## Some Commutative Algebra and a bit of Geometry

Before moving on to morphisms of varieties, it would do us some good to talk about a bit of commutative algebra and to do a little bit more geometry of individual objects. The central notion of algebra that we’ll be needing is that of a commutative ring with identity.

So a commutative ring with identity is a collection of objects along with two operations, called addition and multiplication. We require that both be associative, commutative and have identities, and also that addition has inverses (like negative numbers) and that multiplication distributes over addition. The idea is to generalize both the integers and polynomials to get a bit more flexibility to work with and for language.

So now we want to redefine the notion of ideal. Now, we want to start with a ring $R$ and define an ideal $I\subseteq R$ to be a subset which is closed under addition and multiplication by arbitrary elements of $R$. That is, if $f,g\in I,a\in R$ then $f+g\in I$ and $af\in I$. The real reason for using ideals and general rings other than just polynomials is that we can now define the quotient ring $R/I$ for a ring $R$ and an ideal $I$. We define first the notion of equivalence classes. We say that $f\equiv g$ modulo $I$ if $f-g\in I$. For any two elements of a ring, either they are equivalent in this way or they are not, everything is equivalent to itself, and if two things are each equivalent to a third thing, then they are equivalent.

Thus, we can break up the ring $R$ into equivalence classes, each one being a collection of elements of the ring that are equivalent. Because $I$ is an ideal, addition and multiplication of equivalence classes is well defined, that is, we get the same equivalence class regardless of what representatives we work with. And so we define the ring $R/I$ to be the equivalence classes of $R$ modulo $I$ with addition and multiplication.

Why is this relevant, you might ask? Well, recall that we first defined ideals in the context of affine algebraic sets, so as subsets of the polynomial ring. Given a variety $V$, we discussed the ideal $I(V)$ of polynomials vanishing on $V$. So if we want to look at the polynomial functions on $V$, we should be able to add or subtract elements of $I(V)$ without changing things, because they’ll have no effect on the values of the functions. So really we aren’t working in $\mathbb{C}[x_1,\ldots,x_n]$ anymore, but rather in $\mathbb{C}[x_1,\ldots,x_n]/I(V)$, which we will denote by $\mathbb{C}[V]$ as a convenient shorthand.

It’s hardly surprising that this ring encodes almost all of the properties of an affine variety (in fact, it encodes all of them, we’ll talk about this more later). This correspondence is why algebraic geometry really works, because we can work either geometrically with the variety or algebraically with the ring (called the affine coordinate ring) to get results and can move back and forth.

What about mimicking this construction for projective varieties? Taking $\mathbb{C}[x_0,\ldots,x_n]/I(V)$ where we’re looking at the ideal generated by the homogeneous polynomials vanishing on $V$ gives us something called the projective coordinate ring. The connection between geometry and algebra here is slightly less tight, because, as we will prove later, having the same affine coordinate ring is the same as saying that two affine varieties are the same, but two projective varieties that are the same might not have the same projective coordinate ring.

A last bit of commutative algebra: a consequence of the Nullstellensatz is that radical ideals in $\mathbb{C}[V]$ (or homogeneous radical ideals in its projective counterpart) will correspond to algebraic sets contained in $V$ and that maximal ideals, that is, ideals which are not contained properly in any ideals other than the whole ring itself, correspond to minimal algebraic sets, that is, points.

Now a bit of algebraic geometry: a variety will correspond to a prime ideal. A prime ideal is one such that $fg\in I$ implies that $f\in I$ or $g\in I$.

We’ll prove this as follows: let $V$ be an algebraic set, and let $R$ be the coordinate ring. We will only consider the affine case, the projective case works in a very similar manner. Then $R=\mathbb{C}[x_1,\ldots,x_n]/I(V)$. If $V$ is not a variety, then there exist $V_1,V_2$ properly contained in $V$ whose union is the whole thing. That is, there are polynomials $f,g$ such that $f$ vanishes everywhere on $V_1$ but not on $V_2$ and $g$ is the opposite. Then $fg=0$ in $\mathbb{C}[V]$. But that means that $fg\in I(V)$ but that $f,g\notin I(V)$ because neither vanishes everywhere on $V$, so $I(V)$ isn’t prime. Now we assume that $I(V)$ isn’t prime. Then there exist $f,g$ such that $f,g\notin I(V)$ but $fg\in I(V)$. Then the sets $V(f),V(g)$ show that $V$ is not irreducible.

This is wonderful, because there is now an algebraic criterion that matches irreducibility. Now we’ll define one more thing that will be useful later and will lend a bit more geometry to the subject, and will eventually allow us to make a connection to differential geometry: the tangent space.

The first thing is that we need to look at the definition of an ideal and note that an ideal is itself a ring (albeit generally without an identity). This turns out to be wonderful news, because it’s just what we need. Now we will make the following notation: $I^2=\{fg|f\in I,g\in I\}$. This is, in fact, an ideal in $I$, and so we can define the quotient ring $I/I^2$. We can think of this, roughly, as the elements of $I$ that aren’t in $I^2$. It’s much like killing off higher order terms. In fact, let’s do an example with polynomials. Say that $I=(x,y)$ in the ring $\mathbb{C}[x,y]$. Then $I^2=(x^2,xy,y^2)$, and so $I/I^2$ is going to just be linear functions in $x,y$.

Some of you with differential geometry background are probably thinking “Linearization! This is wonderful!” You’re very, very right. So now we let $V$ be a variety, affine or projective, and look at $R$ the coordinate ring, also affine or projective. Then we take a point $x\in V$ and look at $\mathfrak{m}_x$, the ideal (maximal, in fact) of all polynomials vanishing at $x$. Now we construct $\mathfrak{m}_x/\mathfrak{m}_x^2$. Now, this doesn’t quite turn out to be the tangent space, for many reasons, but next time when we talk about morphisms, we’ll see that the linearizations of maps would go in the wrong direction, and so we take the dual vector space (note that this IS in fact a vector space). That is, we take the set of all linear maps $\mathfrak{m}_x/\mathfrak{m}_x^2\to \mathbb{C}$. THIS we call the tangent space at $x$, which we denote by $T_xV$.

That’s a lot of ground covered, next time we move on to maps between varieties, and that’s where the fun really begins. ## About Charles Siegel

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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### 6 Responses to Some Commutative Algebra and a bit of Geometry

1. Anonymous says:

I guess in your definition of I² you want to take linear combinations of terms of the form fg, for otherwise this is not necessarily an ideal.

2. Charles says:

Yeah, though rather than saying linear combinations, should say the ideal generated by the elements of the form $fg$ for $f,g\in I$.

3. Landau says:

Wait; you say an ideal is in particular a ring. That can’t be true (given that we’re assuming rings have an identity element): if it contains 1, then it was already the whole ring. E.g. the even integers 2Z is an ideal, but doesn’t contains 1.

So I am wondering what you mean by m/m^2. As m is a maximal ideal, k:=R/m is a field. So I guess we could say m^2 is an (R/m)-module, i.e. a k-vector space. And then m^2 is a k-subspace of m. And then m/m^2 is a k-vector space. And then we take its dual vector space to be the tangent space.

Am I on the right track here?

• Charles Siegel says:

Ahh, I added a parenthetical noting that ideals are rings without identity, and that’s pretty much how we get a vector space structure on the tangent spaces, as you sketched out.