Let $R = \mathbb{Z}_{(2)}$ be the localization of $\mathbb{Z}$ at the prime ideal generated by $2$ in $\mathbb{Z}$. Then prove that the ideal generated by $(2x-1)$ is maximal in $R[x]$.
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user26857
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Digjoy Paul
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Hint. Prove that $R[x]/(2x-1)\simeq R[\frac12]$ and $R[\frac12]=\mathbb Q$.
user26857
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I like the example and explanation. It bugged me so much since the ideal $(2x -1)$ is a principal maximal ideal in a two dimensional regular ring. It'd be nice if someone can explain a bit more on this phenomenon. – Youngsu Apr 15 '14 at 04:53
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@Youngsu You can find more details here. – user26857 Apr 15 '14 at 11:19
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The above hint is correct. Sending $x$ to $1/2$ we get a surjective map from $R[x]$ to $\mathbb Q$ and then look kernel of this homomorphism that will be exactly $(2x-1)$. So we are done.
user26857
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Digjoy Paul
- 87