The problem says that prove that a nonzero homomorphic image of a local ring is a local ring.
Could you give me a scratch of a proof for this or maybe a full answer if you don't mind?
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The reason is that for every ring $R$ and a proper ideal $I$ there's a one on one correspondence between ideals of $R$ containing $I$ and ideals of $R/I$.
Since for every homomorphism $f\colon R \to S$ we have that $\text{Im} f \cong R/\ker f$, we have that the ring $\text{Im}f$ have to have as much maximal ideals as the maximal ideals of $R$ which contain $I=\ker f$. Since $R$ has just one maximal ideal also $\text{Im}f$ have a unique maximal ideal, hence it's local.
Giorgio Mossa
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In the second paragraph, you mean the kerenel of $f$ by $I$. Right? – le4m Oct 22 '13 at 20:41
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@julypraise yes, my bad :) – Giorgio Mossa Oct 22 '13 at 20:48