Does an injective endomorphism of a finitely-generated free R-module have nonzero determinant?

Alternately, let $$M$$ be an $$n \times n$$ matrix with entries in a commutative ring $$R$$. If $$M$$ has trivial kernel, is it true that $$\det(M) \neq 0$$?

This math.SE question deals with the case that $$R$$ is a polynomial ring over a field. There it was observed that there is a straightforward proof when $$R$$ is an integral domain by passing to the fraction field.

In the general case I have neither a proof nor a counterexample. Here are three general observations about properties that a counterexample $$M$$ (trivial kernel but zero determinant) must satisfy. First, recall that the adjugate $$\text{adj}(M)$$ of a matrix $$M$$ is a matrix whose entries are integer polynomials in those of $$M$$ and which satisfies $$M \text{adj}(M) = \det(M).$$

If $$\det(M) = 0$$ and $$\text{adj}(M) \neq 0$$, then some column of $$\text{adj}(M)$$ lies in the kernel of $$M$$. Thus:

If $$M$$ is a counterexample, then $$\text{adj}(M) = 0$$.

When $$n = 2$$, we have $$\text{adj}(M) = 0 \Rightarrow M = 0$$, so this settles the $$2 \times 2$$ case.

Second observation: recall that by Cayley-Hamilton $$p(M) = 0$$ where $$p$$ is the characteristic polynomial of $$M$$. Write this as $$M^k q(M) = 0$$

where $$q$$ has nonzero constant term. If $$q(M) \neq 0$$, then there exists some $$v \in R^n$$ such that $$w = q(M) v \neq 0$$, hence $$M^k w = 0$$ and one of the vectors $$w, Mw, M^2 w,\dots, M^{k-1} w$$ necessarily lies in the kernel of $$M$$. Thus if $$M$$ is a counterexample we must have $$q(M) = 0$$ where $$q$$ has nonzero constant term.

Now for every prime ideal $$P$$ of $$R$$, consider the induced action of $$M$$ on $$F^n$$, where $$F = \overline{ \text{Frac}(R/P) }$$. Then $$q(\lambda) = 0$$ for every eigenvalue $$\lambda$$ of $$M$$. Since $$\det(M) = 0$$, one of these eigenvalues over $$F$$ is $$0$$, hence it follows that $$q(0) \in P$$. Since this is true for all prime ideals, $$q(0)$$ lies in the intersection of all the prime ideals of $$R$$, hence

If $$M$$ is a counterexample and $$q$$ is defined as above, then $$q(0)$$ is nilpotent.

This settles the question for reduced rings. Now, $$\text{det}(M) = 0$$ implies that the constant term of $$p$$ is equal to zero, and $$\text{adj}(M) = 0$$ implies that the linear term of $$p$$ is equal to zero. It follows that if $$M$$ is a counterexample, then $$M^2 \mid p(M)$$. When $$n = 3$$, this implies that $$q(M) = M - \lambda$$

where $$\lambda$$ is nilpotent, so $$M$$ is nilpotent and thus must have nontrivial kernel. So this settles the $$3 \times 3$$ case.

Third observation: if $$M$$ is a counterexample, then it is a counterexample over the subring of $$R$$ generated by the entries of $$M$$, so

We may assume WLOG that $$R$$ is finitely-generated over $$\mathbb{Z}$$.

• Out of curiosity, how did this question come up? – Mike B Jun 22 '12 at 6:40
• @Mike: it came up in the linked math.SE thread. At first I was under the impression that this could be proven using abstract nonsense but after playing around with the question a bit this seemed not to be the case. – Qiaochu Yuan Jun 22 '12 at 6:45
• I should have read the comments more closely. Lang has a section on determinants, chapter 13, section 4 in Algebra. It looks like his rings are only assumed commutative, but he has a sneaky habit of assuming more at a random point and not mentioning it after. – Mike B Jun 22 '12 at 6:54
• @QiaochuYuan, I think that Ex. 5.23A in Lam's Exercises in modules and rings, which deals with McCoy's rank, does it for you. – Mariano Suárez-Álvarez Jun 22 '12 at 6:59
• @Mariano: unfortunately I do not have access to this text. If you'd like to quote it in an answer I'll accept that. – Qiaochu Yuan Jun 22 '12 at 7:02

Yes, such an injective morphism has non-zero determinant.

Actually, if $M$ is a finitely generated free module over the commutative ring $R$ and $u:M\to M$ is an endomorphism, one has the precise equivalence: $$u \;\text {is injective}\iff \det(u) \; \text {is not a zero divisor in}\; R.$$ The proof is based on the fact that elements $m_1,m_2, \ldots ,m_n\in M$ form a linearly independent set iff there exists a non-zero $0\neq \lambda\in R$ with $\lambda (m_1\wedge m_2\wedge \ldots\wedge m_n)=0\in \Lambda^nM$.

