## Tag Archives: module

### A fact about the rank of a module over an integral domain

Let $R$ be an integral domain with quotient field $F$ and let $M$ be any (left, unital) $R$-module. Prove that the rank of $M$ equals the dimension of $F \otimes_R M$ over $F$.

Recall that the rank of a module over a domain is the maximal number of $R$-linearly independent elements.

Suppose $B \subseteq M$ is an $R$-linearly independent set, and consider $B^\prime \subseteq F \otimes_R M$ consisting of the tensors $1 \otimes b$ for each $b \in B$. Suppose $\sum \alpha_i (1 \otimes b_i) = 0$. For some $\alpha_i \in F$. Clearing denominators, we have $\sum r_i ( 1 \otimes b_i) = 0$ for some $r_i \in R$. Now $\sum 1 \otimes r_ib_i = 0$, and we have $1 \otimes \sum r_ib_i = 0$. By this previous exercise, there exists a nonzero $r \in R$ such that $\sum rr_ib_i = 0$. Since $B$ is linearly independent, we have $rr_i = 0$ for each $i$, and since $r \neq 0$ and $R$ is a domain, $r_i = 0$ for each $i$. Thus $\alpha_i = 0$ (since the denominators of each $\alpha_i$ are nonzero). So $B^\prime$ is $F$-linearly independent in $F \otimes_R M$. In particular, we have $\mathsf{rank}_R(M) \leq \mathsf{dim}_F(F \otimes_R M)$.

Now note that if $b_i^\prime = \sum \alpha_{i,j} \otimes m_{i,j} \in F \otimes_R M$, is a linearly independent set, with $\alpha_{i,j} = r_{i,j}/s_{i,j}$, then ‘clearing denominators’ by multiplying $b_i^\prime$ by $\prod_j s_{i,j}$, we have a second linearly independent set with the same cardinality whose elements are of the form $1 \otimes m_i$ for some $m \in M$. Suppose there exist $r_i \in R$ such that $\sum r_im_i = 0$; then $\sum r_i(1 \otimes b_i) = 0$, and (since the $1 \otimes b_i$ form a basis) we have $r_i = 0$. So the $m_i$ are $R$-linearly independent in $M$. Thus $\mathsf{rank}_R(M) \geq \mathsf{dim}_F(R \otimes_R M)$.

### Properties of the rows of a relations matrix

Let $R$, $M$, $\varphi$, $x_j$, $y_i$, and $A$ be defined as in this previous exercise.

1. Show that interchanging the generators $y_a$ and $y_b$ has the effect of interchanging the $a$th and $b$th rows of the relations matrix $A$.
2. Show that, for any $r \in R$, replacing $y_a$ by $y_a + ry_b$ $a \neq b$) gives another generating set $S^\prime$ for $\mathsf{ker}\ \varphi$. Show further that the matrix with respect to this generating set is obtained from $A$ by adding $r$ times the $b$th row to the $a$ th row.

It is clear that interchanging $y_a$ and $y_b$ in $S$ interchanges the $a$th and $b$th rows of $A$, since the $i$th row of $A$ is merely the $B$-expansion of $y_i$.

Our proof in the previous exercise that adding a multiple of one element in a generating set to another element does not change the generated submodule holds here as well, so the altered set $S^\prime = \{y_1, \ldots, y_a + ry_b, \ldots, y_b, \ldots, y_m\}$ is indeed a generating set for $\mathsf{ker}\ \varphi$. Moreover, it is clear that the new $a$th row is just the old $a$th row plus $r$ times the old $b$th row.

### Properties of the columns of a relations matrix

Let $R$ be a Euclidean domain and let $M$ be a finitely generated (left, unital) $R$-module. By this previous exercise, there is a surjective module homomorphism $\varphi : R^n \rightarrow M$ for some natural number $n$. By this previous exercise, $\mathsf{ker}\ \varphi \subseteq R^n$ is finitely generated as an $R$-module. If we let $B = \{x_1,\ldots,x_n\}$ be an (ordered) basis for $R^n$ and let $S = \{y_1, \ldots, y_m\}$ be an (ordered) generating set for $\mathsf{ker}\ \varphi$, then for each $y_i$, we have unique elements $a_{i,j} \in R$ such that $y_i = \sum_{j=1}^n a_{i,j} x_j$.

