Tag Archives: integral domain

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).

Exhibit a nonzero torsion module over an integral domain whose annihilator is trivial

Exhibit an integral domain R and a nonzero torsion module M over R such that \mathsf{Ann}(M) = 0. Prove that if N is a finitely generated torsion R-module (R a domain), then \mathsf{Ann}(N) \neq 0.


First consider the direct sum M = \bigoplus_{\mathbb{N}^+} \mathbb{Z}/(2^k) as a \mathbb{Z}-module in the usual way. Every element \alpha of M is annihilated by 2^t, where t is the largest nonzero component of \alpha. However, for all k \in \mathbb{Z}, there exist elements not annihilated by k. For example, if 2^t \leq k < 2^{t+1}, then the element with 1 in the t+1 component an 0 elsewhere is not annihilated by k.

Suppose now that N is a finitely generated torsion R-module; say N = (A)_R, with A = \{a_i\}_{i=1}^n. For each a_i, there exists a nonzero element r_i \in R such that r_ia_i = 0, since N is torsion. Let r = \prod r_i. Since R is an integral domain, r \neq 0, and certainly rN = 0. Thus \mathsf{Ann}(N) \neq 0.

Nonprincipal ideals of an integral domain have rank 1 but are not free

Let R be an integral domain, and let M \subseteq R be a nonprincipal ideal of R, considered as a (left, unital) R-module. Prove that M has rank 1 over R but is not free as an R-module.


Let \{x,y\} \subseteq M be a subset containing two nonzero elements. Since x \cdot y + (-y) \cdot x = 0, \{x,y\} is not linearly independent. Thus the rank of M as an R-module is at most 1. Since R is an integral domain, every (nonzero) singleton set is linearly independent; hence M has rank 1.

Suppose M is free. Then M must have free rank 1, since any free generating set is linearly independent. That is, M = R\alpha for some \alpha. But then M is principal, a contradiction.

Over an integral domain, rank N + rank M/N = rank M

Let R be an integral domain, let M be a (left, unital) R-module, and let N \subseteq M be a submodule. Prove that \mathsf{rank}(N) + \mathsf{rank}(M/N) = \mathsf{rank}(M).


Suppose S = \{s_i\} \subseteq N is a cardinality-maximal linearly independent set, and choose T = \{t_i\} \subseteq M such that \overline{T} \subseteq M/N is a cardinality-maximal linearly independent set and T and \overline{T} have the same cardinality. Note in particular that S \cap T is empty. We claim that S \cup T is linearly independent in M. To see this, suppose \sum r_i s_i + \sum r_i^\prime t_i = 0. Mod N, we see that \sum r_i^\prime \psi(t_i) = 0, so that r_i^\prime = 0. But then \sum r_is_i = 0, so that r_i = 0. Thus S \cup T is linearly independent in M. In particular, \mathsf{rank}(M) \geq \mathsf{rank}(N) + \mathsf{rank}(M/N).

Now let A \subseteq M be the submodule generated by S \cup T. We claim that M/A is torsion. Suppose to the contrary that M/A is not torsion; that is, that there exists y \in M/A such that if ry \in A, then r = 0. In other words, if ry + \sum r_i s_i + \sum r_i^\prime t_i = 0, then r = 0, and hence r_i = r_i^\prime = 0. Thus S \cup T \cup \{y\} is linearly independent. If y \in N, then we violate the maximality of S, and if y \notin N, then we violate the maximality of T. So M/A must be torsion. By this previous exercise, \mathsf{rank}(M) = \mathsf{rank}(N) + \mathsf{rank}(M/N).

Over an integral domain, the rank of a direct sum is the sum of the ranks

Let R be an integral domain and let A and B be (left, unital) R-modules. Prove that \mathsf{rank}(A \oplus B) = \mathsf{rank}(A) + \mathsf{rank}(B), where the rank of a module is the largest possible cardinality of a linearly independent subset.


