# The Structure Sheaf of a Variety

Alright, so I’m still taking this really round about way to the Nullstellensatz, but someday I’ll get there.

For those of you that know about sheaves, some of the things I’ve been talking about should be looking vaguely familiar. We haven’t fully gotten there yet, but that is what today is about.

I won’t explicitly define what a general sheaf is, but of course there is always wikipedia or a textbook if you really want to know.

Let’s think back to what we had before. We define what we called $k[V]$ the coordinate ring on the algebraic set $V$. So now we do the natural thing, we look at the field of fractions of $k[V]$ which we will denote $k(V)$. You should say, “Wait a minute!” at this point, since we might have some “zero denominators.” So let’s hold off on actually defining this until we’ve built the way to work around the problem.

So as a set, $f \in k(V)$ is something of the form $f=g/h$, where $g, h \in k[V]$. So it is a fraction of polynomials, or a rational function. The problem is that it is not defined at zeros of $h$. Luckily, zeros of polynomials are all we’ve been studying and talking about for awhile.

Call $f \in k(V)$ regular at a point $P \in V$ if there is a representation $f=g/h$ such that $h(P) \neq 0$. In fact for any $h \in k[V]$ we can define a set corresponding to where it can be in the denominator, i.e. $V_h=\{P \in V : h(P) \neq 0\}$. Note that this is just the principal open set we defined earlier for the Zariski topology, but now it seems to have vital use.

Let’s now define the local ring of V at P to be $\mathcal{O}_{V, P}=\{f \in k(V) : \ f \ regular \ at \ P\}$. Clearly this is a subring of $k(V)$. The not as obvious fact is that it is actually local. If you want to check, the unique maximal ideal is the set of elements of the form f/g where $f(P)=0$ and $g(P) \neq 0$. So now some things are shaping up, since we have an object defined for sets and have a ring of functions at a point.

What would really be exciting is if this construction which seemed ad hoc by taking everything in the field of fractions and throwing out things that don’t work, actually turned out to be a nice localization of the ring. Define the ideal $\overline{M}_P=\{ f \in k[V] : f(P)=0\}$. So this is technically what we were calling $\overline{I({P})}$ before. (The line meaning that we aren’t in $k[x_1, \ldots, x_n]$ anymore, we’re in $k[V]=k[x_1, ldots , x_n]/I(V)$. So this is is a maximal ideal and hence prime, so we can localize at it.

Exactly what we were hoping for actually does happen, i.e. $k[V]_{\overline{M}_P}=\mathcal{O}_{V, P}$. In words, the localization of the coordinate ring at $\overline{M}_P$.

Now for any open set $U \subset V$ we define $\mathcal{O}(U)=\{ f \in k(V) : f \ regular \ on \ U\}$. And for convenience $\mathcal{O}_V(\emptyset)={0}$. So not only is $\mathcal{O}_V(U)$ a ring, it is a k-algebra. This set of rings with the restrictions we defined last time form the structure sheaf $\mathcal{O}_V$, and the local ring $\mathcal{O}_{V, P}$ is the stalk of the sheaf at P with the elements as the germ of functions at P.

So I’ll leave you with a nice way to rephrase some older posts: we should now think of $k[V]=\mathcal{O}(V)$, and $\mathcal{O}(V_h)=k[V][h^{-1}]=k[V]_h$.

Severely edited: Sorry, some weird bug took out every backslash of this post rendering it incomprehensible. I’m really glad I decided to glance at it randomly.