# A surjective endomorphism (of a Noetherian ring) is injective.

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.

• I have the right solution for that somewhere, but I wanted to first remove this wrong approach from my head. – 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.
• Great example, thanks. So the isomorphism theorem here looks like $R/S \cong R$ ? – Kristina Aug 26 '15 at 13:56
• Yes, indeed.  – Orat Aug 26 '15 at 13:58