Suppose A has rank n and B has rank m. By the previous exercise, there exist free submodules A_1 \subseteq A and B_1 \subseteq B having free ranks n and m, respectively, such that the quotients A/A_1 and B/B_1 are torsion. Note that A_1 \oplus B_1 \subseteq A \oplus B is free. By this previous exercise, we have (A \oplus B)/(A_1 \oplus B_1) \cong_R (A/A_1) \oplus (B/B_1). Note that since R is an integral domain, finite direct sums of torsion modules are torsion. Thus (A \oplus B)/(A_1 \oplus B_1) is torsion. Since A_1 \oplus B_1 is free and has free rank n+m, by this previous exercise, A \oplus B has rank n+m.

A characterization of rank in modules over an integral domain

Let R be an integral domain and let M be a (left, unital) R-module.

  1. Suppose M has rank n, and that S = \{m_i\} is a linearly independent subset of M having maximal cardinality. Let N be the submodule of M generated by S. Prove that N is a free R-module having free rank n. Prove also that M/N is torsion.
  2. Conversely, prove that if there exists a submodule N \subseteq M which is free and has free rank n and such that M/N is torsion, then M has rank n.

  1. Note that N is free since its generating set S is R-linearly independent, and that S is a free generating set. Thus N has free rank n. Now let x+N \in M/N. Note in particular that x and S = \{m_i\} are linearly dependent in M, since S was a linearly independent set of maximal cardinality. Thus there exist s,r_i \in R such that at least one of the s,r_i is nonzero and sx + \sum r_im_i = 0. Suppose s = 0; then in fact \sum r_i m_i = 0, so that r_i = 0 for all i– a contradiction. Thus s \neq 0. Now sx \in N, so that s(x+N) = 0 in M/N. Thus x+N is a torsion element. Since x+N was arbitrary, M/N is torsion.
  2. Let S \subseteq N be a free generating set (of cardinality n). It is certainly the case that S is linearly independent in M, so that the rank of M is at least n. Now let T = \{t_i\}_{i=1}^{n+1} be a set of n+1 elements in M. Since M/N is torsion, for each t_i, there exists a nonzero element r_i \in R such that r_it_i \in N. If we have r_it_i = r_jt_j for some i and j, then T is linearly dependent. Suppose this is not the case, so that \{r_it_i\} \subseteq N is a subset having cardinality n+1. Since N is free of free rank n, there exist nonzero s_i such that \sum s_ir_it_i = 0. In particular, T is linearly dependent in M. Hence the rank of M is at most n. Thus the rank of M is precisely n.

Over an integral domain, the modules R and R/Tor R have the same rank

Let R be an integral domain and let M be a (left, unital) R-module. Recall that the rank of a module is the largest possible cardinality of a linearly independent subset. \mathsf{Tor}\ M is the set of all m \in M such that rm = 0 for some nonzero r \in R.

  1. Prove that \mathsf{Tor}\ M has rank 0.
  2. Prove that M and M/\mathsf{Tor}\ M have the same rank.

Let S \subseteq \mathsf{Tor}\ M be a nonempty set. Note that for each m \in S, there exists a nonzero element r \in R such that rs = 0. In particular, S cannot be linearly independent. Thus \mathsf{Tor}\ M has rank 0.

Let S = \{m_i\}_I \subseteq M be a linearly independent set of maximal order. Note in particular that no element of S can be in \mathsf{Tor}\ M (each subset of S is linearly independent, in particular the singleton subsets are linearly independent.) We claim that the subset S + \mathsf{Tor}\ M \subseteq M/\mathsf{Tor}\ M is linearly independent. To see this, suppose \sum r_i m_i \in \mathsf{Tor}\ M for some r_i \in R. Then there exists a nonzero element a \in R such that \sum ar_im_i = 0 in M. Since S is linearly independent, we have ar_i = 0 in R. Since R is an integral domain, and a \neq 0, we have r_i = 0 for all i. Thus S + \mathsf{Tor}\ M is linearly independent. We claim also that S and S + \mathsf{Tor}\ M have the same cardinality. It suffices to show that for all i and j, m_i - m_j \notin \mathsf{Tor}\ M. To that end, suppose a \in R such that am_i - am_j = 0. Since m_i and m_i are linearly independent, a = 0. Thus S and S + \mathsf{Tor}\ M have the same cardinality. In particular, the rank of M/\mathsf{Tor}\ M is at least as great as the rank of M.

