We have this strong version of Hilbert's Nullstellensatz:

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

Now I have two questions:

1) Why this is equivalent to say that any field extension of $k$ that is finitely generated as ring is finitely generated as a module?

2) And why this form implies the weak form of Nullstellensatz? (If $k$ is an algebraically closed field, then the maximal ideals of $k[x_1, \dots, x_n]$ are precisely those ideals of the form $(x_1-\alpha_1, \dots, x_n -\alpha_n)$).

  • 3
    $\begingroup$ You appear to be going through, one-by-one, the problems in Vakil Part II, chapter 1&2. Anyways, for example, to answer your first question, if you are f.g. as a $k$-algebra, and a field, you look like $k[x_1,\ldots,x_n]/M$ for some maximal ideal $M$. But, the assumption says that you are then a finite $k$-space, which is what you wanted. For the second part, use the fact that necessarily $k[x_1,\ldots,x_n]/M$ is $k$ (being a finite extension of $k$) and then chase where the $x_i$ go to try and get which $x_i-a_i$ generate $M$. $\endgroup$ Sep 17, 2013 at 10:31

1 Answer 1


A lot of this just echoes what Alex Youcis has already said; I thought it would be good to have an answer.

What does it mean for a $B$-algebra $B \to A$ to be finitely generated (or "finite type")? The translation between this statement and "$A$ is a quotient of a polynomial ring $B[x_1, \dots, x_n]$ in finitely many variables by an ideal $I$" must become second nature. Of course, explicitly writing $A$ out in this way gives something slightly more: a choice of generators and a realization of $\operatorname{Spec} A$ inside $\mathbf{A}^n_B$ as a "closed subscheme". Now, $A$ is a field if and only if $I$ is maximal.

If $k$ is algebraically closed then you should be able to show that it has no nontrivial finite field extensions. Maybe it's good to turn things around and see where ideals like $\mathfrak{m} = (x_1 - a_1, \dots, x_n - a_n)$ turn up. What follows applies to any $k$ and describes the "$k$-rational points" of $\mathbf{A}^n_k$. The difference will be that there are closed points not of this form, e.g., the maximal ideal $(x^2 + 1, y - x)$ of $\mathbf{Q}[x, y]$.

Say I have a point $(a_1, \dots, a_n)$ in $k^n$. This induces a morphism of $k$-algebras $\xi\colon k[x_1, \dots, x_n] \to k$, $f \mapsto f(a_1, \dots, a_n)$ and the kernel is $\mathfrak{m}$. Work this out! Now show yourself that you can recover the point of $k^n$ from $\xi$.


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