G-Ideals, Maximal Ideals, and The Nullstellensatz

Let $ R$ be an integral domain and $ K$ is fraction field. If $ K$ is finitely generated over $ R$ then we say that $ R$ is a $ G$-domain, named after Oscar Goldman. This innocuous-looking definition is actually an extremely useful device in commutative algebra that pops up all over the place. In fact, the $ G$-domain usually comes up in the context of quotienting by a prime ideal, so let's call a prime $ P$ in $ R$ a $ G$-ideal if $ R/P$ is a $ G$-domain. In this post, we shall see a few applications of this concept, following Kaplansky's book Commutative Rings" for the theory and some standard examples for illustrations. At the end, we shall see a short paragraph proof of the Nullstellentsatz assuming the theory of $ G$-ideals.

Why is this concept so useful? Perhaps because of the following result: an ideal $ I$ in $ R$ is a $ G$-ideal if and only if it is the contraction of a maximal ideal in $ R[x]$ (although we won't use it, it's interesting to note that the nilradical is actually the intersection of all the $ G$-ideals). It's worth looking at one direction of a proof of this result since it's so short. First, the reader should prove that $ K$ can be generated by one element over $ R$ if and only if it is finitely generated, as an exercise.

Now, it suffices to prove that $ R$ is a $ G$-domain if and only if $ M\cap R = 0$ for some maximal ideal $ M$ of $ R[x]$. If $ R$ is a $ G$-domain, then $ K = R[u^{-1}]$ for some $ u\in R$, and so $ R[x]\to K$ obtained by $ x\mapsto u^{-1}$ is surjective with kernel $ M$ a maximal ideal, and clearly $ M\cap R = 0$ (why clearly?).

That's the easy direction! What's the harder" direction?

Here is a sketch: if $ M$ is a maximal ideal in $ R[x]$ and $ M\cap R = 0$, then $ R\to R[x]\to R[x]/M$ is an injection of $ R$ into a field generated by one element over $ R$. Now we would like to claim that $ K$ is finitely generated over $ R$ too. This follows from two facts:

  1. If $ R$ is an integral domain the polynomial ring $ R[x]$ is not a $ G$-domain
  2. If $ R\subseteq T$ are commutative rings and $ T$ is algebraic over $ R$ and finitely generated as a ring over $ R$, then $ T$ is a $ G$-domain if and only if $ R$ is.

At any rate, we return to the most important for $ G$-ideals: an ideal in $ R$ is a $ G$-ideal if and only if is the contraction of a maximal ideal in $ R[x]$. Let's call a commutative ring a Hilbert ring if every $ G$-ideal is maximal. Here's another fact about $ G$-ideals that make them of central value in conceptual thinking: a ring $ R$ is a Hilbert ring if and only if $ R[x]$ is a Hilbert ring. Hence all polynomial rings (in finitely many variables!) over fields are Hilbert rings (what about infinitely many variables???). Trivial exercise: the homomorphic image of a Hilbert ring is a Hilbert ring.

How can we use this? Now we'll see some applications.

Fiber Product of Function Fields

Let $ k(u)$ and $ k(v)$ be two purely transcendental extensions of $ k$ of transcendence degree one. A nice example in algebraic geometry is to show that $ \mathrm{Spec}(A)$ is infinite and one-dimensional where $ A = k(u)\otimes_kk(v)$. Since $ A$ is actually isomorphic to the localisation of $ k[u,v]$ at the multiplicatively closed set $ f(u)g(v)$ where $ f\in k[u]$ and $ g\in k[v]$, showing that any maximal ideal of $ k[u,v]$ contains some nonzero $ f(u)\in k[u]$ is actually a main piece of this puzzle (the rest follows almost immediately).

However, this follows easily from our discussion on $ G$-ideals: if $ M$ is maximal in $ k[u,v]$ then $ M\cap k[u]$ is a $ G$-ideal, so that $ k[u]/(M\cap k[u])$ is a $ G$-domain, showing that $ M\cap k[u]\not= 0$.

Varieties and Maps of Closed Points

Another typical example of the application of $ G$-ideals, consider a homomorphism of finitely generated $ k$-algebras $ f:A\to B$ where $ k$ is a field. We claim the induced map on affine schemes takes closed points to closed points. Back in algebraic language, this just means the inverse image of a maximal ideal is maximal!

How can we use $ G$-ideals to show this? Let $ M$ be a maximal ideal of $ B$. Then there is an induced injective map $ A/\varphi^{-1}(M)\to B/M$ where $ B/M$ is a finite extension of $ k$, showing that in particular, $ \varphi^{-1}(M)$ is a $ G$-ideal. Since $ A$ is a finitely generated $ k$-algebra, $ \varphi^{-1}(M)$ must be maximal!

The Nullstellensatz

Once the machinery of $ G$-ideals is set up, Hilbert's Nullstellensatz is not far off. Let's prove the weak version", which states that for an algebraically closed field $ k$, any maximal ideal of $ k[x_1,\dots,x_n]$ is of the form $ (x_1-a_1,\dots,x_n-a_n)$ for $ a_i\in k$. Hey, we're trying to prove something about a maximal ideal of a polynomial ring again! So chances are we can use $ G$-ideals.

Let's first observe something that has nothing to do with fields, and whose proof is immediate: if $ M$ is a maximal ideal in $ R[x]$ and $ M\cap R = N$ is maximal in $ R$, then $ M$ can be generated by $ N$ and one more element $ f\in R[x]$, where $ f$ is a monic polynomial which is irreducible in $ R/N[x]$. Obviously, if $ R/N$ is algebraically closed then $ f$ must be of the form $ x-a$ where $ a\in R$.

Now we prove the Nullstellensatz. We use induction on $ n$, the case $ n=1$ being clear. Set $ R = k[x_1,\dots,x_{n-1}]$, and $ M$ maximal in $ R[x_n] = k[x_1,\dots,x_n]$. Then $ M\cap R = N$ is a $ G$-ideal, hence maximal, so it is of the form $ N = (x_1 – a_1,\dots,x_{n-1}-a_{n-1})$ by the induction hypothesis. Since $ R/N\cong k$, it is algebraically closed, so using the fact in the previous paragraph we see $ M = (x_1-a_1,\dots,x_n-a_n)$. .

Again, the main property we seem to keep using is that the contraction of a $ G$-ideal is maximal. Here, we combined it with the property of Hilbert rings" being closed under polynomial extensions.

Can you think of more applications of $ G$-ideals???

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