Regular Morphism From Affine Line 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!
 A: 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. 
