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Working on some Bezout's theorem examples I arrived at a point where I need $$(\mathbb{K}[x]/(x^2))_{(x)} = \mathbb{K}[x]/(x^2)$$ (i.e. localize don't do anything)($\mathbb{K}$ alg. closed and nice field). This is true (you can prove it removing the denominator in the localization) but I wonder if there is a more general proof in more general cases.

If $ \mathbb{K}[x]/(x^2) \setminus (x)$ consists of units only then the localization obviously don't do anything (in this case are elements of the form $c + bx, c,b \in \mathbb{K}, c \neq 0$, which are units) and that is the case.

Another example: $$(\mathbb{K}[x]/(x^2(x-1)))_{(x-1)} = \mathbb{K}[x]/(x-1) \cong \mathbb{K}$$ Does anyone know a nice proof of this (a short argumentation)? I mean not the specific case but some rule that permit to simplify localization of quotients.

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Hint: Chinese Remainder Theorem. –  Konstantin Ardakov Aug 8 '12 at 8:11
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The ring $K[x]/(x^2)$ is local with maximal ideal $(x)$... –  Keenan Kidwell Aug 8 '12 at 23:51
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