We now construct the $m \times n$ matrix $A = [a_{i,j}]_{i=1,j=1}^{m,n}$ whose $i$th row is simply the coefficients of $y_i$. This $A = \mathsf{Mat}^B_S(\varphi)$ is called a relations matrix with respect to the homomorphism $\varphi$, the basis $B$, and the generating set $S$.

1. Show that interchanging $x_a$ and $x_b$ in the basis $B$ has the effect of interchanging the $a$th and $b$th columns of $A$.
2. Show that, for any $r \in R$, replacing the element $x_b$ in $B$ by $x_b - rx_a$ ($a \neq b$) results in another basis. Show further that this has the effect of adding $r$ times the $b$th column to the $a$th column of $A$.

It is clear that interchanging $x_a$ and $x_b$ in $B$ merely interchanges the $a$th and $b$th columns of $A$, since the rows of $A$ are precisely the (ordered) expansions of the $y_i$ in terms of the $x_j$.

Now we will show that replacing $x_b$ by $x_b - rx_a$ results in a new basis. Recall that a basis is a linearly independent generating set. To see that $B^\prime = \{x_1, \ldots, x_a, \ldots \ldots, x_b - rx_a, \ldots, x_n\}$ is linearly independent, note that if $0 = \sum_{j \neq a,b} s_jx_j + s_ax_a + s_b(x_b - rx_a) = \sum_{j \neq a,b} s_jx_j + (s_a - rs_b)x_a + s_bx_b$, then $s_j = 0$ for all $j \neq a$ since $B$ is linearly independent, and then $s_a = 0$. Now if $\alpha \in R^n$, then since $B$ is a generating set, we have $\alpha = \sum s_jx_j$ for some $s_j \in R$. Certainly then $\alpha = \sum_{j \neq a,b} s_jx_j + (s_a + rs_b)x_a + s_b(x_b - rx_a)$, so that $B^\prime$ is a generating set, and thus a basis, for $R^n$.

Our argument that $B^\prime$ is a generating set now shows that the matrix for $B^\prime$ is obtained from that for $B$ by adding $r$ times column $b$ to column $a$.

### Finitely generated modules over R are precisely the module-homomorphic images of Rⁿ

Let $R$ be a ring. Prove that $M$ is a finitely generated module over $R$ if and only if $M$ is a module-homomorphic image of $R^n$.

Suppose first that $M$ is a finitely generated $R$-module – say by $A = \{a_1,\ldots,a_n\}$. Let $e_i$ denote the $i$th standard basis vector in $R^n$, and define $\varphi : R^n \rightarrow M$ by $e_i \mapsto a_i$, and extend linearly. Certainly then $\varphi$ is a surjective module homomorphism.

Conversely, suppose $\varphi : R^n \rightarrow M$ is a surjective module homomorphism, and for each $e_i$ let $a_i = \varphi(e_i)$. Since $\varphi$ is surjective, and since the $e_i$ generate $R^n$, the set $A = \{a_1, \ldots, a_n\}$ generates $M$ over $R$ as desired.

### If R is a Noetherian ring, then Rⁿ is a Noetherian R-module

Let $R$ be a Noetherian ring. Prove that $R^n$, as an $R$-module in the usual way, is also Noetherian.

Recall that a ring is Noetherian if every ideal is finitely generated, and a module is Noetherian if every submodule is finitely generated. (That is, a ring is Noetherian (as a ring) if it is Noetherian (as a module) over itself.)

We will proceed by induction on $n$.

For the base case, $n = 1$, $R^1$ is certainly Noetherian as a module over $R$.

For the inductive step, suppose $R^n$ is Noetherian. Now let $M \subseteq R^{n+1}$ be a submodule; our goal is to show that $M$ must be finitely generated. To that end, let $A = \{r \in R \ |\ (m_i)_0 = r\ \mathrm{for\ some}\ (m_i) \in M \}$. That is, $A$ is the collection (in $R$) of all zeroth coordinates of elements in $M$.

We claim that $A \subseteq R$ is an ideal. If $a,b \in A$, then there exist elements $(m_i)$ and $(n_i)$ in $M$ such that $m_0 = a$ and $n_0 = b$. Now since $M \subseteq R^{n+1}$ is a submodule, we have $(m_i) + r(n_i) \in M$ for all $r \in R$, so that $a+rb \in A$. We clearly have $0 \in A$, so that by the submodule criterion, $A$ is an ideal of $R$.

