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Sorry if this is an ignorant question. I am studying algebraic geometry. This isn't an exercise problem. It is an assumption I can use to prove something else. I think it must be obvious, but I don't know how to prove it.

My attempt:

Let $V=(\{F_i\}_{i\in I})$ be an algebraic set. Let $f$ be a map. Then I couldn't get to anywhere.

I tried an example: $V=\{(x-1)(x-2), (x-1)(x-2)(x-3)\}=\{1,2\}$. Let $f(x)=x^2$. Then $f(V)=\{1,4\}$. They are zero loci of $\{(x-1)(x-4), (x-1)(x-4)(x-3)\}$. But that would change the $F_i$'s randomly. And for higher dimensions, you don't have a finite set to deal with.

Please help.

KittyL
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    Your question is very good, you are on the right track. In fact the image is, in general, given by a combination of equalities and non-equalities. For instance, what is the image of ${ xy=1}$ under $(x,y)\mapsto x$? – orangeskid Jan 28 '15 at 10:58
  • Just to add to the other helpful comments, a keyphrase you can search for is "constructible set": this is the fancy name for the kind of sets @orangeskid is describing. –  Jan 28 '15 at 11:10
  • @orangeskid: hmm, so you are saying that it doesn't have to be so? because the map you gave maps the algebraic set to a non-algebraic set. – KittyL Jan 28 '15 at 11:12
  • @AsalBeagDubh: Thank you. I'll look for that. – KittyL Jan 28 '15 at 11:13
  • @KittyL: Correct. For algebraically closed fields, like $\mathbb{C}$ there is a theorem that describes the image in similar terms. Like the comment from Asal Beag Dubh. But it makes for an interesting theory nevertheless. Now that you are at it, you could try to find the projection of a curve $C\colon F(x,y)=0$ on the first axis. – orangeskid Jan 28 '15 at 11:24
  • @orangeskid: That will be very interesting. In real field, it could be the whole field $R$, or an infinite set. I wonder if it is possible to get a finite set. It shouldn't be, right? – KittyL Jan 28 '15 at 11:38
  • @KittyL: Over the field $\mathbb{R}$ the curve itself $C$ could be a finite set. For instance $(x^2-x)^2 +(y^2-y)^2 =0$ And it gets even more interesting over fields like the rationals $\mathbb{Q}$. Consider the curves $x^2+y^2=1$ and $x^3+y^3=1$. Can you find rational points on the first one ? On the second one ? – orangeskid Jan 28 '15 at 12:17
  • @KittyL: I recommend using some software (a computer algebra system) for graphing, and also for computations. – orangeskid Jan 28 '15 at 12:20
  • @orangeskid: I see. Didn't think of that one. Those two are related to Fermat's Last Theorem right? In fact, I was trying to study "elliptic curves" then I found that I need to learn some algebraic geometry to understand it. :) – KittyL Jan 28 '15 at 23:39
  • @KittyL: Yes, there is an ellptic curve there. I recommend to start with the book by Silverman and Tate, make sure that you have a functional computer algebra system and experiment. Especially adding points in ellptic curves to get the feel. You do not need much algebraic geometry to get you started. – orangeskid Jan 31 '15 at 20:23
  • @orangeskid: Thank you for your help! I will look into that book. – KittyL Jan 31 '15 at 22:17

1 Answers1

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Let $\phi\colon k^n \to k^m$ be given by polynomials $\phi_1,\ldots,\phi_m\in k[x_1,\ldots,x_n]$. You want to show that for any algebraic set $V\subseteq k^n$, the set $\phi(V)$ is again algebraic?

If so, then I have to disappoint you. This only holds under additional assumptions. A counterexample would be the hyperbola $V=Z(1-xy)\subseteq k^2$ (assume $k$ algebraically closed to avoid problems) and the projection map $\phi:k^2\to k$ given by $(x,y)\mapsto x$. You can see that the image of $V$ is $\phi(V)=k\setminus\{0\}$, which is not an algebraic set because it is not finite.

Jesko Hüttenhain
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