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I recently learned about intersection multiplicity and tried to calculate a simple example. Unfortunately, I am having difficulty.

Consider the two curves in $\mathbb{C}^2$ given by $y=0$ and $y-x^n=0$. If there is any justice in the world, the intersection multiplicity at $(0,0)$ should be $n$.

By definition, the intersection number is the dimension as a $k$-module of

$$\left( \frac{k[x,y]}{(y,y-x^n)} \right)_{(x,y)}.$$

This is clearly equivalent to

$$\left( \frac{k[x,y]}{(y,x^n)} \right)_{(x,y)}.$$

Before localization, the quotient ring is all elements of the form


(polynomials in $x$ of degree less than $n$). This already has dimension $n$. Wouldn't taking the localization make the ring much bigger, with all sorts of nasty fractions, and make the dimension larger as well?

My question: Why is the dimension of the localized ring $n$?

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As an aside, that you may be interested in - the "justice in the world" you refer to is Bézout's theorem - en.wikipedia.org/wiki/Bezout%27s_theorem –  Matt Pressland Sep 27 '12 at 22:40
The localization in this case gives you all kinds of nasty fractions... with only elements of $k$ on the denominator. The variables $x$ and $y$ are not used for denominating, since they are in the ideal you're localizing over. Since $k$ is already a field, no change. –  Arthur Sep 27 '12 at 22:53
+1 for "if there is any justice in the world" –  Andrew Sep 28 '12 at 1:03

1 Answer 1

up vote 5 down vote accepted

The ring $A=\frac{k[x,y]}{(y,x^n)}=\frac{k[x]}{(x^n)} =k[\xi] $ is already a local ring with maximal ideal $\mathfrak m=(\bar x, \bar y)=(\bar x)=(\xi)$.
Localizing a local ring at its maximal ideal does not change it (a result worth remembering !).
So here localizing the local ring $A$ at $\mathfrak m$ does nothing to it: $A_\mathfrak m=A$.
That solves your problem and indeed $A=A_\mathfrak m$ has dimension $n$ over $k$.

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Why is it true that localizing a local ring at its maximal ideal does not change it? –  Potato Sep 28 '12 at 0:40
@Potato Because the localization is universal with respect to inverting the elements of a given multiplicative subset. –  Andrew Sep 28 '12 at 1:02
@Andrew Thanks. I have figured out an alternate way for those reading along: just note that everything in $A-m$, where $A$ is the local ring and $m$ is the ideal, is a unit. To see this, take $x\in A$, and note $xA$ must be $A$ or $m$. In the first case it is a unit, and in the second it is in $m$. –  Potato Sep 28 '12 at 1:18

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