In particular, since $R$ is Noetherian, the ideal $A$ is finitely generated – say by $\{a_0, a_1, \ldots, a_k\}$. We will let $\alpha_i$ be an element of $M$ whose zeroth coordinate is $a_i$. Now let $m = (m_0,\ldots,m_{n+1}) \in M$. Now $m_0 = \sum c_i a_i$ for some $c_i \in R$, and so $m - \sum c_i m_i = (0,t_1,\ldots,t_{n+1})$ is an element of $M$ whose first coordinate is 0. In particular, we have $M = (m_0,\ldots,m_k) + B$, where $B = M \cap (0 \times R^n)$. (We showed the $(\subseteq)$ direction, and the $(\supseteq)$ inclusion is clear.) Now $M \cap (0 \times R^n)$ is an ideal of $0 \times R^n \cong_R R^n$, which is Noetherian by the induction hypothesis. So $B$ is finitely generated as an $R$-module, and thus $M$ is finitely generated over $R$.

Since $M$ was an arbitrary submodule of $R^{n+1}$, $R^{n+1}$ is Noetherian as a module.

### Characterize the irreducible torsion modules over a PID

Let $R$ be a principal ideal domain and let $M$ be a torsion $R$-module. Prove that $M$ is irreducible if and only if $M = Rm$ for some nonzero element $m \in M$ where the annihilator of $m$ in $R$ is a nonzero prime ideal $P$.

Suppose $M$ is an irreducible, torsion $R$-module. By this previous exercise, we have $M \cong_R R/I$ where $I \subseteq R$ is a maximal ideal. Since $R$ is a principal ideal domain, in fact $I = (p)$ for some prime element $p \in R$. Let $\psi : R/(p) \rightarrow M$ be an isomorphism, and let $m = \psi(1 + (p))$. Now $M = Rm$, and moreover $\mathsf{Ann}_R(m) = \mathsf{Ann}_R(1 + (p)) = (p)$.

Conversely, consider the module $M = Rm$, where $\mathsf{Ann}_R(m) = (p)$ is a prime ideal. Define $\psi : R \rightarrow (m)_R$ by $r \mapsto r \cdot m$. Certainly we have $\mathsf{ker}\ \psi = \mathsf{Ann}_R(m) = (p)$. By the First Isomorphism Theorem for modules, we have $\overline{\psi} : R/(p) \rightarrow M$ an isomorphism. Again by this previous exercise, $R/(p)$ is irreducible.

### Describe M/Tor(M) for a finitely generated module M over a PID

Let $M$ be a finitely generated module over a principal ideal domain $R$. Describe the structure of $M/\mathsf{Tor}(M)$.

We begin with a lemma.

Lemma: Let $R$ be a domain, and let $M$ and $N$ be (left, unital) $R$-modules. Then $\mathsf{Tor}(M \oplus N) = \mathsf{Tor}(M) \oplus \mathsf{Tor}(N)$. Proof: $(\subseteq)$ Let $(m,n) \in \mathsf{Tor}(M \oplus N)$. Then there exists a nonzero element $r \in R$ such that $r(m,n) = (0,0)$. Then $(rm,rn) = (0,0)$, and so $rm = 0$ and $rn = 0$. So $m \in \mathsf{Tor}(M)$ and $n \in \mathsf{Tor}(N)$. $(\supseteq)$ Let $(m,n) \in \mathsf{Tor}(M) \oplus \mathsf{Tor}(N)$. Then there exist nonzero elements $r,s \in R$ such that $rm = 0$ and $sn = 0$. Now $rs$ is nonzero in $R$ (since it is a domain), and we have $rs(m,n) = (s(rm), r(sn)) = 0$. So $(m,n) \in \mathsf{Tor}(M \oplus N)$. $\square$

Now by the Fundamental Theorem of Finitely Generated Modules over a PID (FTFGMPID), if $M$ is such a module, we have $M \cong_R R^t \oplus \bigoplus R/(p_k^{e_k})$ for some primes $p_k \in R$ and natural numbers $e_k$. Now $\mathsf{Tor}(M) = \mathsf{Tor}(R^t) \oplus \mathsf{Tor}(\bigoplus R/(p_k^{e_k}))$ $= \bigoplus R/(p_k^{e_k})$, since $R^t$ is torsion-free (since $R$ is a domain) while… the stuff inside the big sum is all torsion.

