Let $K$ be an algebraically closed field and let $k$ be a subfield of $K$ such that the field extension $K \mid k$ is algebraic. Let $B$ be the polynomial ring $K [x_1, \ldots, x_n]$ and let $A$ be the subring $k [x_1, \ldots, x_n]$. It is clear that any $k$-subalgebra of $K$ is a field, so it follows that the intersection of $A$ and any maximal ideal of $B$ is a maximal ideal of $A$. Moreover, every maximal ideal of $A$ arises in this way.

Question. What are the maximal ideals of $A$, described in terms of the maximal ideals of $B$?

It is well-known that the maximal ideals of $B$ are all of the form $(x_1 - a_1, \ldots, x_n - a_n)$ for some $n$-tuple $(a_1, \ldots, a_n)$ of elements of $K$. Thus, for each maximal ideal $\mathfrak{m}$ of $A$, we may define the set $$P(\mathfrak{m}) = \{ (a_1, \ldots, a_n) \in K^n : (x_1 - a_1, \ldots, x_n - a_n) \cap A = \mathfrak{m} \}$$ and this set is finite (because it is a 0-dimensional closed subset of $\mathbb{A}^n_K$). Consider $G = \mathrm{Aut}(K \mid k)$. Of course, $G$ acts on $B$, and every element of $A$ is invariant under the action of $G$, so it follows that $P (\mathfrak{m})$ is closed under the action of $G$ in $K^n$ as well, i.e. $P (\mathfrak{m})$ is the union of $G$-orbits.

Conjecture. $G$ acts transitively on $P (\mathfrak{m})$.

This is certainly true if $k$ is a perfect field and $n = 1$ – after all, in that case, $\mathfrak{m}$ is a principal ideal generated by a polynomial in one variable irreducible over $k$. In fact this is true even without the assumption that $k$ is perfect: we know that the residue field $\kappa (\mathfrak{m}) = A / \mathfrak{m}$ is a finite field extension of $k$ and that the right $G$-set of $k$-algebra embeddings $\kappa (\mathfrak{m}) \to K$ is isomorphic to the right $G$-set of zeros in $K$ for the generator of $\mathfrak{m}$; but the homomorphism extension property of $\kappa (\mathfrak{m}) \to K$ implies that any two embeddings can be connected by an isomorphism of $K$, so $G$ acts transitively on the set of $k$-algebra embeddings $\kappa (\mathfrak{m}) \to K$.

What about the general case where $n > 1$? I had always taken for granted that the indicated conjecture was true, but I have not been able to find a proof in the usual places. By descent theory, one can show that $$\operatorname{Spec} B \otimes_k B \rightrightarrows \operatorname{Spec} B \to \operatorname{Spec} A$$ is a coequaliser diagram in the category of schemes, so $\operatorname{Spec} A$ is a scheme-theoretic quotient of $\operatorname{Spec} B$ by an internal equivalence relation, but it is not clear to me what this internal equivalence relation is when $n > 0$. (For $n = 0$ and $k$ perfect, something quite remarkable happens: $\operatorname{Spec} K \otimes_k K$ has a canonical continuous free transitive action of $\mathrm{Gal}(K \mid k)$!)


1 Answer 1


It is well-known that the set of maximal ideals of $k[x_1,\dotsc,x_n]$ corresponds to $K^n / \mathrm{Aut}(K/k)$. In fact, every maximal ideal $\mathfrak{m}$ has a residue field which is finite over $k$ (Corollary of Noether normalization), hence has an embedding into $K$, so $\mathfrak{m} = \{f \in k[x_1,\dotsc,x_n] : f(a)=0\}$ for some $a \in K^n$. When $a,b$ are conjugate under the action of $\mathrm{Aut}(K/k)$, the maximal ideals clearly coincide. Conversely, if $\{f : f(a)=0\} = \{f : f(b)=0\}$, then $k(a_1,\dotsc,a_n) \cong k(b_1,\dotsc,b_n)$ via $a_i \mapsto b_i$, and this extends to an automorphism of the algebraic closure $K$. Hence $a$ and $b$ are conjugate.

Hence, $\mathrm{Spm}(k[x_1,\dotsc,x_n]) \cong \mathrm{Spm}(K[x_1,\dotsc,x_n]) / \mathrm{Aut}(K/k)$ as sets.

More generally, if $R \subseteq R'$ is an integral extension of normal domains such that $Q(R')/Q(R)$ is a normal field extension, then $\mathrm{Spm}(R) \cong \mathrm{Spm}(R') / \mathrm{Aut}(R'/R)$ and the same with $\mathrm{Spec}$ (Cohen-Seidenberg).


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