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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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