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.
Best Answer
Let $S$ be a non-zero ring and take infinity many direct
sumproduct $R = S^{\mathbb{N}}$. 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.