# 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 $M = (x_1-a_1,\dots,x_n-a_n)$. QED.

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???