# Injective ring homomorphism between finite direct product of rings

All rings below are commutative with unity.

Let $R$ be a Noetherian ring such that there is an injective ring homomorphism from $R^m$ to $R^n$ , then is it true that $m\le n$ ? If not true in general then can we impose any condition (apart from Artinian) on $R$ to make the statement true ?

( NOTE : I know the statement to be true for any ring if we consider injective module homomorphism ... )

• Be careful, in the preamble, the affirmation 'A similar argument , remembering the fact than an injective ring endomorphism of Artinian ring is isomorphism' is not correct; it is enough to take an endomorphism of a field that is not an automorphism: $\eta:\mathbb{Q}(\pi)\to\mathbb{Q}(\pi)$, with $\eta(\pi)=\pi^2$. – arienda Jul 30 '17 at 18:24
• @arienda : true . I have removed my remarks . I was making the mistake that $Im f$ is an ideal , which it is not . Thank you – user Jul 30 '17 at 18:27

Suppose $\alpha:R^n\times R\to R^n$ is an injective ring homomorphism. Extend the codomain to get a (non-unital) endomorphism $\beta:R^n\times R\to R^n\times R$, where $\beta(x)=(\alpha(x),0)$.

Consider the idempotent element $e_0=(0,1)\in R^n\times R$ and its images $e_n=\beta^n(e_0)$ under powers of $\beta$. Then $\{e_n\vert n\geq0\}$ is an infinite set of orthogonal idempotents ($e_ie_j=0$ for $i\neq j$) since $e_0\beta(e_0)=0$.

The ideal generated by $\{e_n\vert n\geq0\}$ is infinitely generated, so $R^n\times R$ is not noetherian. Since a finite direct product of noetherian rings is noetherian, $R$ is not noetherian.

• From what I can see , the injectivity of $\beta$ was not fully needed right ? All was needed is that $\beta$ sends a non-zero idempotent to a non-zero idempotent ... right ? – user Jul 28 '17 at 14:30
• @users Yes, that's right. – Jeremy Rickard Jul 28 '17 at 14:41

You can easily reduce to the Artinian case by localizing at a minimal prime. Localization is flat, so you still get an injective homomorphism and the ring is now Artinian.

• the example here math.stackexchange.com/questions/223895/… suggests that localization need not preserve injectivity of ring homomorphism ... – user Jul 27 '17 at 16:55
• Injectivity is preserved under localizations since it is flat as I said. Easy to prove. The example in the post there is not just localization. There, the map $A_P\to B_P$ would still be injective, but further localization $B_P\to B_Q$ may not be. – Mohan Jul 27 '17 at 17:40
• I'm not quite sure what you mean by localizing here. The injective ring homomorphism $R^m\to R^n$ may not be an $R$-module homomorphism. – Jeremy Rickard Jul 27 '17 at 23:00