Let $X \subset \mathbb A^n$ be an affine variety, say the zero set of polynomials $P_1,\dots, P_m$. Now, for an arbitrary polynomial $Q$, we can introduce an equivalence relation on $X$: two points $x_1,x_2\in X$ are equivalent (write $x_1\sim x_2$) if $$Q(x_1) = Q(x_2)$$

Question: In what conditions is the quotient $X/\sim$ still a variety?

For example, let's say $X=\mathbb A^2$ and $Q=x_1$ then $X/\sim$ is isomorphic to $ \mathbb A^1$. But is there a more general theory about this story?

  • 1
    $\begingroup$ Have you tried other examples? A reasonable conjecture would be that if $X = \operatorname{Spec} k[x_1,\ldots,x_n] / (P_1,\ldots,P_n)$, then $X / \sim$ is $\operatorname{Spec} k[x_1,\ldots,x_n] / (P_1,\ldots,P_n,Q)$. $\endgroup$ – Sofie Verbeek Sep 5 '18 at 15:31
  • $\begingroup$ More generally, if you're interested in quotients of varieties see geometric invariant theory: en.wikipedia.org/wiki/Geometric_invariant_theory $\endgroup$ – leibnewtz Sep 5 '18 at 15:32
  • $\begingroup$ @leibnewtz Thanks. I know a little bit GIT, but it seems that there is no group action in our situation, although we also consider quotients. $\endgroup$ – Hang Sep 5 '18 at 15:34
  • $\begingroup$ @Hang Yeah wasn't saying there was, it just seemed related $\endgroup$ – leibnewtz Sep 5 '18 at 15:53

I can positively answer your question when the base field $k$ is algebraically closed. If this assumption is unacceptable for you, let me know - but I fear the answer may be significantly more complex and potentially negative. That's just a feeling, though.

Since $Q$ defines a regular function on your variety, it can be seen as a morphism $q:X\to\Bbb A^1$. You are now looking for a way to give the set of fibers of this morphism the structure of an affine variety. Now, you already have a pretty good candidate for that variety: The image $Y\subseteq\Bbb A^1$ of $q$.

It is known that the image $Y$ of $q$ is a constructible set, i.e. it contains an open and dense subset $U$ of its closure. We are lucky that the Zariski topology of $\Bbb A^1$ is relatively easy when $k$ is algebraically closed: Every finite set is closed, and that's all of them (except for $\Bbb A^1$ itself, of course). Now from this we can conclude that the image of $q$ is open or closed: If the dimension of $U$ is zero, it must be a discrete set, hence $Y=U$, and it is closed. Otherwise, the dimension of $U$ is equal to one and it can only exclude a finite number of points from $\Bbb A^1$. In that case, $Y$ as well can only exclude a finite number of points from $\Bbb A^1$ and must be open. Of course, it could happen that $Y=\Bbb A^1$ in this case which would mean that $Y$ is open and closed.

Now, all the open subsets of $\Bbb A^1$ are also affine because when $Y=\Bbb A^1 \setminus \{ a_1, \ldots, a_n \}$ then we may take $f(t):=\prod_{i=1}^n (t-a_i)$ and observe that $Y=D(f)$, hence $Y=\{ (x,y) ~\mid~ f(x)\cdot y -1 = 0 \}\subseteq\Bbb A^2$ is a realization of $Y$ as a closed subvariety of $\Bbb A^2$.

In summary, the image $Y:=q(X)$ can always be given the structure of an affine variety and this affine variety models your quotient $X/\sim$.

  • $\begingroup$ Thank you. I notice that your answer relies on some special property of $\mathbb A^1$, so I was wondering if we can generalize it to the case there are more than one polynomials $Q_1,\dots Q_k$. $\endgroup$ – Hang Sep 5 '18 at 20:29
  • $\begingroup$ Dear @Hang, that will already be much more tricky. For example, the image of $\Bbb A^2\to \Bbb A^2$ mapping $(x,y)\mapsto(x,xy)$ is neither open nor closed. Note that all these problems completely vanish if you are willing to switch from affine to projective, because the image of a projective morphism is always closed. $\endgroup$ – Jesko Hüttenhain Sep 6 '18 at 7:24

Your Answer

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

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