Geometric interpretation of Hilbert's Nullstellensatz via gluing of Galois conjugates

Question

In Vakil's The Rising Sea, he gives the following version of Hilbert's Nullstellensatz:

Theorem 3.2.6: If $k$ is any field, every maximal ideal of $k[x_1, \dots, x_n]$ has residue field a finite extension of $k$.

He later says the following:

If the field is not algebraically closed, the traditional [closed] points are glued together into clumps by Galois conjugation ... This is a geometric interpretation of Hilbert's Nullstellensatz 3.2.6.

I am trying to understand how this is a geometric interpretation, and I think I am stuck because he hasn't yet explicitly laid out the geometric interpretation of residue fields. My understanding from a few examples is that the closed points of $k[x_1, \dots, x_n]$ are the maximal ideals, and from the Nullstellensatz we get that the quotient $k[x_1, \dots, x_n] / \mathfrak{m}$ is a finite (hence algebraic) extension of $k[x_1, \dots, x_n]$. Somehow this should then tell us that the closed point $\mathfrak{m}$ is generated by irreducible polynomials and therefore corresponds to gluing together Galois conjugates, but I'm not really able to fill in the details.

Could anyone share a clear explanation of how the Nullstellensatz gets interpreted geometrically in terms of gluing Galois conjugates?

Answer 1

Maybe you can think of this example, hope it’s not a duplicate either : take the affine line $ \mathbf{A}^1$ over a a perfect field, not closed. Then any point $P \in \mathbf{A}^1$ corresponds to a maximal ideal $m \subset k[t]$ given by an irreducible polynomial $f(t)$. This polynomial splits in the separable closure $k’$ as a product $\Pi_i(t-a_i)$ and if you apply an element of $G=Gal(k’/k)$ to this splitting, you see that the roots $a_i$ are conjugate under $G$. So you can think of $P$ as a collection of points $a_1, \dots, a_n$ conjugate under $G$.

Answer 2

Let $L/K$ be a finite Galois extension. We have a morphism of $K$-schemes $ \Bbb A^n_{L}\to\Bbb A^n_K$ which induces a map:

$$L^n=\Bbb A^n_{L}(L) \to \Bbb A^n_K(L)$$

This map is surjective: any $K$-algebra homomorphism $K[x_1, \dots, x_n] \to L$ can be factored $K[x_1, \dots, x_n] \to L[x_1, \dots, x_n] \to L$. We have an action of $G:=\mathrm{Gal}(L/K)$ on $\Bbb A^n_{K}(L)$. It's not hard to see that elements in the same orbit get mapped to the same element in $\Bbb A^n_K(L)$

The ring extension $K[x_1, \dots, x_n] \hookrightarrow L[x_1, \dots, x_n]$ is integral (even finite) becaues $K \hookrightarrow L$ is algebraic (finite). Now let $a,b \in L$ such that their associated maximal ideals $\mathfrak{m}_a$ and $\mathfrak{m}_b$ satisfy $k[x_1, \dots, x_n] \cap \mathfrak{m}_a= k[x_1, \dots, x_n] \cap \mathfrak{m}_b$. Take any $f \in \mathfrak{m}_a$. Then by Galois theory, we have

$$\prod_{\sigma \in G} \sigma(f) \in K[x_1, \dots, x_n] \cap \mathfrak{m}_a= K[x_1, \dots, x_n] \cap \mathfrak{m}_b \subset\mathfrak{m}_b$$

Because $\mathfrak{m}_b$ is a prime ideal, we get $\sigma(f) \in \mathfrak{m}_b$ for some $\sigma$, or equivalently $f \in \sigma^{-1}(\mathfrak{m}_b)$. Because this holds for any $f$, we get $$\mathfrak{m}_a \subset \bigcup_{\sigma \in G} \sigma(\mathfrak{m}_b)$$ The prime avoidance lemma implies that $\mathfrak{m}_a \subset \sigma(\mathfrak{m}_b)$ for some $\sigma$ and by incomparability of prime ideals in integral extensions, we get $\mathfrak{m}_a = \sigma(\mathfrak{m}_b)$. So we have shown that elements in the same fiber are in the same Galois orbit. We can sum this up in a commutative diagram:

$$\require{AMScd} \begin{CD} L^n @>{\cong}>> \Bbb{A}^n_L(L)\\ @VVV @VVV\\ L^n/G @>>{\cong}> \Bbb{A}^n_K(L) \end{CD}$$

Now if $\overline{K}/K$ is a fixed algebraic closure (and we assume for simplicity that $K$ is perfect), then we want to take a colimit of this diagram in a suitable sense. This works because of the Hilbert Nullstellensatz you mention: any element of $\Bbb A^n_K(\overline{K})$ is a closed point and hence has residue field a finite extension of $K$ and hence lies in $\Bbb A^n_K(L)$ for some finite Galois extension $L/K$. We obtain $\Bbb A^n_K(\overline{K})=\displaystyle \bigcup \Bbb A^n_K(L)$. We also have that $\overline{K}^n = \displaystyle \bigcup L^n$.

Thus to define a map $\overline{K}^n/\mathrm{Gal}(\overline{K}/K) \to \Bbb A^n_K(\overline{K})$, we can proceed as follows: for an orbit $[x]$ take any representative $x$, then this lies in some $L^n$, then using the diagram at finite level we can map this $L^n \to \Bbb A^n_L(L) \to \Bbb A^n_K(L) \subset \Bbb A^n_K(\overline{K})$. Using the fact that we have a bijection $L^n/G \to \Bbb A^n_K(L)$ for all $L$ and that the commutative diagrams are compatible for varying $L$, one obtains a commutative diagram:

$$\require{AMScd} \begin{CD} \overline{K}^n @>{\cong}>> \Bbb{A}^n_\overline{K}(\overline{K})\\ @VVV @VVV\\ \overline{K}^n/\mathrm{Gal}(\overline{K}/K) @>>{\cong}> \Bbb{A}^n_K(\overline{K}) \end{CD}$$

Again by the version of the Nullstellensatz in the OP, we see that every closed point of $\Bbb A^n_K$ is in fact $\overline{K}$-rational (and conversely every $\overline{K}$-rational point is closed).