# Intersection multiplicity as the dimension of a vector space

I'm trying to solve the following problem in Dino Lorenzini's book on arithmetic geometry:

Let $f,g\in k[x,y]$ be coprime and assume that $P=(0,0)\in V(f)\cap V(g)$ is a nonsingular point of $f$. Show that $V=k[x,y]/(f,g)$ is a finite dimensional $k$-vector space and that the intersection multiplicity of $f$ and $g$ at $P$ equals $\dim_k V$. Here $k$ is assumed to be algebraically closed.

The finite dimensional part is trivial and I can solve it in more than one way. The shortest is probably noting that if $A=k[x]$ then computing the resultant of $f,g$ in $A$, we get

$\textrm{Res}_A(f,g)=uf+vg\in (f,g)$,

so the resultant, being an element of $k[x]$, gives a nontrivial relation of $x$ in $k[x,y]/(f,g)$. The same can be repeated for $y$ by setting $A=k[y]$.

To show that this dimension equals the intersection multiplicity, let $M=(x,y)$. Then assuming the multiplicity is $n$, we have that $g\in M^n\setminus M^{n+1}$ in $B=k[x,y]/(f)$. Furthermore, we know that $B$ is a Dedekind domain, since $P$ is a nonsingular point.

I can't seem to figure out what to do next. We can assume that $MB_M=(x)$, so that $ug=vx^n$ for some $u,v\not\in M$. This shows that $vx^n=0$ in $k[x,y]/(f,g)$. However, this doesn't seem to give the relation I want for $x$, since $v$ is not a constant.

Any ideas?

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## 1 Answer

I think the question as you've written it contains a small error. I think you should prove that the intersection multiplicity is equal to $\dim_k W,$ where $W=V_P$ is the local ring of the intersection point. Otherwise, since the ring $V$ may have support on more than one point, we might not be computing the correct number.

The ring $B_M=(k[x,y]/(f))_M$ is a one-dimensional regular local ring, thus is a DVR, with principal maximal ideal $MB_M=(t).$ Assuming that the definition of intersection multiplicity you are working with takes the maximal $n$ such that $g\in M^nB_M\setminus M^{n+1}B_M,$ we have that $g=ut^n\in B_M$ for some unit $u\in B_M.$ (That is, there exist elements $a,b,h\in B\setminus M$ such that $h(ag-bt^n)=0,$ which implies $ug=vt^n$ for some $u,v\in B\setminus M,$ as you say.)

Now we should try to compute $\dim_k W.$ In particular, we have $V=k[x,y]/(f,g)=\dfrac{k[x,y]/(f)}{(f,g)/(f)}=B/(g)$ by the second isomorphism theorem, which tells us that $W=V_P=B_M/(g)_M.$ But we that $g=ut^n\in B_M.$ Hence, the intersection multiplicity equals $\dim_k B_M/M^nB_M =n.$

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OK, this explains a lot. Looking at the book website, I seem to have missed the fact that the typo in this problem is listed there... Thanks. – asdf Nov 25 '12 at 22:14
Dear @asdf, you're welcome! – Andrew Nov 25 '12 at 22:58