You can find the details in Bourbaki, Algebra, III, §7, Proposition 3, page 524.

• Oh, this is a very nice argument! Exactly the kind of thing I was looking for. – Qiaochu Yuan Jun 22 '12 at 7:34
• Thanks, I'm glad you like it, Qiaochu. – Georges Elencwajg Jun 22 '12 at 7:46
• I'm sure I'm just being silly, but can you explain the very last step of the proof of Proposition 12 on page 520? I'm having trouble seeing why the existence of $f$ implies that $\mu x_1$ is a linear combination of etc. – Qiaochu Yuan Jun 22 '12 at 8:01
• Dear Qiaochu, I think the Master was a little absent-minded and He meant to use Corollary 3 of §7.4, page 512. That corollary (in which you have to replace $x_2$ by $\lambda x_2$) implies that $f(\lambda x_2,...)x_1=\mu x_1= f(x_1,x_3,...)\lambda x_2-f(x_1, \lambda x_2, x_4,...)x_3+...$ Since $\mu\neq 0$ we immediately deduce hat $x_1,x_2,...,x_n$ are linearly dependent. – Georges Elencwajg Jun 22 '12 at 8:56
• Thanks for the insightful comment! I like how Jacobson just casually throws in this exercise in his Basic Algebra 1. I think proving the $\Leftarrow$ implication is not trivial, although you and Anonymous below have provided excellent proofs. – Juan Carlos Ortiz May 28 at 16:12

Lam's Exercises in modules and rings includes the following:

which tells us that your determinant is not a zero-divisor.

The paper where McCoy does that is [Remarks on divisors of zero, MAA Monthly 49 (1942), 286--295] If you have JStor access, this is at http://www.jstor.org/stable/2303094

There is a pretty corollary there: a square matrix is a zero-divisor in the ring of matrices over a commmutative ring iff its determinant is a zero divisor.

• Many thanks, Mariano! I'm going to follow my own advice and wait a little before accepting an answer in case someone can come up with a different proof. – Qiaochu Yuan Jun 22 '12 at 7:15
• McCoy's paper is quite nice, in any case: a ten paper page which starts by defining what a ring is and manages to get to quite interesting stuff is not often spotted in nature! :) – Mariano Suárez-Álvarez Jun 22 '12 at 7:18
• I would love to read it but I won't have JSTOR access for another week or so... – Qiaochu Yuan Jun 22 '12 at 7:31

The argument from Bourbaki referenced in Georges Elencwajg's answer is exactly the kind of argument I was hoping would work so let me record it here, with some simplifications.

Proposition: Let $$m_1, ... m_n$$ be elements of some $$R$$-module $$M$$ which are linearly dependent. Then there exists nonzero $$r \in R$$ such that $$r (m_1 \wedge ... \wedge m_n) = 0$$ in $$\Lambda^n(M)$$.

Proof. If $$\sum r_i m_i = 0$$ is a linear dependence, assume WLOG that $$r_1 \neq 0$$. Then $$r_1 m_1 = - \sum_{i \ge 2} r_i m_i$$, so by basic properties of the wedge product we find that $$(r_1 m_1) \wedge m_2 \wedge ... \wedge m_n = 0$$, so we can take $$r = r_1$$.

Proposition: If $$M$$ is free then the converse holds.

Proof. We proceed by induction on $$n$$. If $$n = 1$$ this is clear. In general, suppose that there exists a nonzero $$r$$ such that $$r (m_1 \wedge ... \wedge m_n) = 0.$$

If $$r (m_2 \wedge ... \wedge m_n) = 0$$, then by the inductive hypothesis $$m_2, ..., m_n$$ are linearly dependent and we are done, so suppose otherwise. Then $$r (m_2 \wedge ... \wedge m_n) \neq 0$$. Hence, using the fact that $$\Lambda^{n-1}(M)$$ is also free (this is crucial!), we conclude that there exists an alternating $$n-1$$-form $$f : M^{n-1} \to R$$ such that $$f(r m_2, ..., m_n) = s \neq 0$$ (because of the following fact: if $$g$$ is a nonzero vector in a free $$R$$-module $$F$$, then there exists some $$R$$-linear map $$\alpha : F \to R$$ such that $$\alpha(g) \neq 0$$).

But since $$m_1 \wedge (rm_2) \wedge ... \wedge m_n = 0$$, the alternating $$n$$-form $$x_1 f(x_2, ..., x_n) - x_2 f(x_1, x_3, ..., x_n) \pm ...$$ necessarily vanishes on it, so $$m_1 f(rm_2, ..., m_n) = m_1 s = rm_2 f(m_1, m_3, ..., m_n) \mp ...$$ and we conclude that the $$m_i$$ are linearly dependent.

