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Let $R$ be a commutative ring, $\mathfrak{m}\subset R$ a maximal ideal and $f$ a monic polynomial in $R[x]$. I want to show that $A:=\frac{R[x]}{\mathfrak{m}[x]+(f)}$ is a semilocal ring, where $(f)$ means the ideal generated by $f$.

One step towards the solution might be to show that $A\cong \frac{(R/\mathfrak{m})[x]}{(\overline{f})}$, where $\overline f$ means $f$ modulo ($\mathfrak{m}[x]+(f))$.

Then as $R/\mathfrak{m}$ is a field, $(R/\mathfrak{m})[x]$ is a PID and thus we have a PID modulo a nonzero ideal...

As I can't show the isomorphism nor the actual claim I am greatful for every answer, hint or advise!

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    $\begingroup$ Sure, the isomorphism holds (apply the universal properties). And a nontrivial quotient of a PID is semilocal because the (maximal) ideals are just the (irreducible) divisors of the polynomial we mod out. $\endgroup$ – Martin Brandenburg Sep 4 '13 at 14:13
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Sure, you are looking at $$\frac{R[x]}{\mathfrak{m}[x]+(f)}\cong \frac{\frac{R[x]}{\mathfrak{m}[x]}}{\frac{\mathfrak{m}[x]+(f)}{\mathfrak{m}[x]}}=\frac{(R/\mathfrak{m})[x]}{(\overline{f})}$$ where $\overline{f}$ is the polynomial coefficients mod $\mathfrak m$.

As you noted, the last ring is a quotient of a PID, and so the complete list of ideals containing $(\overline{f})$ is furnished by the divisors of $\overline{f}$, of which there are only finitely many.

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  • $\begingroup$ Thanks. The isomorphism is clear to me now. But I still don't understand why there are only finetly many ideals. Why are the ideals of $R/\mathfrak{m}[x]$ furnished by $(\overline{f})?$ $\endgroup$ – Heffalump Sep 4 '13 at 15:49
  • $\begingroup$ Dear @Heffalump : as you noticed, $(R/m)[x]$ is a PID, so each of the ideals corresponds to a single polynomial that generates it. In particular, $(f(x))\subseteq (g(x))$ iff $g$ divides $f$. Since PID's are UFDs, $\overline{f}$ factors into finitely many prime powers, and each ideal containing $\overline{f}$ has to be generated by a divisor of $\overline{f}$. There are only finitely many divisors, hence finitely many maximal ideals. $\endgroup$ – rschwieb Sep 4 '13 at 16:36
  • $\begingroup$ Thanks for the good explanation! $\endgroup$ – Heffalump Sep 5 '13 at 7:33

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