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Suppose $k$ is a field that is finitely generated as a ${\mathbb Z}$-algebra. (That is, $k$ is a quotient of ${\mathbb Z}[X_1,\dots,X_n]$ for some $n$). Does it follow that $k$ is finite?

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    $\begingroup$ A quick glance at another post makes me think you should be able to first show that $K$ is of prime characteristic. Perhaps once that's established, you can determine the degree of $K$ over its prime subfield is finite? $\endgroup$ – vgty6h7uij May 23 '12 at 13:04
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Yes!
Consider the morphism $f:\mathbb Z\to k$ and the ideal $\mathfrak m=f^{-1}(0)\subset \mathbb Z$.
Since $\mathbb Z$ is a Jacobson ring and $(0)\subset k$ is maximal, $\mathfrak m$ is maximal too and we obtain a morphism $\bar f:\mathbb F_p\to k$.
Since $k$ is finitely generated over $\mathbb F_p$ and is a field, it is actually a finite extension ("Zariski's version of the Nullstellensatz") and thus $k$ is (set-theoretically!) finite.

Edit
Considering the comments below , I had better state explicitly the theorem I have used.
Theorem
Let $f:A\to B$ be a finitely generated $A$-algebra.
If $A$ is a Jacobson ring, , then $B$ is also Jacobson and for every maximal ideal $\mathfrak m\subset B$ the ideal $f^{-1}(\mathfrak m)\subset A$ is also maximal.

New Edit
Here is a proof in the style of algebraic geometry, due to Akaki Tikaradze.
If $k$ has characteristic $p$ we conclude as above, using Zariski.
If $\operatorname{char}k=0$, the morphism $f:\mathbb Z\to k \:$ is injective, so $A$ has no $\mathbb Z$-torsion and is thus $\mathbb Z$-flat.
But then $\operatorname{Spec}(k)\to\operatorname{Spec}(\mathbb Z)$ is open (flat+finite presentation $\implies$ open), i.e. $(0)\in\operatorname{Spec}(\mathbb Z)$ is open. Contradiction.
(This proof is absurdly sophisticated but it will probably appeal to scheme-theory addicts. Take it as some kind of joke...)

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  • $\begingroup$ Georges: My apologies....I thought this made perfect sense this morning but now I suddenly don't see it anymore. ${\mathbb Q}$ is also a Jacobson ring, but the inverse image of (0) under the map ${\mathbb Z}\rightarrow {\mathbb Q}$ is not a maximal ideal. What additional property are you using? $\endgroup$ – WillO May 24 '12 at 3:21
  • $\begingroup$ Dear Will, I'm using that $k$ is finitely generated over $\mathbb Z$ (and that $\mathbb Z $ is Jacobson).This is crucial as your counterexample correctly shows. An online reference is De Jong and collaborators' Stack Project, [Chapter 7: Commutative Algebra, Proposition 32.18 ](math.columbia.edu/algebraic_geometry/stacks-git/algebra.pdf) $\endgroup$ – Georges Elencwajg May 24 '12 at 5:58
  • $\begingroup$ Another reference is the generalization of the nullstellensatz stated here: en.wikipedia.org/wiki/Nullstellensatz . $\endgroup$ – Justin Young May 24 '12 at 12:12
  • $\begingroup$ Georges: Yes, as I said, I understood this the first time, but somehow when I reread it late at night I lost track of the main point. I remember why it made sense now! Thanks for being patient with me. $\endgroup$ – WillO May 24 '12 at 12:43
  • $\begingroup$ PS: You wrote "Since "k" is a Jacobson ring" but I think you meant "Since ${\mathbb Z}$ is a Jacobson ring"; I think you need this to apply the Nullstellensatz. $\endgroup$ – WillO May 24 '12 at 14:28
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In case you are interested, here is a proof which also uses Zariski's lemma but no difficult theorems about Jacobson rings.

Write $R$ for the image of the unique ring homomorphism $\mathbb{Z} \to k$, so that $k$ is a finitely generated $R$-algebra and hence a finite extension of the fraction field of $R$ by Zariski's lemma. Thus it suffices to show that $R = \mathbb{F}_p$, which is to say $k$ has positive characteristic. If $R = \mathbb{Z}$, meaning $k$ has characteristic zero, then $k$ is a number field which is a finitely generated ring. But this is impossible: if we write $k = \mathbb{Z}[\alpha_1,\dots,\alpha_r]$, then one can choose $n \in \mathbb{Z}$ so that all the denominators of coefficients in the minimal polynomials over $\mathbb{Q}$ of $\alpha_1,\dots,\alpha_r$ divide $n$. This implies that $k$ is integral over $\mathbb{Z}[1/n]$. Then $\mathbb{Z}[1/n]$ must be a field, which is absurd.

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When I read the proofs suggested in the answers half a year ago, I had the problem that I did not understand them because of a lack of algebra-knowledge. That's why I would like to share a (quite detailed) proof of the theorem which I finally understood:

I will use Zariski's Lemma: Let $\mathbb{K}$ be a field and $A$ be a finitely generated $\mathbb{K}$-algebra. If $A$ is a field, then it is a finite degree extension of $\mathbb{K}$.