Corollary: Let $$f : M \to N$$ be an injective map of free modules. Then the induced map $$\Lambda(f) : \Lambda(M) \to \Lambda(N)$$ is also injective.

Proof. Suppose otherwise. Let $$e_i, i \in I$$ be an ordered basis of $$M$$. For a finite subset $$S$$ of $$I$$, let $$e_S$$ denote the wedge of all of the elements of $$S$$ (in the order determined by the order of $$I$$). If $$\Lambda(f)$$ is not injective, then let $$\sum c_S e_S$$ be some element of its kernel. Since the kernel is a two-sided ideal (because $$\Lambda(f)$$ is a ring homomorphism), we may freely take exterior products on either side (this is also crucial!). Now, if $$\sum c_S e_S$$ has at least two nonzero terms in it, then there exists $$i$$ such that $$e_i$$ appears in some term but not every term, so it follows that $$\sum c_S e_S \wedge e_i \in \text{ker}(\Lambda(f))$$

as well. This is an element of the kernel with strictly fewer nonzero terms. Hence an element of the kernel with the minimal number of nonzero terms necessarily has a single nonzero term $$c_S e_S$$. Writing this as $$c_S e_1 \wedge ... \wedge e_k$$

and applying $$\Lambda(f)$$ we get $$c_S f(e_1) \wedge ... \wedge f(e_k) = 0.$$

By the above, it follows that $$f(e_1), ... f(e_k)$$ are linearly dependent, but this contradicts $$f$$ injective.

Corollary: Let $$f : R^n \to R^n$$ be an endomorphism of a free module. Then $$f$$ is injective if and only if $$\det(f)$$ is not a zero divisor.

Proof. By the above, if $$f$$ is injective then $$\Lambda^n(f) : \Lambda^n(R^n) \to \Lambda^n(R^n)$$ is also injective. Since it acts by multiplication by $$\det(f)$$, we conclude that $$\det(f)$$ is not a zero divisor. If $$f$$ is not injective then some $$\lambda_1 e_1 + \lambda_2 e_2 + \cdots + \lambda_n e_n$$ lies in the kernel of $$\Lambda(f)$$, where not all $$\lambda_i$$ are $$0$$; thus, by taking exterior products we conclude that $$\lambda_i e_1 \wedge ... \wedge e_n$$ also lies in the kernel of $$\Lambda(f)$$ for all $$i$$, and therefore $$\lambda_i \det(f) = 0$$ for all $$i$$.

• @ Georges Elencwajg :Since this answer is based on your reference may I ask you some doubt ? In proposition 2 . why we will get an alternating (n-1)-multilinear form which is non zero at $(rm_2,\cdots,m_n)$ ! We know that $\{rm_2,\cdots,m_n\}$ is an linearly independent set in the free module $M.$ – user371231 Apr 11 '18 at 11:18
• @user371231: Added some details to that argument. – darij grinberg Nov 14 '18 at 2:16

Here is an elementary proof of the fact that if the determinant $$D$$ of an $$n \times n$$ matrix M is a zero-divisor, then there is a nonzero vector $$X$$ such that $$MX = 0$$.

Let $$a$$ be a nonzero scalar such that $$aD = 0$$. Let $$M'$$ be a square submatrix of $$M$$ of maximum size $$r$$ such that $$a \det M' \ne 0$$. (If there is no such submatrix, let $$r = 0$$.) We have $$r < n$$ by the definition of $$a$$. After permuting the rows and columns of $$M$$ if necessary, we may assume that $$M'$$ is located in the top left corner of $$M$$.

Let $$M''$$ be the $$r \times (r + 1)$$ matrix in the top left corner of $$M$$, and let $$d_j$$, for $$j = 1, \dots, r+1$$, be the minor of $$M''$$ obtained by deleting its $$j$$th column.

Now define $$X = aX_0$$, where $$X_0 = (d_1, -d_2, \dots, (-1)^r d_{r+1}, 0, \dots, 0).$$ We have $$ad_{r+1} = a\det M' \ne 0$$, so $$X \ne 0$$.

I claim that $$MX = 0$$. If $$i > r$$, then the $$i$$th coordinate of $$MX_0$$ is, up to sign, the minor of $$M$$ obtained from its first $$r + 1$$ columns and rows $$1, 2, \dots, r, i$$. By the definition of $$r$$, this minor becomes zero upon multiplication by $$a$$. When $$i \leq r$$, the the $$i$$th coordinate of $$MX_0$$ is zero because it is, up to sign, the determinant obtained by extending $$M''$$ with its own $$i$$th row. This proves the claim.