6

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?

Gobi
  • 7,458
  • 5
    For example, in $\Bbb C[x]$ the element $1+x$ is not a unit. – Andrew Sep 22 '12 at 16:16
  • Thanks, I missed something. – Gobi Sep 22 '12 at 16:21
  • 1
    The ring $\mathbf{C}[x]$ has infinitely many prime ideals, while $\mathbf{C}[x]_{(x)}$ just has two: $(0)$ and the ideal generated by $x$. The localization is a discrete valuation ring. – Keenan Kidwell Sep 22 '12 at 17:46
  • 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. – Matt Sep 22 '12 at 21:17

2 Answers2

8

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 $(x)$).

Andrew
  • 11,179
1

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

rschwieb
  • 153,510