Cfrgtkky

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?

Yes!

Consider the morphism
$f:mathbb Zto k$ 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 $bar f:mathbb F_pto k$.

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 $f:Ato B$ be a finitely generated $A$-algebra.

If $A$ is a Jacobson ring, , then $B$ is also Jacobson and for every maximal ideal $mathfrak msubset B$ 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 $f:mathbb Zto k :$ is injective, so $A$ has no $mathbb Z$-torsion and is thus $mathbb Z$-flat.

But then $operatorname{Spec}(k)tooperatorname{Spec}(mathbb Z)$ is open (flat+finite presentation $implies$ open), i.e. $(0)inoperatorname{Spec}(mathbb Z)$ is open. Contradiction.

(This proof is absurdly sophisticated but it will probably appeal to scheme-theory addicts.
Take it as some kind of joke...)

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.

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. $Box$

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 $Asubset Bsubset C$ 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.