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!

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.