# 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.

# Lying Over and Going Up Part II

I realized there was one more result I probably should have included last time. Oh well. Here goes:

Let $R^*/R$ be integral, $\frak{p}$ a prime ideal in R and $\frak{p}^*, \frak{q}^*$ prime ideals in $R^*$ lying over $\frak{p}$. If $\frak{p}^*\subset \frak{q}^*$, then $\frak{p}^*=\frak{q}^*$.

Proof: Recall that $S^{-1}R^*$ is integral over $S^{-1}R$ by last time for any multiplicative set, and also that prime ideals are preserved in rings of fractions. Thus the hypotheses still hold if we localize at $\frak{p}$. Thus $R_\frak{p}^*$ is integral over $R_\frak{p}$, and $\frak{p}^*R_\frak{p}^*\subset \frak{q}^*R_\frak{p}^*$ are prime ideals. Thus we can WLOG replace $R^*$ and $R$ by their localizations and hence assume they are local. So now $\frak{p}$ is a maximal ideal in $R$. Thus by last time $\frak{p}^*$ is maximal. Since $\frak{p}^*\subset \frak{q}^*$, we have $\frak{p}^*=\frak{q}^*$.

Now we are ready for the two big theorems. Here is the “Lying Over” Theorem. Let $R^*/R$ be an integral extension. If $\frak{p}$ is a prime ideal in R, then there is a prime ideal $\frak{p}^*$ in $R^*$ lying over $\frak{p}$, i.e. $\frak{p}^*\cap R=\frak{p}$.

Proof: First note that $R \stackrel{i}{\longrightarrow} R^* \stackrel{h^*}{\longrightarrow} S^{-1}R^*$ and $R \stackrel{h}{\longrightarrow} R_\frak{p} \stackrel{j}{\longrightarrow} S^{-1}R^*$ form the two sides of a commutative diagram. By last time $S^{-1}R^*$ is integral over $R_\frak{p}$. Choose a maximal ideal $\frak{m}^*$ in $S^{-1}R^*$. Thus $\frak{m}^*\cap R_\frak{p}$ is maximal in $R_\frak{p}$. But $R_{\frak{p}}$ is local with unique max ideal $\frak{p}R_\frak{p}$, so $\frak{m}^*\cap R_\frak{p}=\frak{p} R_\frak{p}$. But the preimage of a prime ideal is prime, so $\frak{p}^8=(h^*)^{-1}(\frak{m}^*)$ is a prime ideal in $R^*$.

Now we just diagram chase: $(h^*i)^{-1}(\frak{m}^*)=i^{-1}(h^*)^{-1}(\frak{m}^*)=i^{-1}(\frak{p}^*)=\frak{p}^*\cap R$. And also: $(jh)^{-1}(\frak{m}^*)=h^{-1}j^{-1}(\frak{m}^*)=h^{-1}(\frak{m}^*\cap R_\frak{p})=h^{-1}(\frak{p}R_\frak{p})=\frak{p}$.

Thus $\frak{p}^*$ lies over $\frak{p}$.

Our other big theorem is the one about “Going Up”: If $R^*/R$ is an integral extension and $\frak{p}\subset \frak{q}$ are prime in R, and $\frak{p}^*$ lies over $\frak{p}$, then there is a prime ideal $\frak{q}^*$ lying over $\frak{q}$ with $\frak{p}^*\subset \frak{q}^*$.

Proof: By last time $(R^*/\frak{p}^*)/(R/\frak{p})$ is an integral extension where $R/\frak{p}$ is embedded in $R^*/\frak{p}^*$ as $(R+\frak{p}^*)/\frak{p}^*$. Now we just replace $R^*$ and R by these rings so that both $\frak{p}^*$ and $\frak{p}$ are $\{0\}$. Now we just apply the Lying Over Theorem to get our result.

So as we see here integral extensions behave extremely nicely. These theorems guarantee that se always have prime ideals lying over ones in the lower field. This has some important applications to the Krull dimension that we’ll start looking at next time.

