# Zariski topology analogue for non-algebraically closed fields

Let $k$ be a field and $\bar{k}$ its algebraic closure. The set $X$ of $n$-tuples over $\bar{k}$ can be given the Zariski topology in which the closed sets are the sets of zeros of sets of polynomials in $\bar{k}[x_1,...,x_n]$. Now, the subset $Y$ of $X$ of $n$-tuples over $k$ can be given the subspace topology on the one hand, and on the other, the topology in which the closed sets are the sets of zeroes of sets of polynomials in $k[x_1,...,x_n]$. Do those two coincide? (the first is clearly stronger than the second)

I think that for a perfect $k$ they do, since you can replace each polynomial by the product of its Galois conjugates.

It is also interesting to ask a similar question about transcendental extensions.

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+1. What you have described is not quite the Affine $n$-space over $k$ (you don't claim that it is). The underlying set consists of $n$-tuples in $\bar{k}$, where we identify $(a_1,\ldots ,a_n)$ with $(b_1,\ldots ,b_n)$ if there is an automorphism $\sigma$ of $\bar{k}/k$ with $\sigma(a_i)=b_i$ for all $i$. The topology on this set is as you described. – Brett Frankel Jan 15 '13 at 22:56
This is a really excellent question ! – Georges Elencwajg Jan 16 '13 at 9:15
Thank you Georges. – KotelKanim Jan 16 '13 at 9:43

You are right with $\bar{k}$ and you essentially gave a proof.

Let $K$ be an algebraic field extension of $k$ (e.g. $K=\bar{k}$). Then the Zariski topology on $k^n$ is induced by that of $K^n$.

Proof. The trivial direction: any closed subset of $k^n$ is the intersection of a closed subset of $K^n$ with $k^n$. This is true for any extension $K/k$.

Conversely, let $Z(F)$ be a closed subset of $K^n$ defined by a polynomial $F$ with coefficients in $K$. Then $F$ has coefficients in some finite extension $L/k$ (take the extension generated by the coefficients of $F$!). If $L/k$ is separable, we let $f$ be the product of all conjugates (in an obvious sense) of $F$. Then $f$ has coefficients in $k$ and $Z(F)\cap k^n=Z(f)$ is closed.

If $L/k$ is not separable, then $k$ has positive characteristic $p$, and some power $F^{p^r}$ has coefficients in the separable closure $L_s$ of $k$ in $L$ and we are reduced to the separable extension case. So $Z(F)\cap k^n$ is closed in $k^n$. As any closed subset of $K^n$ is an intersection of $Z(F)$, the claim is proved.

In the transcendental case, surprisingly (for me) this is still true. The extension $K/k$ is algebraic over some purely transcendental extension of $k$. By the algebraic extension case we can suppose $K$ itself is purely transcendental and even of finite type: $K=k(s_1, \dots, s_m)$. Let $F\in K[T_1, \dots, T_n]$. Chasing the denominators in $K$ (which doesn't change $Z(F)\subseteq K^n$), we can suppose $F\in k[s_1, \dots, s_m][T_1, \dots, T_n]$, so we can write $$F(T_1, \dots, T_n)=\sum_{\nu\in \mathbb N^m} f_\nu(T_1,\dots, T_n)s^\nu, \quad f_v\in k[T_1, \cdots, T_n].$$ Let $(a_1, \cdots, a_n)\in k^n$. Then $F(a_1,\dots, a_n)=0$ if and only if $$f_\nu(a_1, \cdots, a_n)=0, \quad \forall \nu\in \mathbb N^m.$$ So $Z(F)\cap k^n=\cap_{\nu} Z(f_\nu)$ is closed in $k^n$.

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Thank you for this very complete answer. – KotelKanim Jan 16 '13 at 9:44
Dear QiL, I tried to solve the problem (in the case of an algebraic extension $k\subset K$) more abstractly by using that the morphism $\mathbb A^n_K\to \mathbb A^n_k$ is closed, since it is integral as a base change of $Spec (K) \to Spec(k)$ by $\mathbb A^n_k\to Spec(k)$. However I got confused and didn't convince myself that this idea works. Does it? – Georges Elencwajg Jan 16 '13 at 9:53
Dear @GeorgesElencwajg: I also thought about this possibility and got probably the same problem as you. The image of closed susbset $Z\subseteq \mathbb A^n_K$ is $\mathbb A^n_k$ is a closed subset $Z_0$, and $Z(K)\cap k^n\subseteq Z_0(k)$. But the converse is unclear for me, even though it is probably correct. – user18119 Jan 16 '13 at 22:40
Dear QiL, thanks a lot for your answer: it is reassuring to be in such excellent company! – Georges Elencwajg Jan 17 '13 at 3:55