MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 13 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.

share|cite|improve this answer
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

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.

share|cite|improve this answer
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

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.

share|cite|improve this answer
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.