# A closer look at Spec

Let’s think about what is going on in a different way. So now let’s think of $f \in R$ elements of the ring as functions with domain $Spec(R)$. We define the value of the function at a point in our space $f(P)$ to be the residue class in $R/P$. This looks weird at first, since the image space depends on the point that you are evaluating the function.

Before worrying about that too much, let’s see if we can get this notion to match up with what we did yesterday. We have the nice property that $f(P)=0$ if and only if $f \in P$. (Remember that even though we think of f as a function, it is really an element of the ring).

Define for any subset of the ring S the zero set: $Z(S)=\{P\in Spec(R): f(P)=0, \forall f \in S\}$. Now from what I just noted in the previous paragraph, we get that these are just precisely the elements of $Spec(R)$ that contain S, i.e. the closed sets of the Zariski topology. Thus we can define our basis for the Zariski topology to be the collection of $D(f)=Spec(R)\setminus Z(f)$.

We also will want what is “an inverse” to the zero set. We want the ideal that vanishes on a subset of Spec. So given $Y\subset Spec(R)$, define $I(Y)=\{f \in R : f(P)=0, \forall P\in Y\}$. Now this isn’t really an inverse, but we get close in the following sense:

If $J\subset R$ is an ideal, then $\displaystyle I(Z(J))=\sqrt{J}$. Taking the ideal of the zero set is the radical of the ideal. And the radical has two equivalent definitions: $\displaystyle \sqrt{J}=\cap_{P\in Spec(R), P\supset J} P=\{a\in R : \exists n\in \mathbb{N}, a^n\in J\}$.

If we take the ideal and zero set in the other order we get that $Z(I(Y))=\overline{Y}$ : the closure in the Zariski topology.

We can abstract one step further and put a sheaf on $D(f)$. Note that for any $f\in R$ we have that $\{1, f, f^2, \ldots\}$ is a multiplicative set, so we can localize at it. Since I haven’t talked at all about sheaves, I’m not sure if I want to go any further with this, so maybe I’ll do some more examples next time and possibly start to scratch this surface.

# More on Primality

I want to wrap up some loose ends on the greatness of prime ideals before moving on in the localization theme. So. Recall that we formed the ring of quotients just like you would form the field of quotients. Only this time your “denominator” can be an arbitrary multiplicative set and this construction only gets us a ring. Moreover, this ring is not necessarily local. If we do the construction on a ring R with and the multiplicative set $R\setminus P$ where P is a prime ideal, then we do get a local ring and we call this the localization.

Definition. Unique Factorization Domain (UFD): An integral domain in which every non-zero non-unit element can be written as a product of primes. (Note that there are equivalent definitions other than this one).

Quick property: Every irreducible element is prime.

Thus, it is instructive to look at some properties of prime ideals. First off, let’s look at the special case of UFD’s. It turns out that if R is a UFD, then for a multiplicative set S, $S^{-1}R$ is also a UFD. This mostly has to do with the fact that $R\hookrightarrow S^{-1}R$ is an embedding and anything in $S^{-1}R$ is associate to something in $R$. This makes a nice little exercise for the reader.

So what’s so special about prime ideals in UFD’s? Well every nonzero prime ideal contains a prime element.

Proof: Suppose $P\neq 0$ and P prime. Then there exists $a\in P$, $a\neq 0$ such that $a=up_1\cdots p_n$ where $u$ a unit and $p_i$ prime. Thus $u\notin P$. But this means that $p_1\cdots p_n\in P$ and since it is prime we have some $p_j\in P$.

Theorem: If R is not a PID, and P an ideal which is maximal with respect to the property of not being principal, then P is prime (and will always exist).

Sketch of existence: Zorn’s Lemma. The proof of this contains lots of nitty gritty element-wise computation and a weird trick, so I don’t see it as beneficial. What is beneficial is that we get this great corollary: A UFD is a PID if and only if every nonzero prime ideal is maximal.

