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 ath and bth 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 bth row to the a th row.

It is clear that interchanging y_a and y_b in S interchanges the ath and bth rows of A, since the ith 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 ath row is just the old ath row plus r times the old bth 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 ith 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 ath and bth 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 bth column to the ath column of A.

It is clear that interchanging x_a and x_b in B merely interchanges the ath and bth 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 ith 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?