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I don't know what's the difference between $\mathbb{C}[x]_{(x)}$ and $\mathbb{C}[x]$.

Isn't the localization is just equal to the original ring? Then why the first presentation is used?

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    The localization is a local ring, but by Nullstellensatz we know that $(x-a)$ for any $a\in \mathbb{C}$ is a maximal ideal in $\mathbb{C}[x]$ and hence $\mathbb{C}[x]$ does not have a unique maximal ideal.2012-09-22

2 Answers 2

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The ring $\Bbb C[x]_{(x)}$ is strictly larger. It is the localization of $\Bbb C[x]$ at the multiplicative subset $S=\Bbb C[x]\setminus (x).$

Since $\Bbb C[x]$ is a domain, $S$ has no zerodivisors, which implies that the natural homomorphism $\Bbb C[x]\to\Bbb C[x]_{(x)}$ is injective. In fact, since we obtain $\Bbb C[x]_{(x)}$ by formally inverting elements of $S,$ we can view both rings as subrings of $\Bbb C(x),$ the fraction field of $\Bbb C[x].$ Then $\Bbb C[x]_{(x)}$ consists of those $\dfrac{f(x)}{g(x)}$ such that $g(0)\neq 0,$ meaning the meromorphic function $\dfrac{f(x)}{g(x)}$ is locally regular at $x=0$ (which is exactly the closed point of $\operatorname{Spec}(\Bbb C[x])$ defined by the ideal $(x)$).

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Since $(x)$ is prime, $\Bbb C[x]_{(x)}$ is a local ring. $\Bbb C[x]$ is of course not a local ring.

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    As a comicbook character once quipped: 'Nuff said.2012-09-22