Now the proof: There exist elements $a_1,...,a_n$ such that $R \cong \mathbb{Z} [a_1,...,a_n]$. In order to simplify our notation we may assume that $R = \mathbb{Z} [a_1,...,a_n]$. Why? - If $\phi \colon R \to \mathbb{Z} [a_1,...,a_n]$ is the isomorphism (that means $\phi(\frak{m}) \subseteq \mathbb{Z} [a_1,...,a_n]$ is a maximal ideal) and $\pi \colon \mathbb{Z} [a_1,...,a_n] \to \mathbb{Z} [a_1,...,a_n]/(\phi(\frak{m}))$ denotes the canonical projection, then $\pi \circ \phi \colon R \to \mathbb{Z} [a_1,...,a_n]/(\phi(\frak{m}))$ is a surjective ring homomorphism with $\ker(\pi \circ \phi) = \frak{m}$. Hence the fundamental homomorphism theorem tells us that $R/\frak{m} \cong \mathbb{Z} [a_1,...,a_n]/(\phi(\frak{m}))$. So the ring of the left side is a finite field iff the ring on the right side is a finite field.

Let $i$ be the only possible ring homomorphism $i \colon \mathbb{Z} \to R/ \frak{m}$ and let $S := i( \mathbb{Z}) = \{z + m \mid z \in \mathbb{Z} \}$. Note that "$S$" is a ring. Since $\mathbb{Z}[a_1,...,a_n] = \{ f(a_1,...,a_n) \mid f \in \mathbb{Z}[x_1,...,x_n] \}$, it is easy to see that $R/\frak{m} = \mathbb{Z}[a_1,...,a_n] / \frak{m} =$ $S[\bar a_1, ..., \bar a_n]$, where $\bar a_i = a_i + \frak{m}$, is a finitely generated $S$-algebra, hence a finite degree extension of the quotient field of $S$ (according to Zariski's lemma). Let's assume that $i$ is injective $\Longleftrightarrow \mathbb{Z} = S$. Thus $R/ \frak{m}$ is a finite degree extension of $\mathbb{Q}$. From this we conclude that the elements $\bar a_1,..., \bar a_n$ with $R/ \mathfrak{m} = \mathbb{Z}[\bar a_1,..., \bar a_n]$ are algebraic over $\mathbb{Q}$ which means that they are zeros of certain monic polynomials with coefficients in $\mathbb{Q}$. Now let $k$ be the product of the denominators of the coefficients of the minimal polynomials of the $\bar a_i$ over $\mathbb{Q}$. As a consequence, the coefficients of the minimal polynomials can be considered as elements of $\mathbb{Z}[ \frac{1}{k}]$, hence the $\bar a_i$ are integral over $\mathbb{Z}[ \frac{1}{k}]$ and so the whole $R/ \mathfrak{m}$ is integral over $\mathbb{Z}[ \frac{1}{k}]$.

Let $a \in \mathbb{Z}[ \frac{1}{k}], a \neq 0$ arbitrary, thus $a \in \mathbb{Q} \subseteq R/ \frak{m}$. Since $R/ \frak{m}$ is (a field and) integral over $\mathbb{Z}[ \frac{1}{k}]$, there exist $m \in \mathbb{N}, c_0,...,c_{m-1} \in \mathbb{Z}[ \frac{1}{k}]$ such that $$a^{-m} + c_{m-1} a^{-(m-1)}+\cdots+c_0 = 0,$$ and thus $$a^{-1}=-(c_{n-1}+\cdots+c_0 a^{m-1}) \in \mathbb{Z}[ \frac{1}{k}].$$ Hence $\mathbb{Z}[ \frac{1}{k}]$ is a field, but that's a contradiction, since any prime number $q$, which does not divide $k$, cannot be invertible in $\mathbb{Z}[ \frac{1}{k}]$, because there is no element like $\frac{1}{q}$. So $i$ cannot be injective.

Thus there exists a $p \in \mathbb{N}$ prime such that $S = \mathbb{Z}/ p \mathbb{Z}$, hence $S$ is a finite field. From above we conclude that $R/ \frak{m}$ is a finite degree extension of a finite field, thus a finite field. $\Box$

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This is a mild generalization of Justin Campbell's answer, and the proof is almost the same as that of Proposition 7.8 in Atiyah and MacDonald's Introduction to Commutative Algebra. (This proposition is sometimes called the Artin-Tate Lemma.) The notation has been designed to make the analogy as clear as possible.

Let $A\subset B\subset C$ be a tower of domains such that $B$ is the field of fractions of $A$, and $C$ is a field which is finitely generated as an $A$-algebra. Then (a) $C$ is a finite extension of $B$, and (b) $B$ is a finitely generated $A$-algebra.

As (a) results immediately from Zariski's Lemma, it suffices to prove (b). The following statement is ironically stronger and easier to prove than (b):

Assume that $C=A[x_1,\dots,x_m]$; that $y_1,\dots,y_n$ is a $B$-basis of $C$ with $y_1=1$; and that we have $$x_i=\sum_jb_{ij}y_j,\quad y_iy_j=\sum_kb_{ijk}y_k$$ with $b_{ij}$, $b_{ijk}$ in $B$. Then the $A$-algebra $B$ is generated by the $b_{ij}$ and the $b_{ijk}$.

The proof is straightforward, and left to the reader.

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