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If a ring $R$ is commutative, I don't understand why if $A, B \in R^{n \times n}$, $AB=1$ means that $BA=1$, i.e., $R^{n \times n}$ is Dedekind finite.

Arguing with determinant seems to be wrong, although $\det(AB)=\det(BA ) =1$ but it necessarily doesn't mean that $BA =1$.

And is every left zero divisor also a right divisor ?

user26857
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Theorem
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    Try to use the classical adjoint in order to find $B$ from $AB=1$. –  Feb 09 '13 at 16:29
  • For the case of fields, see http://math.stackexchange.com/questions/3852 – Martin Brandenburg Feb 09 '13 at 16:53
  • What is the answer to "is every left zero divisor also a right divisor ?" finally? I see no helpful answer... –  Jul 20 '17 at 15:19
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    Regarding zero-divisors - It is true if the ring is finite. Indeed in that case $A^i=A^{i+k}$ for some $i,k\geq1$, so $A^i(1-A^k)=0$. Take the minimal such $i$, so $B=A^{i-1}(1-A^k)\neq0$. Then we have $AB=BA=0$. (This excludes the case where $(1-A^k)=0$, which would make $A$ invertible and not a zero-divisor.) – mr_e_man Oct 23 '23 at 20:45
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    And it's true if $n=2$, since the adjugate matrix $A'$ has the same components as $A$, just rearranged or negated. We have the identity $$AA'=A'A=(\det A)I.$$ If $A\neq0$ and $\det A=0$ then we can take $B=A'\neq0$ to get $AB=BA=0$. If $\det A\neq0$ but $b\det A=0$ for some scalar $b\neq0$, then we can take $B=bA'$, unless this is $0$; but if $bA'=0$ then also $bA=0$, so we can take $B=bI$. If $\det A\neq0$ and $b\det A\neq0$ for all scalars $b\neq0$, then $A$ is not a zero-divisor; if $AB$ or $BA=0$ then multiplying by $A'$ gives $(\det A)B=0$ and thus $B=0$. – mr_e_man Oct 23 '23 at 21:23
  • So any counter-example needs to have $R$ infinite and $n\geq3$. – mr_e_man Oct 23 '23 at 21:26

4 Answers4

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Instead of working with matrices, let us with endomorphisms of finite free modules, or more generally finitely generated modules. In the following, $R$ is a commutative ring.

Lemma. Every surjective endomorphism $f : M \to M$ of a finitely generated $R$-module $M$ is an isomorphism.

Proof. $M$ becomes an $R[x]$-module, where $x$ acts by $f$. By assumption, $M=xM$. Nakayama's Lemma implies that there is some $p \in R[x]$ such that $(1-px)M=0$. This means $\mathrm{id}=p(f) f$. Hence, $f$ is injective. $\square$

Corollary: If $f,g$ are endomorphisms of a finitely generated $R$-module satisfying $f \circ g=\mathrm{id}_M$, then also $g \circ f=\mathrm{id}_M$.

This follows from the lemma, since $f$ is surjective, hence bijective, and then $g$ is the inverse map of $f$.

About the question regarding zero divisors: I am pretty sure that this is not true. One can show that $A \in M_n(R)$ is left regular iff its columns are linearly independent, and that $A \in M_n(R)$ is right regular iff its rows are linearly independent. There should be no connection here. I will add a specific counterexample when I find it.

Martin Brandenburg
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I believe arguing with the determinant works, as $1 = A B$ implies $1 = \det(A B) = \det(A) \det(B)$, so $\det(A) \in R$ is invertible, and $A$ is.

PS I believe this argument is implicit in @YACP comment to the original post.

Andreas Caranti
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Here's a slightly different argument.

The poster's initial argument was by using determinants, which as said above, implies that both $\det(A), \det(B) \in R^\times.$ Hence, both matrices have bilateral inverses, i.e. $\operatorname{adj}(A^T) \det(A)^{-1}$ and resp. for $B$ - recall that the formula $$A\operatorname{adj}(A^T)= \operatorname{adj}(A^T)A= (\det A) I$$ is universal.

Now, for two maps (of sets) $f:S\to T$ and $g:T\to S$, if $f$ or $g$ is bijective and $fg=id_T$, then clearly $gf=id_S.$ This settles the first question.

Zero-divisor question: Suppose that $AB=0$ for $A, B \neq 0$ square matrices; then $\det(A) \det(B)=0.$ Will there be a square matrix $C\neq 0$ such that $CA=0$

If $\det(A)$ is a zero divisor (and by definition, non-zero), we're done, since for $0\neq b\in R$ satisfying $\det(A)b=0$, the matrix $C=\operatorname{adj}(A^T)b$ is by the above a zero-divisor on both sides!

The case I still haven't worked out is the following. If $\det(A)=0$ and $\operatorname{adj}(A^T)\neq 0$, then $C=\operatorname{adj}(A^T)$ satisfies $CA=AC=0$, but what if $\operatorname{adj}(A)=0$?

So, the case that remains (so far) is when $\operatorname{adj}(A)=0$, which is equivalent to the condition $\Lambda^{n-1}A=0$ (exterior product). We'll come back later.

user26857
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Theon Alexander
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If you'd like to use the determinant, the following argument works:

$$\det(A) \det(B) = \det (AB) = \det (1) = 1$$

so $\det(A)$ is a unit in $R$ and hence $A$ is invertible. Let $C$ denote the inverse of $A$, so that $AC = CA = 1$. Then

$$ BA = 1(BA) = (CA)(BA) = C(AB)A = C (1) A = CA = 1.$$

Mark
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  • This is already in @AndreasCaranti's answer. – mr_e_man Oct 19 '23 at 22:38
  • The existing answers show that if $AB=1$ then $BA=1$. What remains is to show that if $AB=0$ for some $B\neq0$ then $CA=0$ for some $C\neq0$. – mr_e_man Oct 19 '23 at 22:44
  • I saw your bounty. It would be better to create a new question for it. 1) Then the question will be easier to find. 2) It will be seen by more people. 3) The answer for the 2nd problem can be accepted as well (this doesn't work here). – Martin Brandenburg Oct 20 '23 at 05:27