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The argument for the category of all rings works just as well for the category of Noetherian rings, since $\mathbb{Z}[x]$ is Noetherian.

However, $\mathbb{Z}[x]$ is not Artinian. So, is it still true that monomorphisms in the category of (left) Artinian rings are injective?

If the answer is "no", then perhaps there should be a non-injective monomorphism $f:A \to B$ with $A$ and $B$ commutative (which would then also answer the question for right Artinian rings, rings that are both left Artinian and right Artinian, and commutative Artinian rings).

If the answer is "yes", then perhaps the answer should still be "yes" if "left Artinian rings" were replaced with "right Artinian rings", "rings that are both left Artinian and right Artinian", or "commutative Artinian rings".

Geoffrey Trang
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Suppose $f:A\to B$ is a morphism of Artinian rings, and form the kernel pair $A\times_f A=\{(x,y)\in A\times A:f(x)=f(y)\}$ in the category of rings. I claim that $A\times_f A$ is in fact Artinian (so it is also a kernel pair in the category of Artinian rings). Indeed, note that $A\times_f A$ is an $A$-submodule of $A\times A$, and so is Artinian as an $A$-module and therefore also as a module over itself (since the action of $A$ is just via the diagonal subring of $A\times_f A$).

So, if $f$ is a monomorphism in the category of Artinian rings, this in particular implies the two projections $A\times_f A\to A$ must be equal, since they become equal after composing with $f$. It follows that $f$ is injective.

Eric Wofsey
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