Conversely, let S = \{m_i\} and suppose S + \mathsf{Tor}\ M is a linearly independent subset of maximal order. We claim that S is linearly independent in M. To see this, suppose \sum r_i m_i = 0. Now \sum r_i \overline{m_i} = 0 in M/\mathsf{Tor}\ M. Thus r_i = 0, and so S \subseteq M is linearly independent. Certainly the cardinality of S is at least as great as the cardinality of S + \mathsf{Tor}\ M. Thus the rank of M is at least as great as the rank of M/\mathsf{Tor}\ M.

Hence M and M/\mathsf{Tor}\ M have the same rank.

Exterior powers of subdomains of a field of fractions

Let R be an integral domain, and let F be its field of fractions.

  1. Considering F as an R-module in the usual way, prove that \bigwedge^2(F) = 0.
  2. Let I \subseteq F be a submodule. Prove that \bigwedge^k(I) is torsion for all k \geq 2.
  3. Exhibit an integral domain R and an R-submodule I \subseteq F such that \bigwedge^k(I) \neq 0 for all k \geq 0.

Let \frac{a}{b} \otimes \frac{c}{d} \in \mathcal{T}^2(F) be a simple tensor. Note that \frac{a}{b} \otimes \frac{c}{d} = \frac{ad}{bd} \otimes \frac{cb}{bd} = abcd(\frac{1}{bd} \otimes \frac{1}{bd}) \in \mathcal{A}^2(F). In particular, we have \frac{a}{b} \wedge \frac{c}{d} = 0 in \bigwedge^2(I).

Suppose \frac{a_1}{b_1} \wedge \frac{a_2}{b_2} \wedge \cdots \wedge \frac{a_k}{b_k} \in \bigwedge^k(I) be nonzero; then the a_i are nonzero, and certainly the b_i are nonzero. Now a_1a_2b_1b_2 \neq 0 in R, and evidently a_1a_2b_1b_2(\frac{a_1}{b_1} \wedge \frac{a_2}{b_2} \wedge \cdots \wedge \frac{a_k}{b_k}) =  \frac{a_1a_2}{1} \wedge \frac{a_1a_2}{1} \wedge \cdots \wedge \frac{a_k}{b_k} = 0. So every element of \bigwedge^k(I) is torsion, and thus \bigwedge^k(I) is torsion as an R-module.

Now consider R = \mathbb{Z}[x_1,\ldots,x_n], and let I = (x_1,\ldots,x_n). We claim that if \sum \alpha_ix_i = \sum \beta_ix_i \in I, then there exist h_i \in I such that \alpha_i - \beta_i = h_i. To see this, choose some j and consider \alpha_jx_j - \beta_jx_j = \sum_{i \neq j} (\beta_i - \alpha_i)x_i. Since R is a domain, x_j divides the right hand side of this equation; say \sum_{i \neq j} (\beta_i - \alpha_i)x_i = x_jh_j. Note that every term on the left hand side is divisible by some x_i other than x_j. Hence every term in h_j is divisible by some x_i other than x_j. In particular, h_j \in I, and \alpha_i - \beta_i = h_i.

Now consider the elements of \prod^k I as “column vectors”- that is, write \sum \alpha_i x_i as [\alpha_1\ \alpha_2\ \ldots\ \alpha_n]^\mathsf{T}. Note that this does not give a unique representation of the elements of I. However, if two column vectors A and B represent the same element in I, then there exists a third column vector H such that A = B+H, and moreover the entries of H are in I.

