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Let $R$ be an integral domain which is not a field. Then is it true that $\langle f \rangle$ cannot be a maximal ideal of $R[x]$ for any non-constant polynomial $f(x) \in R[x]$ ?

I know that it holds in case of $R=\mathbb Z$ and I can adapt the proof to draw similar conclusion in case $R$ is a UFD with infinitely many mutually non-associated prime elements. But I don't know what happens in general. If the answer in general is not true then does it at least hold if we also assume $R$ is Noetherian ?

user26857
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    How about $R = \mathbb{Z}_{(p)} = { a/b \text{ such that } a, b\in \mathbb{Z} \text{ and } \operatorname{gcd}(b, p) = 1}$ and $f(x) = px-1$, with $R[x]/f(x) = \mathbb{Q}$. – anon Aug 29 '16 at 15:32
  • Also here: http://math.stackexchange.com/questions/99343/is-xy-1-a-maximal-ideal-in-kxy?noredirect=1&lq=1 – user26857 Sep 01 '16 at 05:59
  • And here: http://math.stackexchange.com/questions/752342/checking-the-maximality-of-a-principal-ideal-in-rx – user26857 Sep 07 '16 at 14:39

1 Answers1

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Let $K$ be a field, and let $R=K[[y]]$. Consider the polynomial $f=xy-1$ in $R[x]$. Then the quotient $R/\langle f \rangle$ is isomorphic to $K[[y]][x]/\langle xy-1\rangle$, which is a field (it is isomorphic to the field of fractions of $K[[y]]$).

Thus the ideal $\langle f \rangle$ can be maximal.

Pierre-Guy Plamondon
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