I’ve been kind of stingy on the examples, so I’ll leave you with a pretty common example of a ring of fractions. These are usually called dyadic rational numbers. Take your ring to be $\mathbb{Z}$. Then take your multiplicative set to be $S=\{1, 2, 2^2, 2^3, \ldots\}$. Now $S^{-1}\mathbb{Z}$ are just the rational numbers with denominator a power of 2.

More generally we can form the p-adic integers (although that term is laden with many meanings, so I hesitate to actually use it). Let $R=\times_{i=1}^\infty \mathbb{Z}/p^i$. Where we have the restriction $a\in R$ iff $a=(a_1, a_2, \ldots )$ satisfies $a_i\cong a_{i+1} \mod p^i$. So  elements of the ring are sequences. (Note $\mathbb{Z}$ embeds naturally since $i\mapsto (i, i, i, \ldots)$ satisfies that relation). This is a ring with no zero divisors, so we can take it to be the multiplicative set and we get the field of fractions $\mathbb{Q}_p$. The multiplicative group has a nice breakdown as $\mathbb{Q}_p^{\times}\cong p^{\mathbb{Z}}\times \mathbb{Z}_p^{\times}$.

Next time: Why Noetherian is important. How primality relates to it. And possibly another example.

# Localization 2

Let’s figure out what “local” means and see if our construction somehow makes a local ring, i.e. is a “localization.”

Local: A ring is called local if there is a unique maximal ideal. This seems like a rather silly term, but it actually makes sense when you look at how rings arise in algebraic geometry or manifold theory. We won’t go there, though.

Sadly, it turns out that $S^{-1}R$ is not always a local ring. But this is where primality comes into play. If $P\subset R$ is a prime ideal then $S=R\setminus P$ is a multiplicative set. Suppose it weren’t, then there would be two elements $x,y\in S$ such that $xy\notin S$, i.e. $xy\in P$, but this is impossible, since by definition either $x\in P$ or $y\in P$. We now denote the localization of $R$ at $P$, to be $S^{-1}R=(R\setminus P)^{-1}R$ which we denote with the shorthand $R_P$. This does turn out to be local since by the property listed last time of the embedding $\phi^{-1}(S^{-1}P)=P$, so $S^{-1}P=\{r/s : r\in P, \ s\notin P\}$ is the unique maximal ideal in $R_P$.

Proof: Suppose $x\in R_P$, then $x=r/s$ with $r\in R$ and $s\notin P$. If $r\notin P$, then $r/s$ is a unit in $R_P$. So all nonunits are in $S^{-1}P$. Now if I is any ideal in $R_P$ that contains an element $r/s$ with $r\notin P$, then $I=R_P$. Thus every proper ideal in $R_P$ is contained in $S^{-1}P$. So $R_P$ is local with unique max ideal $S^{-1}P$. For notational purposes outside of this blog, people usually write the prime ideal as $\mathfrak{p}$ and the unique maximal ideal of $R_\mathfrak{p}$ as $\mathfrak{p}R_{\mathfrak{p}}$.

I guess I’ve been rather sparse on the examples. The first one that comes to mind is surely to take $\mathbb{Z}=R$. Then our prime ideals are just the principal ideals generated by the primes, so take $P=p\mathbb{Z}$ for some prime p. Then $\mathbb{Z}_P=\mathbb{Z}_{(p)}$.

I guess the importance of prime ideals leads us to explore some properties of prime ideals that could be useful.

Property 1: If $S\subset R$ is any multiplicative set (not containing 0) and if $P\subset R\setminus S$ is a maximal ideal, then $P$ is prime. Also, any ideal $I\subset R\setminus S$ is contained in such a $P$. I’ll omit proving this. The first part is fiddling with things until it works and the second statement uses Zorn’s Lemma.

OK, well I thought I had some other properties, but I can’t seem to find them/think of them now. I’m not sure where I’m going next. I’ll either move on to some related things to get at this better like the nilradical, or I’ll generalize this one more time to modules and do it using the categorical construction. If anyone has suggestions on which of these paths to take, just post. You probably have a few days as I’ll get busy again.

