# $\operatorname{Spec}(\mathbb{C}[[x,y]])$.

Let $\mathbb{C}[[x,y]]$ be the ring of the formal series with coefficients in $\mathbb{C}$. I have to find $\operatorname{Spec}(\mathbb{C}[[x,y]])$.

I think that it is a local ring because it should represents germs of holomorphic functions.

If then I consider a point $f \in \mathbb{C}[[x,y]]$ what can I say about the Spec of localization: $\operatorname{Spec}(\mathbb{C}[[x,y]]_f)$?

• – Martin Brandenburg Sep 21 '13 at 10:05
• @MartinBrandenburg Is it true that $Spec(\mathbb{C}[[x]])=\{(0),(x)\}$? How can I prove it? – ArthurStuart Sep 21 '13 at 10:40
• You can check directly that the ideals of $k[[x]]$ are given by $(0)$ and $(x^n)$ for $n \in \mathbb{N}$. Hint: $f \in k[[x]]^*$ is invertible iff $f(0) \in k^*$. The prime ideals are of course just $(0)$ and $(x)$. – Martin Brandenburg Sep 21 '13 at 10:50
• @MartinBrandenburg I read that is very difficult to descrive the specrum of formal series. But if I localize them in a point is it easier? – ArthurStuart Sep 21 '13 at 10:55

$\mathbb{C}[[x,y]]$ is a regular local ring of dimension $2$. As such, it is a UFD, so the prime ideals are $0$, $(f)$ for $f$ an irreducible element of $\mathbb{C}[[x,y]]$, and the maximal ideal $(x,y)$. Note that being irreducible as a power series is substantially stronger than being irreducible as a polynomial: e.g. $y^2 - x^2 - x^3$ is an irreducible polynomial in $\mathbb{C}[x,y]$ (by Eisenstein's criterion), but is not irreducible as a power series (cf. Hartshorne I.5.6.3).
If $0 \ne f \in \mathbb{C}[[x,y]]$, factor $f$ into irreducibles $f = f_1^{e_1}...f_k^{e_k}$. Then $\text{Spec}(\mathbb{C}[[x,y]]_f)$ can be identified with $\text{Spec}(\mathbb{C}[[x,y]]) \setminus \{(x,y), (f_1), \ldots, (f_k) \}$.