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Suppose $A$ and $B$ are two algebraic varieties, and $f:A\to B$ is a morphism of algebraic varieties. I guess it is true that $\text{im}(f)$ is itself an algebraic variety. But how to prove it?

Abramo
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  • What is your definition of an algebraic variety? If X and Y are integral schemes of finite type over a field K, the image f(X) can be viewed as an integral subscheme of Y: it is irreducible, by continuity of $f$, and can be given the reduced subscheme structure induced from Y; if $f$ is closed (e.g. proper), f(X) is also of finite type over K (since closed immersions are of finite type, as are compositions of morphisms of finite type). –  Mar 26 '13 at 12:27
  • I'm working now with a very basic definition of an affine variety: An irreducible closed subset of k^n, with the Zariski topology. So basically the zero locus of a prime ideal in k[x_1,...,x_n] – Abramo Mar 26 '13 at 13:50
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    Now that you know that the result you're asking about is not true, you maybe want to read about "constructible subsets" (e.g. http://en.wikipedia.org/wiki/Constructible_set_(topology)). – Nils Matthes Mar 26 '13 at 14:18

2 Answers2

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You can't prove it because it is not true!

Consider the (dominant) morphism $f\colon \mathbb C^2\to \mathbb C^2\colon (x,y)\mapsto (x,xy)$.
Its image is the subset $Im(f)=\{(u,v)\in \mathbb C^2 | u\neq 0 \} \cup \{(0,0)\}$.
This set is not locally closed in $\mathbb C^2$ and so $Im(f)$ is not a subvariety of $\mathbb C^2$.
Feel free to soup up the example by introducing arbitrary fields, schemes,... :-)

Georges Elencwajg
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  • Dear @Georges, how do you see that $Im(f)$ is not locally closed? Here is how i see it: since $Im(f)$ is dense, it enough to prove that its complement is not closed. Towards a contradiction, suppose that $Z(u) \cap (\mathbb{C}^2-(0,0)) = Z(f_1,\dots,f_s)$. Then $f_i(0,a) = 0$ for every $a \neq 0$ and so $f_i(0,v)=0$. Hence $f_i(u,v) = u g_i(u,v)$. But then we see that $(0,0)$ is in fact inside $Z(f_1,\dots,f_s)$, contradiction. Any comments on my proof? – Manos Jun 17 '14 at 21:35
  • Dear @Manos, a subset is locally closed iff it is open in its closure. Here the closure of $Im f$ is $\mathbb C^2$ and obviously $Im f$ is not open in $\mathbb C^2$. Your proof seems correct but a bit complicated. – Georges Elencwajg Jun 17 '14 at 22:36
  • Dear @Georges, why is $Imf$ "obviously" not open? How can you see that directly? – Manos Jun 19 '14 at 00:15
  • @Manos It is not even open in the classical topology of $\mathbb C^2$: no ball of center $(0,1)$ and radius $\epsilon$ is included in Im$f$. – Georges Elencwajg Jun 19 '14 at 08:19
  • @GeorgesElencwajg Do you mean $(0,0)$? $(0,1)$ is not in $\text{Im}(f)$, so of course there is no ball! – user2154420 Dec 23 '17 at 18:10
  • Dear @user2154420: yes, you are perfectly right, my comment to Manos contained a typo . The point of the image with no neighbourhood included in the image is indeed $(0,0)$, as you correctly suggest. Thanks for your vigilance. – Georges Elencwajg Dec 24 '17 at 12:20
  • Hello, I don't understand why $Imf$ is not open in $\mathbb C^2$, and I'm not sure to see why the classical topology of $\mathbb C^2$ tells us something about the Zariski Topology. If for example we take $\mathbb K$ instead of $\mathbb C$ how are we sure that $Imf$ is not open, that is the complement of a zero locus ? – raisinsec Sep 22 '22 at 07:46
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Ill put this here for future refrence.

$\textbf{Theorem:}$ If $\phi: V \to W$ is an onto morphism of affine algebraic sets and if $V$ is a variety, so is $W$.

$\textbf{Proof}$

Suppose by way of contradiction that $W$ is reducible. I.e., $$W=U_1 \cup U_2$$ where $U_1,U_2 \subsetneq W$ and $U_1,U_2$ are closed in $W$. We know $$\phi:V \to W$$ is such that $\phi(V)=W$ and is a morphism and thus continuous in the Zariski topology, and as $U_1,U_2 \subset W$ are closed in the Zariski topology, then by continuity, $$\phi^{-1}(U_1),\phi^{-1}(U_2)$$ are closed in $V$ and thus algebraic sets. Moreover, we know the are proper subsets as $\phi$ is onto such that \begin{align*} \phi^{-1}(U_1) \cup \phi^{-1}(U_2) &=\phi^{-1}(U_1 \cup U_2)\\ &=\phi^{-1}(W)\\ &=V \end{align*} contradicting irreducibility of $V$ thus $W$ is irreducible and thus an affine variety. $\blacksquare$

homosapien
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