Let $\mathbb K$ be a field, $X\subseteq \mathbb K^n$ a subset, and $\ f:\mathbb K^n\rightarrow \mathbb K^m$ an affine map. Using the standard notation $V$, $I$ for algebraic geometry, it should hold that $$V(I(f(X)))=f(V(I(X))),$$ i.e. that the Zariski closure $VI$ commutes with the affine transformation $f$. Why is that?

(To specify the notation more: for a set $Y$, we mean by $I(Y)$ the set of all polynomials vanishing on $Y$; and for a set of polynomials $J$, we mean by $V(J)$ the set of all zeros common to all polynomials in $J$.)

Note that I do NOT assume $f$ to be bijective, and that $X$ does not need to be an algebraic set.

The equality obviously doesn't hold for every polynomial mapping $\ f$, as that would mean that an image of Zariski closed set under polynomial mapping is always Zariski closed, which is not true (e.g. $f(x)=x^2:\mathbb R\rightarrow \mathbb R$ and $X=\mathbb R$).

EDIT: As pointed out by reuns, using the general topology fact that $\overline{f(X)}\supseteq f(\overline{X})$ for continuous maps, it is enough to show that $f(V(I(X)))$ is Zariski closed (i.e. an algebraic set).

  • $\begingroup$ The reformulation of the question would be "if $f$ is $\textbf{affine}$ then $\overline{f(X)}=f(\overline{X})$". The equality does not hold for $f$ continuous in general, as my example shows. If $f(x)=x^2$, we have $\overline{f(R)}=\overline{[0,\inf)}=R \neq [0,\inf)=f(R)=f(\overline{R})$. But you are right that $\supseteq$ holds in general, so that shows one direction. $\endgroup$ – OnDragi Jan 21 at 13:26
  • $\begingroup$ In what sense is $f : \mathbb{R} \to \mathbb{R}, x \to x^2$ continuous, is it continuous for the topology induced by $V(I(X))$ ? $\endgroup$ – reuns Jan 21 at 13:40
  • $\begingroup$ Yes, it is a polynomial map, and polynomial maps are continuous w.r.t. Zariski topology. $\endgroup$ – OnDragi Jan 21 at 13:42
  • $\begingroup$ $\overline{f(X)} = f(\overline{X})$ is true whenever $f$ maps closed sets to closed sets (that's part of your definition of affine right ?). Here you are saying $f^{-1}(U)$ is open whenever $U$ is open (this would make $f$ continuous) but $f(X)$ isn't closed whenever $X$ ? Isn't the situation different for polynomial maps $f : \mathbb{C} \to \mathbb{C}$. $\endgroup$ – reuns Jan 21 at 13:53
  • $\begingroup$ Indeed, that is the case. I am not sure whether the situation is different for algebraically closed fields. $\endgroup$ – OnDragi Jan 21 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.