Does localization preserves dimension?

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.

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@Matt: ah, I misread the question. My apologies. –  Qiaochu Yuan Jun 21 '12 at 6:00
@QiaochuYuan: Dear Qiaochu, No worries; following your lead, I deleted my comment. Cheers, –  Matt E Jun 21 '12 at 6:11

You have the $k[x,y]$-module $k[x]/(x^3)$, on which $y$ acts as $0$. If you're not able to compute the localization of this very concrete module at the ideal $(x,y)$, then you will need to practice with localizations.

The general answer to your question is, in any case, "not always" (although in your particular example it does).

You want to try some one-variable cases first.

E.g. consider $M = k[x]/(x^2)$ and $M = k[x]/(x^2 -x)$. The former is its own localization at the prime ideal $(x)$, and so the dimension doesn't change after we localize at that prime. The latter module has one-dimensional localization at $(x)$.

Now to give another two-dimensional example: $V(x^2 - x) \cap V(y)$ correspond to the ring $k[x,y]/(x^2 - x,y) \cong k[x]/(x^2 - x)$. This is two-dimensional, but the localization at $(x,y)$ is just one-dimensional. Try drawing a picture to see if you can understand why this holds geometrically.

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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 $R=k\oplus k\bar x\oplus k\bar x^2$).

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Fulton's book Algebraic Curves gives an easy way to compute the intersection number in theorem 3 in section 3.3.

For you curve for example, we can use the algorithmic method provided by Fulton. Let $P = (0,0)$. Then \begin{eqnarray} I(P, D \cap C ) = I(P,Y \cap C) = I(P,Y\cap C(X,0)) = I(P,Y\cap -X^3) = 3. \end{eqnarray}

So the intersection multiplicity of $C$ and $D$ at the origin is $3$.

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