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Given a field $\mathbb{K}$ which is algebraically closed and of characteristic 0, we can say exactly what the maximal ideals of $\mathbb{K}[x_1,\dots,x_n]$ are and they correspond to points in $\mathbb{K}^n$ (thanks to the Nullstellensatz).

Can I say anything about the maximal ideals of $\mathbb{K}$ if char$\mathbb{K}\neq0$ or if it is not algebraically closed?


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The nullstellensatz does not require characteristic zero. Generalizations can be found here: . – Justin Young May 1 '12 at 14:48
+1,also, For it motivates the nice answer. :) – wxu May 1 '12 at 15:50
Even Hilbert's Nullstellensatz says something far more interesting than "maximal ideals = points": it says a polynomial function $f$ vanishes on the zero locus of an ideal $I$ of $k[x_1, \ldots, x_n]$ if and only if a power of $f$ is contained in $I$. This really and honestly does require the field to be algebraically closed. – Zhen Lin May 1 '12 at 16:56

1) Characteristic $0$ is irrelevant: if $K$ is an algebraically closed field of any characteristic the maximal ideals of $K[x_1,\ldots,x_n]$ are the ideals $I_a=(x_1-a_1,...,x_n-a_n)$ consisting of the polynomials $f\in K[x_1,\ldots,x_n]$ vanishing on $a=(a_1,\ldots,a_n)\in K^n$.

2) For an arbitrary, not necessarily algebraically closed, field $K$ the maximal ideals of $K[x_1,\ldots,x_n]$ correspond to the closed points of affine space $\mathbb A^n_K$.
A rough description of these points is as follows:
Choose an algebraic closure $K^a$ of $K$. There is a canonical morphism of schemes $\mathbb A^n_{K^a} \to \mathbb A^n_K$, dual to the inclusion $K[x_1,\ldots,x_n]\to K^a[x_1,\ldots,x_n]$. The closed points of $\mathbb A^n_K$ are the images of the closed points of $\mathbb A^n_{K^a}$.
Here are a few facts surrounding/interpreting that geometric description :

a) A prime ideal $\mathfrak p\subset K[x_1,\ldots,x_n]$ is maximal $\iff $ the extension $K\subset Frac(A/\mathfrak p)$ is finite.
b) An ideal $I\subset K[x_1,\ldots,x_n]$ is maximal $\iff $ there exists $a\in (K^a)^n$ such that $I$ is the zero set $$I_a= \lbrace f\in K[x_1,\ldots,x_n]\mid f(a)=0\rbrace $$ c) Given $a,b \in (K^a)^n$ , the corresponding maximal ideals $I_a, I_b$ are equal $\iff$ there exists a $K$-automorphism $s\in Aut(K^a/K)$ such that $s(a_i)=b_i$. (Note that $K^a/K$ is not assumed Galois.)

Geometrically b) says that $\mathbb A^n_{K^a} \to \mathbb A^n_K$ is surjective and c) describes the fibers of that morphism as the orbits of $Aut(K^a/K)$ acting on $(K^a)^n$

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+1,Nice answer! – wxu May 1 '12 at 15:43
Thanks, wxu ${}$. – Georges Elencwajg May 1 '12 at 15:44
Thank you! So I guess I only need to know what happens if the field is both of positive characteristic and it is not algebraically closed! – user11428 May 1 '12 at 17:18
Dear user, yes and no: if the field $K$ is non algebraically closed of characteristic zero and $n=1$, the determination of the maximal ideals of $K[x]$ amounts to that of the irreducible monic polynomials in $K[x]$. Even if there are irreducibility algorithms and criteria , the problem remains non-trivial and is even more difficult for $n\gt 1$. – Georges Elencwajg May 1 '12 at 18:31

If the field has characteristic $0$ (or, more generally, is perfect) then you can use Galois theory: the maximal ideals correspond to orbits of points in $\overline{K}^n$ by ${\rm Gal}(\overline{K}/K)$. Given a point $P=(a_1,\dots,a_n)$ in $\overline{K}^n$, evaluation of polynomials in $K[x_1,\dots,x_n]$ at $P$ is a $K$-alg. hom from polynomials to $\overline{K}$ whose kernel is a maximal ideal of $K[x_1,\dots,x_n]$. All maximal ideals arise in this way, and two points $P$ and $Q$ give rise to the same maximal ideal iff $Q = \sigma(P)$ for some $\sigma$ in ${\rm Gal}(\overline{K}/K)$.

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