TITLE: Does localization preserves dimension?
QUESTION [7 upvotes]: Does localization preserves dimension?
Here's the problem:
Let $C=V(y-x^3)$ and $D=V(y)$ curves in $\mathbb{A}^{2}$. I want to compute the intersection multiplicity of $C$ and $D$ at the origin.
Let $R=k[x,y]/(y-x^3,y)$. The intersection multiplicity is by definition the dimension as a $k$-vector space of $R$ localized at the ideal $(x,y)$.
Note now that:
$R \cong k[y]/(y) \otimes _{k} k[x]/(x^3)$
Clearly $k[y]/(y) \cong k$ so the above tensor product is isomorphic to $k[x]/(x^3)$.
Therefore $R$ has dimension $3$.
However I want to compute the dimension of $R$  localized at the ideal $(x,y)$. Does localization preserves dimension or how do we proceed? I would really appreciate if you can please explain this example in full detail.

REPLY [6 votes]: Since $R$ is already a local ring with maximal ideal $\mathfrak m=(\bar x,\bar y )$ , localizing changes nothing!
In other words $R\cong R_{\mathfrak m}$ and thus  $dim_kR_{\mathfrak m}=dim_kR=3$ (since $R=k\oplus k\bar x\oplus k\bar x^2$).