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I already got a proof for the fact that if a polynomial map is surjective then it is also injective. However, I used the invariant dimension of a ring and I want a simpler proof. Bravo for any try. For preciseness, the statement of the fact is as follows:

Statement: Consider two polynomial rings $k[x_1,...,x_n], k[y_1,...,y_n]$. Let $\Phi: k[x_1,...,x_n] \rightarrow k[y_1,...,y_n]$ be a $k$-algebra homomorphism. If $\Phi$ is surjective then $\Phi$ is also injective.

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Dear Martin, thanks for your comment. However, I think you misread our statement here. The Ax-Grothendieck theorem says that if a polynomial map $\Phi: \mathbb{C}^n \rightarrow \mathbb{C}^n$ is injective then it is also surjective. Here we state the other way around over any field. –  mr.bigproblem Jul 25 '11 at 14:49
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4 Answers

up vote 12 down vote accepted

If $A$ is any Noetherian ring, then any surjective homomorphism $\varphi: A\to A$ is injective. One has the ascending chain of ideals $\ker \varphi\subseteq \ker \varphi^2\subseteq \cdots$. Thus $\ker \varphi^n=\ker \varphi^{n+1}$ for some $n$. Let $a\in \ker \varphi$. Since $\varphi^n$ is surjective, we can write $a=\varphi^n(b)$ for some $b\in A$. The $0=\varphi(a)=\varphi^{n+1}(b)$. So $b\in \ker \varphi^{n+1}=\ker \varphi^n$. Thus $a=\varphi^n(b)=0$ and so $\varphi$ is injective.

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Thanks Benjamin! So elementary and so cool! –  mr.bigproblem Jul 26 '11 at 0:55
... and a solution to a well-known exercise ;). –  Martin Brandenburg Aug 5 '11 at 9:51
@Martin, I agree and certainly claim no originality here. But I think that this was the answer the OP was looking for. Since the other responses used more complicated and less general methods, I thought it worth adding. –  Benjamin Steinberg Aug 5 '11 at 17:33
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In general, let $\phi \colon A \to B$ be a ring homomorphism and set $X= \operatorname{Spec}(A)$ and $Y=\operatorname{Spec}(B)$.

Then $\phi$ induces a mapping $\phi^{*} \colon Y \to X;$ moreover, if $\phi$ is surjective than $\phi$ is an isomorphism of $Y$ into the closed subset $V(\ker \phi) \subset X$ [Atiyah-Macdonald, Ex. 21 of Chapter 1].

In your case, $X=Y=\mathbb{A}_k^n$, the affine $n$-space over $k$.

Since the only closed subset of $\mathbb{A}_k^n$ isomorphic to $\mathbb{A}_k^n$ is $\mathbb{A}_k^n$ itself, it follows $V(\ker \phi)=\mathbb{A}_k^n$. Then $\ker \phi=\emptyset$, i.e. $\phi$ is injective.

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How do you prove the fact that the only closed subset of $\mathbb{A}^n_k$ isomorphic to $\mathbb{A}^n_k$ is itself? To prove the similar algebraic fact for polynomial rings, I had to use dimension. – mr.bigproblem 0 secs ago –  mr.bigproblem Jul 25 '11 at 8:27
You are right. But really only the definition of dimension sufficies to prove this statement. For a short proof, see [Shafarevich, Algebraic Geometry 1, Chapter I, Section 6, Theorem 1]. By the way, also Jack Huizenga's nice proof uses some kind of "dimension argument": in fact $M/M^2$ can be seen as the cotangent space of $\mathbb{A}^n$ at $(0, \ldots, 0)$. –  Francesco Polizzi Jul 25 '11 at 8:41
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Denote by $\Psi : k^n\to k^n$ the map of affine spaces corresponding to $\Phi$, and without loss of generality assume $\Psi(0) = 0$. Putting $M = (x_1,\ldots,x_n)$ and $N = (y_1,\ldots,y_n)$, this means that $\Phi^{-1}(N) = M$, so $\Phi(M) = N$ since $\Phi$ is surjective. We then get an induced map $\Phi_a:M^a/M^{a+1} \to N^{a}/N^{a+1}$ for any $a\geq 1$. Here both $M^a/M^{a+1}$ and $N^{a}/N^{a+1}$ are $k$-vector spaces of the same dimension, and $\Phi_a$ is thus an isomorphism since it is clearly surjective. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. Choose $a$ so that $f$ lies in $M^a$ but not in $M^{a+1}$ (such an $a$ clearly exists: it is the degree of the lowest degree homogeneous piece of $f$). We then have $\Phi_a(f) = 0$ and $f\notin M^{a+1}$, contradicting that $\Phi_a$ is an isomorphism.

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Dear Jack, how do you imply that $\Phi_*: M/M^2 \rightarrow N/N^2$ is isomorphic? I guess, to verify this, one needs the condition that $Ker \Phi|_M = 0$, which is equivalent to $Ker \Phi = 0$. Moreover, why does it contradict when one has $\Phi_*(f) = 0$? Do you mean that this implies $f \in M^2$ and then using induction implies $f \in M^n$ and finally by Krull's intersection theorem, $f = 0$, a contradiction? –  mr.bigproblem Jul 25 '11 at 7:09
You are right, there were some issues with the original. I think it's been fixed now. –  Jack Huizenga Jul 25 '11 at 7:26
Yes, now it's a correct proof. Thanks! –  mr.bigproblem Jul 25 '11 at 7:41
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The very short proof I have is as follows.

Suppose that $\Phi: k[x_1,...,x_n] \rightarrow k[y_1,...,y_n]$ is surjective then we have an isomorphism $k[x_1,...,x_n]/I \cong k[y_1,...,y_n]$ for some ideal $I$ of $k[x_1,...,x_n]$. This implies that $\mbox{dim}k[x_1,...,x_n]/I = \mbox{dim}k[y_1,...,y_n] = n$. Hence, we can find a maximal chain of primes $0 \subset P_0/I \subset ... \subset P_n/I$ in $k[x_1,...,x_n]/I$. If $I \neq 0$ then we have a longer chain of primes $0 \subset P_0 \subset ... \subset P_n$ in $k[x_1,...,x_n]$, a contradiction. So $I = 0$ and $\Phi$ is injective.

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Your chains should stop at $P_{n-1}$ (to get chains of lengths $n$ and $n+1$ respectively). I don't see how your proof is different from that of Francesco Polizzi. –  Qing Liu Jul 25 '11 at 20:07
Dear Qing Liu, in the first chain, $0/I$ is not counted so the length is $n$. In the second chain $0 \subset P_0 \subset ... \subset P_n$ has length $n+1$. You are right that this proof is just the algebraic version of Francesco's. –  mr.bigproblem Jul 26 '11 at 0:58
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