# Right invertible and left zero divisor in matrix rings over a commutative ring

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 ?

• Try to use the classical adjoint in order to find $B$ from $AB=1$.
– user26857
Commented Feb 9, 2013 at 16:29
• For the case of fields, see math.stackexchange.com/questions/3852 Commented Feb 9, 2013 at 16:53
• What is the answer to "is every left zero divisor also a right divisor ?" finally? I see no helpful answer...
– user171110
Commented Jul 20, 2017 at 15:19
• 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.) Commented Oct 23, 2023 at 20:45
• 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 $B=0$. Commented Oct 23, 2023 at 21:23

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.

• How does the Corollary follow from the Lemma? Commented Aug 26, 2022 at 6:21
• @Filippo See the third paragraph of Theon Alexander's answer. Commented Oct 20, 2023 at 0:54

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.

• "...so $\det A$ and $A$ are invertible" - Why does this tell us that $BA=1$? Commented Aug 26, 2022 at 6:13
• @Filippo once you know that $A$ has a (two-sided) inverse, this coincides with any one-sided inverse. Commented Aug 26, 2022 at 8:12

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.

• In the case where $\det(A)$ is a zero-divisor, it's possible that $\text{adj}(A)b=0$. Commented Oct 19, 2023 at 22:21

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

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

• This is already in @AndreasCaranti's answer. Commented Oct 19, 2023 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 $C\neq0$. Commented Oct 19, 2023 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). Commented Oct 20, 2023 at 5:27