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?

  • 1
    $\begingroup$ I think this has been covered in detail here before (but I'm on my phone and can't search well at the moment). The basic idea is that the map $k[x]\to\overline{k}[x]$ works by having the preimage of any ideal of the form $(x-g(a))$ for a fixed $a\in\overline{k}$ and variable $g\in Gal(\overline{k}/k)$ be the same ideal, $(m_a(x))$, the minimal polynomial of $a$. So one can say that the map glues together galois conjugates. $\endgroup$
    – KReiser
    Nov 3, 2022 at 20:45
  • $\begingroup$ See my answer here: math.stackexchange.com/a/3301566/348926 $\endgroup$ Nov 4, 2022 at 8:37
  • $\begingroup$ Both of these are helpful, but I guess I'm wondering about how this changes in the multivariate case, and how that presumably leads to using this version of the Nullstellensatz? For example, Exercise 3.2.H in The Rising Sea and the paragraph before it hints that we can view $\mathbb{A}_\mathbb{Q}^n$ as $\mathbb{A}_{\overline{\mathbb{Q}}}^n$ with Galois conjugates glued together, and this quote from section 3.3 seems to suggest that the Nullstellensatz is used in getting this same idea to work in affine $n$-space $\endgroup$ Nov 4, 2022 at 15:46

2 Answers 2


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 $G$.


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).


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