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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?

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    $\begingroup$ 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
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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 $X/\sim$.

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  • $\begingroup$ 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

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