# Localization 1

I’m calling this “Localization 1” since it is technically the beginning of the construction that we eventually want. But for now, I don’t even want to define what it means for a ring to be local, or why this may or may not be a (the) “localization.”

Last time we formed the field of fractions, but it came with some restrictions that we don’t need (as long as we don’t care that the end result isn’t a field). So today we will be forming the “ring of fractions.” Let’s think about what made that construction work. Essentially we didn’t want to “divide” by things that could end up as zero. This was the restriction that R be a domain. So let’s throw out that restriction, and just make sure there are no zero divisors. We also didn’t really care about the ring structure at all for the denominator. We only cared that it was closed under multiplication.

Let $S$ be a multiplicative set with 1 (it is closed under mult.) with no zero divisors. Now do the same exact construction as before on $\{(r,s)\in R\times S\}$. We call the set of equivalence classes the ring of fractions and denote it $S^{-1}R$. It is straightforward computation to check that this is still a ring.

Now the only reason I included 1 in S was so that there is a natural embedding $\phi :R\hookrightarrow S^{-1}R$ given by $r\mapsto r/1$. This embedding tends to be really nice for proving certain properties about the ring of fractions. One of these properties is exactly what we would hope: the embedding sends ideals to ideals, i.e. if $I\subset R$ an ideal such that $I\cap S=\emptyset$, then $\phi(I)\subset S^{-1}R$ is an ideal (with this being a bijection between prime ideals).

So the preimage under the embedding turns out to be fairly important. What is it on the element level? Well $\phi^{-1}(S^{-1}I)=\{x\in R : \exists s\in S, \ sx\in I\}$. And without giving too much away, we get the nice result that if $I$ is prime and $I\cap S=\emptyset$, then $\phi^{-1}(S^{-1}I)=I$. Then we get that if $P\subset R$ prime, then $S^{-1}P\subset S^{-1}R$ is prime. So maybe there will be some importance to the notion of prime later.

Now let’s backtrack and try to do this construction one more time (OK, so it will come up again next time, but…). We really, really don’t want to have to restrict the set S to have no zero divisors. This is just too pesky of a condition. Can we somehow get around it?

This time let’s suppose $S$ is multiplicative with 1, and let’s only assume $0\notin S$. There can be zero divisors, though. Let $I=\{x\in R: \exists s\in S, \ sx=0\}$. Then $I$ is an ideal. Clearly if $x,y\in I$ then pick $s,t\in S$ such that $sx=0$ and $ty=0$. Then $st\in S$ and $st(x-y)=stx-sty=0$, so $x-y\in I$. Also $x\in I$ and $r\in R$, (since we’ve assumed commutativity) we clearly have $rx\in I$.

Since we have a perfectly good ideal, and this ideal in a sense mimics the property of zero divisors, we do the natural thing and mod out by it. Thus form $R\to R/I=\overline{R}$, and $S\to \overline{S}$. Then $\overline{S}$ has no zero divisors in $\overline{R}$. Suppose there was one, say $a\in \overline{S}$. Then there exists $r\in \overline{R}$ such that $ra=0$. By definition of the “bar” we have that $a=s+I$ for some $s\in S$ and $r=r_0+I$ for some $r_0\in R$ (and $0=I$), so $(s+I)(r_0+I)=sr_0+I=I$ so $sr_0=0$ which means that s itself was a zero divisor. But trivially all zero divisors are sent to 0 under the mapping, so $a=0\in \overline{S}$.

So we form $S^{-1}R$ by first canonically factoring through $\overline{S}^{-1}\overline{R}$. Why all this trouble? In essence what we are trying to do is form a new minimal ring that contains the old ring such that every element is a unit. If the kernel of the mapping is not {0} (i.e. there were zero divisors in the set), then the mapping is not an embedding and so the new ring doesn’t contain the old ring. The astute reader will note that this almost seems like a universal construction. This is what we’ll do when I actually get to localization for real.