Using this previous exercise, we have that $M/\mathsf{Tor}(M) \cong R^t$.

### Compute a quotient module

Let $R$ be a principal ideal domain, let $a \in R$ be nonzero, and let $M = R/(a)$. Given a prime $p \in R$, say $a = p^nq$, where $p$ does not divide $q$. Prove that $(p^k)M/(p^{k+1})M$ is module-isomorphic to $R/(p)$ if $k < n$ and to $0$ if $k \geq n$.

We begin with some lemmas.

Lemma 1: Let $C \subseteq A,B$ be ideals of a ring $R$, and consider $A/C$ as an $R$-module. Then $B(A/C) = (BA)/C$. Proof: $(\subseteq)$ If $x \in B(A/C)$, then $x = \sum b_i(a_i + C) = \sum (b_ia_i + C)$ $= (\sum b_ia_i) + C$ $\in (BA)/C$. $(\supseteq)$ If $x \in (BA)/C$, then $x = (\sum b_ia_i) + C$ $= \sum (b_ia_i + C)$ $= \sum b_i(a_i+C)$. Thus $x \in B(A/C)$. $\square$

Lemma 2: Let $R$ be a principal ideal domain and let $a,b,c \in R$ be nonzero with $b|c$. Note that $(ac) \subseteq (ab)$. Prove that $(ab)/(ac) \cong_R (b)/(c)$. Proof: Let $\psi: (b) \rightarrow (ab)/(ac)$ be given by $bx \mapsto \overline{abx}$. (This is well-defined since $R$ is a domain, and is clearly an $R$-module homomorphism.) Certainly $\psi$ is surjective. Now if $bx \in \mathsf{ker}\ \psi$, then $abx \in (ac)$, so that $bx \in (c)$. Conversely, if $bx \in (c)$, then $bx = cy$ for some $y$, and so $\overline{abx} = \overline{acy} = 0$. By the First Isomorphism Theorem, $(b)/(c) \cong_R (ab)/(ac)$. $\square$

Lemma 3: Let $R$ be a principal ideal domain and $a,b,c \in R$ such that $b|c$. If $(a,c/b) = (1)$, then $(a)((b)/(c)) = (b)/(c)$. Proof: Say $c = bd$. Note that $ax+dy = 1$ for some $x,y \in R$, so that $abx + cy = b$. Now $b+(c) = abx + (c)$ $= a(bx+(c))$ $\in (a)((b)/(c))$, so that $(a)((b)/(c)) = (b)/(c)$. $\square$

If $k < n$, then $p^{k+1}|a$, so that $(a) \subseteq (p^k), (p^{k+1})$. By Lemma 1, we have $(p^k)(R/(a))/(p^{k+1})(R/(a)) = ((p^k)/(a))/((p^{k+1})/(a))$ $\cong_R (p^k)/(p^{k+1})$ by the Third Isomorphism Theorem. Using Lemma 2, this is isomorphic to $(1)/(p) \cong_R R/(p)$.

If $k \geq n$ with (say) $k = n+t$, then $((p^k)(R/(a)))/((p^{k+1})/(R/(a)) = ((p^t)((p^n)/(p^nq)))/((p^{t+1})((p^n)/(p^nq)))$ $= ((p^n)/(p^nq))/((p^n)/(p^nq))$ $= 0$, using Lemma 3 and the Third Isomorphism Theorem.

### Every torsion module over a principal ideal domain is the direct sum of its p-primary components

Let $R$ be a principal ideal domain and let $N$ be a torsion $R$-module. Prove that the $p$-primary component of $N$ is a submodule for every prime $p \in R$. Prove that $N$ is the direct sum of its $p$-primary components.

We proved this in a previous exercise.

### The quotient of a product is module isomorphic to the product of quotients

Let $R$ be a ring, let $\{A_i\}_{i=1}^n$ be a finite family of (left, unital) $R$-modules, and let $B_i \subseteq A_i$ be a submodule for each $i$. Prove that $(\prod A_i)/(\prod B_i) \cong_R \prod A_i/B_i$.

We did this previously. D&F, why repeat an exercise?