The problem is stated as follows:

"Let $R$ be a Noetherian ring and $\theta$ be a ring homomorphism from $R$ to $R$. Show that if $\theta$ is surjective then it is also injective."

Regardless of the right solution, I don't understand why is the following wrong:

We have $\theta: R\to R$. By the isomorphism theorem $R/\ker\theta\cong\operatorname{Im}\theta$. Since $\operatorname{Im}\theta = R$, it follows $\ker\theta =0$, so it's injective.

Harsh criticism will be appreciated. Thanks.

  • $\begingroup$ I have the right solution for that somewhere, but I wanted to first remove this wrong approach from my head. $\endgroup$ – Kristina Aug 26 '15 at 14:02

Let $S$ be a non-zero ring and take infinity many direct sum product $R = S^{\mathbb{N}}$. Then the homomorphism defined by $$ f \colon R \to R,\ (r_1, r_2, r_3, \dotsc) \mapsto (r_2, r_3, r_4, \dotsc) $$ is clearly surjective but its kernel is $\ker f = S \times 0 \times 0 \times \dotsb \neq 0$.

So naive inference "$f \colon R \to R,\ \operatorname{im}f = R \implies \ker f = 0$" is wrong.

  • $\begingroup$ Great example, thanks. So the isomorphism theorem here looks like $R/S \cong R$ ? $\endgroup$ – Kristina Aug 26 '15 at 13:56
  • $\begingroup$ Yes, indeed. $ $ $\endgroup$ – Orat Aug 26 '15 at 13:58
  • $\begingroup$ TS asked about Noetherian rings. $\endgroup$ – Canis Lupus Sep 20 '16 at 3:25
  • $\begingroup$ @Corvus No. She asked why the same is not true (or more precisely, why the above argument is wrong) without Noetherian assumption. $\endgroup$ – Orat Sep 20 '16 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.