# Categories? Rings?

Well, things can get ultra busy around mid-terms. I don’t think I’ve posted in two weeks. What I really wanted to do next was to post some category theory basics. I’m not sure if I should, though, since so many math blogs have already done this. I then wanted to go on to define the fundamental group purely in categorical language. It turns out to be a really nice construction compared to the tedious typical way.

Instead, I’ve recently become quite interested in rings. There also seems to be a very large lack of “pure” ring theory in the blogosphere. Sure rings pop up and are needed by people doing things with algebraic geometry, say, but using ring theory isn’t the same as developing it.

I’m going to cover the basics quite quickly with the assumption of previous exposure, since I really want to get to some of the more interesting constructions (i.e. localization), then I’ll slow it down.

Ring: We have a set with two operations, we’ll call them addition and multiplication. The addition part forms an abelian group, and the multiplication…well, it puts you back in the set and is associative. We need a way to relate these operations, so we also require that $a(x+y)=ax+ay$ and $(x+y)b=xb+yb$, i.e. there is a distributive law in effect. Note that there is not required in general to be a multiplicative identity, multiplicative inverses, or commuting of the multiplication.

NOTE: Until I say otherwise I will assume the ring is commutative (meaning multiplication) with 1 (meaning having a mult identity). $R$ will always denote this.

Subring: A subset of $R$ that is itself a ring.

Ideal: A subring that “swallows” multiplication. So $I\subset R$ is an ideal if for any $a\in I$ we have that $ra\in I$ for all $r\in R$.

Prime ideal: An ideal $I$ is prime if for element $ab\in I$ we have that either $a\in I$ or $b\in I$.

Principal ideal: An ideal $I$ is principal if it is generated by a single element. So an ideal is generated by “a” if $I=Ra=\{ra: r\in R\}$.

Maximal ideal: A proper ideal $I$ such that there is no other ideal $K$ with the property $I\subset K\subset R$ (where all containments are proper).

Domain: A ring in which the cancellation law holds. i.e. if $ab=ac$ and $a\neq 0$, then $b=c$. Note that no element can divide zero, so if $ab=0$, then either $a=0$ or $b=0$.

We can quotient in the natural way: $R/I$ is the set of cosets of I where our operations are $(a+I)+(b+I)=(a+b)+I$ and $(a+I)(b+I)=ab+I$. We get the nice result that any ideal of $R/I$ is of the form $K/I$ where $K$ is an ideal of $R$ (and $I\subset K\subset R$).

I think that may lay down all the terminology I’ll need to get started. I’m not sure if I’ll really use any of these terms for awhile, though.

Common rings: $\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$. Note that we can get $\mathbb{Q}$ from $\mathbb{Z}$ by taking “quotients.” This can be made precise for any domain. It is called the fraction field of $R$ denoted $Frac(R)$.

Let me take some time to explain this, since it is the motivation for localization. We want to form a field $F$ that contains $R$ as a subring such that the elements of $F$, say $f\in F$ have the form $f=ab^{-1}$ where $b\neq 0$. Note that this “looks” like division, and in fact is division in the case of $\mathbb{Z}$.

To make this process precise takes a bit of work, though. Set up $X=\{(a,b)\in R\times R : b\neq 0\}$. Define (a,b)~ (c,d) iff $ad=bc$. This is done since we want our relation to look like fractions, a/b , so a/b=c/d if we can cross-multiply and get the same thing. It is straightforward to check that this defines an equivalence relation. Now we let $F$ be the set of equivalence classes.

Our operations on $F$ should mimic those of fractions, so our addition is [a,b]+[c,d]=[ad+bc, bd] and [a,b][c,d]=[ac,bd]. These are well-defined and it is just computation to check that the axioms of a field are satisfied. (If you want a hint: the zero is [0,1] and the 1 is [1,1], the additive inverse of [a,b] is [-a,b] and the mult inverse is [b,a]).

Before finishing up, I want to point out how restrictive we had to be. We want a more general way of doing this. We don’t want to require that R be a domain, and we don’t want to have to take fractions with every single element in R. It turns out this general process is extremely useful and it is called localization.