## Gradings of Rings and Modules

Today we do a bit of commutative algebra which we’ll need later. Once we’ve got this, we’ll spend the next post playing with the Hilbert Polynomial (for which all of these definitions are needed) and then we’ll prove one of the coolest theorems in all of mathematics (at least, in my opinion) – Bezout’s Theorem.

We’ll begin with the notion of a grading. We say that a ring $R$ is graded if we can break it up into $\ldots\oplus R_{-2}\oplus R_{-1}\oplus R_0\oplus R_1\oplus R_2\oplus\ldots$. That is, we can write it as a direct sum of a collection of abelian groups $R_i$. We also require that if we take $f_i\in R_i$ and $f_j\in R_j$, then $f_if_j\in R_{i+j}$.

An example of such a ring is $\mathbb{C}[x,y,z]$, where the grading is by degree, and all the negative degrees are zero. Another example is $\mathbb{C}(x,y,z)$ also with degree. The real value is that if we take the quotient of a graded ring by a homogeneous ideal (that is, an ideal which can be generated by elements in the $R_i$, roughly, of pure degree) we get a ring that is also graded. So in fact, the coordinate rings of projective varieties are graded.

Also nice is that if we take a graded ring and invert a bunch of homogeneous elements we again get a graded ring. So every single ring that we’ve encountered involving projective varieties is graded. So it might be good to be able to say some things about graded rings (and we will next time, today we’re mostly focusing on definitions).

We define an $R$-module to be an abelian group such that we can multiply by elements of $R$. That is, we have an operation with $r_1(r_2m)=(r_1r_2)m$ and $(r_1+r_2)m=r_1m+r_2m$ and $r(m_1+m_2)=rm_1+rm_2$. We say that a module is graded if we can break it up as above so that $R_iM_j\subset M_{i+j}$.

We’ll call a module finitely generated if there is a finite collection of elements $m_1,\ldots,m_n\in M$ such that every element $m\in M$ can be written as $\sum_{i=1}^n r_im_i=m$ for some elements of the ring. We note now that some examples of modules are vector spaces over a field, and ideals of rings. In fact, an ideal in a ring $R$ is precisely the same as a submodule of $R$ over itself. We define a ring to be noetherian if every ideal is finitely generated as an $R$-module.

This lets us phrase a stronger version of the Hilbert Basis Theorem that we mentioned before: let $R$ be a noetherian ring. Then $R[x]$ is noetherian.

And, by analogy with vector spaces, a map of modules will just be a map that is linear in precisely the same way, that $f:M\to N$ is a module homomorphism if $f(r_1m_1+r_2m_2)=r_1 f(m_1)+r_2 f(m_2)$.

Now, in addition to gradings on modules, we define a filtration to be a descending sequence $M=M^0\supset M^1\supset\ldots\supset M^r=0$. So now I can state a result that we’ll need.

Let $M$ be a finitely generated graded module over a noetherian graded ring $S$. Then there exists a filtration such that $M^i/M^{i+1}$ is isomorphic $S/\mathfrak{p}_i$ with the grading shifted for some prime ideal $\mathfrak{p}_i$, this filtration is not unique though. Also, though the ordering may change, the collection of prime ideals $\{\mathfrak{p}_i\}$ along with the number of times each appears doesn’t depend on the filtration.

This turns out to be EXTREMELY valuable. Let $\mathfrak{p}$ be a prime ideal in a noetherian graded ring $S$, and let $M$ be an $S$-module. We define the multiplicity of $\mathfrak{p}$ in $M$, denoted by $\mu_{\mathfrak{p}}(M)$, to be the number of times that $\mathfrak{p}$ appears in any filtration as above.

This notion will turn out to be just what we need to define an intersection multiplicity that will make a certain more general version of Bezout’s Theorem work.

That’ll be all for now. Next time, we’ll define the Hilbert Polynomial, and talk a bit about the geometric concepts is extends or rigorizes.

Advertisements ## 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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### 4 Responses to Gradings of Rings and Modules

1. Francisco Palmios says:

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