6
$\begingroup$

Hello, I was looking for an answer to the following question:

Consider an algebraically closed field $K$ and a map $K \rightarrow K^n$ given by $a \mapsto (p_1 (a) , \ldots p_n (a))$ for $p_i \in K[t]$. Is the image of this map an algebraic set?

Certainly this is false for $K^2 \rightarrow K^n$, for example $(x,y) \mapsto (x, xy)$ but I feel like polynomials in $1$ variable aren't complicated enough to give this bad behavior.

Thanks!

Improve this question
$\endgroup$
2
  • 11
    $\begingroup$ Yes. Probably there are many solutions. Here is one. We can assume the map is non-constant, so it extends to a non-constant map $f:\mathbf{P}^1 \rightarrow \mathbf{P}^n$ with closed image (due to completeness) that is a curve. As such it meets the hyperplane at infinity, and that intersection must be the point at infinity on $\mathbf{P}^1$ since $f$ carries $\mathbf{A}^1$ into $\mathbf{A}^n$. It follows that the closed set $f(\mathbf{P}^1)$ in $\mathbf{P}^n$ meets $\mathbf{A}^n$ in exactly $f(\mathbf{A}^1)$, so the latter is closed in $\mathbf{A}^n$. $\endgroup$
    – BCnrd
    Commented Oct 10, 2010 at 0:27
  • $\begingroup$ Very nice and geometric. $\endgroup$
    – Georges Elencwajg
    Commented Oct 10, 2010 at 9:27

1 Answer 1

Reset to default
5
$\begingroup$

Dear Damien, let's show that your morphism $f: \mathbb A^1_K \to \mathbb A^n_K $ is proper, hence closed, hence certainly has closed image.

For that it is enough to prove that each $f_i:\mathbb A^1_K \to \mathbb A^1_K$ is proper. But this follows from the stronger property that $f_i$ is finite or dually that the ring morphism $K[T] \to K[T]: T\to p_i(T)$ is finite. This is elementary: it follows, for example, from the fact that $T$ is (tautologically) integral over $K[p_i (T)]$.

Note that in this proof you needn't assume that the field $K$ is algebraically closed.

Edit: As BCnrd remarks, this proof only works if all polynomials $p_i(T)$ are non-constant. Let me modify the proof to take his judicious comments into account. If all polynomials are constant, your morphism is not proper but its image is clearly closed. If at least one polynomial is non-constant, say the first, then the argument above proves that the corresponding morphism $f_1: \mathbb A^1_K \to \mathbb A^1_K $ is finite.The obvious closed immersion $j: \mathbb A^1_K \to \mathbb A^n_K $ (last $n-1$ coordinates given by the other polynomials) is finite and the composition, which is your morphism $f=j\circ f_1: \mathbb A^1_K \to \mathbb A^n_K $ , is thus also finite.

Improve this answer
$\endgroup$
2
  • 3
    $\begingroup$ The properness of $f_i$ fails when $p_i$ is constant, so one needs a small extra argument to deal with that (though finiteness of a single $f_i$ implies that of $f$). Or put another way, the argument is that $K[T]$ is module-finite over every $K$-subalgebra strictly larger than $K$. $\endgroup$
    – BCnrd
    Commented Oct 10, 2010 at 12:53
  • 1
    $\begingroup$ You're absolutely right BCnrd: nothing escapes your eagle eye! I have accounted for the possibility of some constant polynomials in an edit. $\endgroup$
    – Georges Elencwajg
    Commented Oct 11, 2010 at 0:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .