus, we have \mathsf{Hom}_R(A \times B, M) \cong_R \mathsf{Hom}_R(A,M) \times \mathsf{Hom}_R(B,M).

Now let \pi_A : A \times B \rightarrow A and \pi_B : A \times B \rightarrow B be the left and right coordinate projections; certainly these are R-module homomorphisms. Define \Psi : \mathsf{Hom}_R(M,A \times B) \rightarrow \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B) by \Psi(\varphi) = (\pi_A \circ \varphi, \pi_B \circ \varphi). If \varphi, \psi \in \mathsf{Hom}_R(M,A \times B) and r \in R, then

\Psi(\varphi + r \cdot \psi)  =  (\pi_A \circ (\varphi + r \cdot \psi), \pi_B \circ (\varphi + r \cdot \psi))
 =  (\pi_A \circ \varphi + r \cdot (\pi_A \circ \psi), \pi_B \circ \varphi + r \cdot (\pi_B \circ \psi))
 =  (\pi_A \circ \varphi, \pi_A \circ \psi) + r \cdot (\pi_B \circ \varphi, \pi_B \circ \psi)
 =  \Psi(\varphi) + r \cdot \Psi(\psi).

Thus \Psi is an R-module homomorphism. Now suppose \varphi \in \mathsf{ker}\ \Psi. In particular, if m \in M, then 0 = \Psi(\varphi)(m,m) = ((\pi_A \circ \varphi)(m), (\pi_b \circ \psi)(m)) = (\pi_A(\varphi(m)), \pi_B(\varphi(m)) = \varphi(m). So \varphi = 0, hence \mathsf{ker}\ \Psi = 0, and thus \Psi is injective. Finally, suppose (\alpha,\beta) \in \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B). Define \varphi_{\alpha,\beta} : M \rightarrow A \times B by \varphi_{\alpha,\beta}(m) = (\alpha(m), \beta(m). Certainly \varphi_{\alpha,\beta} is an R-module homomorphism. Moreover, we have \Psi(\varphi_{\alpha,\beta})(m) = ((\pi_A \circ \varphi_{\alpha,\beta})(m), (\pi_B \circ \varphi_{\alpha,\beta})(m)) = (\alpha(m), \beta(m)). Thus \Psi(\varphi_{\alpha,\beta}) = (\alpha,\beta), and so \Psi is surjective. Hence \mathsf{Hom}_R(M, A \times B) \cong_R \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B).

Schur’s Lemma

Let R be a ring with 1. Prove that if M_1 and M_2 are irreducible unital left R-modules, then any nonzero R-module homomorphism \varphi : M_1 \rightarrow M_2 is an isomorphism. Deduce Schur’s Lemma: If M is an irreducible unital left R-module, then \mathsf{End}_R(M) is a division ring.


Suppose \varphi : M_1 \rightarrow M_2 is an R-module homomorphism with M_1 and M_2 irreducible. We showed here that \mathsf{ker}\ \varphi and \mathsf{im}\ \varphi are submodules of M_1 and M_2, respectively. Since \varphi is not the zero homomorphism, its kernel is not all of M_1, and so (since M_1 is irreducible) \mathsf{ker}\ \varphi = 0. Thus \varphi is injective. Similarly, \mathsf{im} \varphi is not the zero submodule, and so must be M_2, hence \varphi is surjective. Thus \varphi is an R-module isomorphism.

Now let M be an irreducible unital left R-module. Recall that \mathsf{End}_R(M) is a ring under pointwise addition and composition. Now if \varphi \in \mathsf{End}_R(M) is nonzero, then by the above argument it is an isomorphism and so has an inverse \varphi^{-1} which is also in \mathsf{End}_R(M). So \mathsf{End}_R(M) is a division ring. Note that \mathsf{End}_R(M) is not necessarily commutative, so it need not be a field.