$\mathbb{Z}[x_{1},\dots,x_{n}]/I$ is a field therefore it's finite

Question

I'd spent much time for this but didn't get any results.. Could u give me only the idea but not a full proof

Answer 1

It is indeed true. Suppose that $I$ is an ideal of $\mathbf{Z}[x_{1},\dots,x_{n}]$ and that $A :=\mathbf{Z}[x_{1},\dots,x_{n}]/I$ is a field (so that $I$ is maximal ideal of $\mathbf{Z}[x_{1},\dots,x_{n}]$) and note $f : \mathbf{Z}\to A$ the unique morphism of rings from $\mathbf{Z}$ to $A$. Its kernel is an strict (because this morphism is not the zero morphism !) ideal of $\mathbf{Z}$ whose ideals are $(0)$ and $(p)$ for $p$ prime. Imagine that we have shown that it is not $(0)$ but a $(p)$ for $p$ prime. Then $A$ is a field extension of $\mathbf{F}_p$ such that $A = \mathbf{F}_p [u_1,\ldots,u_n]$ where $u_i$ is the image of $x_i$ in $A$. This implies that all $u_i$ are algebraic over $\mathbf{F}_p$, and that the extension $A$ is of finite degree that we note $d$, so that $A$ is isomorphic to $(\mathbf{F}_p)^d$ as $\mathbf{F}_p$-vector space, and is of cardinal $p^d$, and is indeed finite. Now, to conclude, we need to show that the kernel cannot be equal to $(0)$. Then $f$ extends to a (necessarily injective) morphism of fields $F : \mathbf{Q}\to A$. Note that if $u_i$ is the image of $x_i$ in $A$, we have $A = \mathbf{Z}[u_{1},\dots,u_{n}] = \mathbf{Q}[u_{1},\dots,u_{n}]$ and as this is a field, it implies that each $u_i$ is in fact algebraic over $\mathbf{Q}$, and $A$ is an extension of finite degree of $\mathbf{Q}$ (through $F$). Now each $u_i$ has a minimal polynomial over $\mathbf{Q}$, write all its coefficient as irreductible fractions, and note $d_i$ the product of all denominators of these fractions, and finally note $d$ the product of the $d_i$'s. Then $A$ is integral over $B=\mathbf{Z}\left[\frac{1}{d}\right]$. Which implies that $\mathbf{Q}$ is integral over $B$, which implies (as $Q$ is a field) that $B$ is a field, which is a contradicticon.

For a more conceptual proof, you can look for the so-called Zariski's lemma. See the discussion here :

If a ring is Noetherian, then every subring is finitely generated?

Answer 2

First we need to show the kernel of $f:\mathbb Z\to K:=\mathbb Z[x_1,\dots,x_n]/I$ contains some prime number $p$. Suppose not, then we have an inclusion of rings $\mathbb Z\subset \mathbb Q\subset K$. Since $K$ is a finitely generated $\mathbb Z$-algebra, it is also a finitely generated $\mathbb Q$-algebra. By Zariski's lemma, $K$ is a finite extension of $\mathbb Q$. In other words, $K$ is a finite $\mathbb Q$-module. Also $K$ is a finitely generated $\mathbb Z$-algebra. This implies $\mathbb Q$ is a finitely generated $\mathbb Z$-algebra (by Proposition 14.2 in Sharp, Steps in Comuutative Algebra), which is nonsense.

Now knowing that $I$ contain some prime number $p$, so we can quotient out $(p)$ to get $K=\mathbb F_p[x_1,\dots,x_n]/\bar I$, which is a finitely generated $\mathbb F_p$-algebra. Apply Zariski's lemma again, we know $K$ is a finite extension of $\mathbb F_p$, which is a finite field.