Now write the elements of R^k as (square) matrices. Let us consider the determinant of such a matrix A, as an element of R, reduced mod I. This is an alternating bilinear map on the set of matrices over R; we claim that this is also well-defined up to our nonunique identification of matrices over R with elements of I^k. To that end, suppose matrices A and B represent the same element of I^k; then we have a matrix H such that A = B+H. Consider the determinant of both sides mod I. Using the combinatorial formula for computing determinants, we have \mathsf{det}(B+H) = \sum_{\sigma \in S_n} \epsilon(\sigma) \prod (\beta_{\sigma(i),i} + h_{\sigma(i),i}). Note that each h_{i,j} is divisible by some x_i, and so goes to zero in the quotient R/I. So in fact \mathsf{det}(A) = \mathsf{det}(B+H) \equiv \mathsf{det}(B) \mod\ I; thus the map \mathsf{det} : I^k \rightarrow R/I is a well-defined alternating bilinear map. Since \mathsf{det}(x_1 \otimes \cdots \otimes x_n) = 1 is nonzero, this map is nontrivial. Thus \bigwedge^k(I) \neq 0 for all k.

Over an integral domain, every maximal ideal is irreducible

Let R be an integral domain. Prove that every maximal ideal in R is irreducible.


We begin with a lemma. This lemma is actually true (with a minor modification) over a much larger class of rings. However, I am not entirely comfortable with the proof of the general version. I am, however, comfortable with this more specific version. I referred to a discussion at MathForum.org for a proof of the lemma.

Lemma: Let R be an integral domain and let I \subseteq R be an ideal which is finitely generated as a \mathbb{Z}-module. If I^2 = I, then either I = 0 or I = R. Proof: Suppose I \neq 0. Let A = \{\alpha_i\}_{i=1}^n be a generating set for A over \mathbb{Z}; in particular, some \alpha_i is nonzero. Since I^2 = I, there exist u_{i,j} \in A such that \alpha_i = \sum_j u_{i,j} \alpha_j for each i. In particular, 0 = \sum_j (u_{i,j} - \delta_{i,j})\alpha_j for each i, where \delta_{i,j} is the Kronecker delta. That is, [\alpha_1\ \cdots\ \alpha_n]^\mathsf{T} is a nontrivial solution to the matrix equation Mx = 0, where M = [u_{i,j} - \delta_{i,j}]. Thinking of R as embedded in its field F of fractions, we have \mathsf{det}(M) = 0. On the other hand, by the Leibniz expansion of \mathsf{det}(M), we have \mathsf{det}(M) = \beta - 1, where \beta \in I (using the fact that I is an ideal). In particular, -1 \in I, so that I = R. \square

Now let M \subseteq R be nonzero and maximal. If M = AB, then M \subseteq A and M \subseteq B, so that A,B \in \{M, R\}. If A = B = M, then M^2 = M. By the lemma, either M = 0 (a contradiction) or M = R (also a contradiction). So either A or B is R. Since R is a unit in the semigroup of ideals of R under ideal products, M is irreducible.

The tensor square of a princopal ideal in a domain is torsion free

Let R be an integral domain and let I \subseteq R be a principal ideal. Prove that the (left) R-module I \otimes_R I has no nonzero torsion elements; that is, if m \in I \otimes_R I and r \in R is nonzero such that r \cdot m = 0 then m = 0.


Say I = (a). Note that every simple tensor (hence every element) of I \otimes_R I can be written in the form ax \otimes ay = xy \cdot (a \otimes a). We claim that I \otimes_R I is free on a \otimes a. To see this, note that by the universal propery of free modules we have a module homomorphism R \rightarrow I \otimes_R I given by r \mapsto r \cdot (a \otimes a).

Now let \psi : I \times I \rightarrow R be given by (xa,ya) = xy. Clearly \psi is bilinear, and so induces a module homomorphism \Psi : I \otimes_R I \rightarrow R with \Psi(xa \otimes ya) = xy.

Note that (\Psi \circ \Phi)(r) = \Psi(ra \otimes a) = r and (\Phi \circ \Psi)(xa \otimes ya) = \Phi(xy) = xa \otimes ya. Thus \Phi and \Psi are mutual inverses, so that I \otimes_R I \cong_R R. Since R is a domain, R is torsion free as an R-module, and thus I \otimes_R I has no nonzero torsion elements.