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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 $w, Mw, M^2 w, ... M^{k-1} w$ 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 $M^2 | p(M)$. 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 $\mathbb{Z}$.

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Out of curiosity, how did this question come up? –  Mike B Jun 22 '12 at 6:40
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@Patrick: sorry, I meant $M^k w = 0$ where $w = q(M) v$. I'll edit. –  Qiaochu Yuan Jun 22 '12 at 6:44
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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
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@Mike B : It is personal taste, but I hate Lang's book just because you can't learn anything from it. It's a good reference, but I wouldn't suggest it for anything else. –  Patrick Da Silva Jun 22 '12 at 6:55
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@Mariano Suarez-Alvarez : Perhaps it would've been a good thing to do. Thanks for calming my anger against Lang's books :) but I can't edit this comment anymore.. –  Patrick Da Silva Jun 22 '12 at 7:10

3 Answers 3

up vote 11 down vote accepted

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.

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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
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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
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Is there no "add to favorites" for answers?! –  rschwieb Jun 22 '12 at 14:59

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

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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.

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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-Alvarez 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 $r \in R$ 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 $r$ such that $$r (m_1 \wedge ... \wedge m_n) = 0.$$

If $r (m_2 \wedge ... \wedge m_n) = 0$, then by the inductive hypothesis $r m_2, ... m_n$ are linearly dependent and we are done, so suppose otherwise. Then 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 $$f(r m_2, .... m_n) = s \neq 0.$$

But since $m_1 \wedge (rm_2) \wedge ... \wedge m_n = 0$, the alternating $n$-form $$x_1 f(x_2, ... x_n) - x_2 f(x_1, ... x_n) \pm ...$$

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 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 $f$ 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 $e_i$ lies in the kernel of $\Lambda(f)$, and by taking exterior products we conclude that $e_1 \wedge ... \wedge e_n$ also lies in the kernel of $\Lambda(f)$.

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