## 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 $\mathsf{Hom}_R(F,R) \cong_R F$.

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.