I've started to learn algebraic geometry this week (so I do not have much knowledge in the subjet) and, after reading the definition of the Zariski closure $V(I(S))$ of a set $S$, I've tried to do the following exercises without much success:

Find the Zariski closure of the following sets:

1) $\{(n^2,n^3): n \in \mathbb{N} \} \subset \mathbb{A}^2(\mathbb{Q})$

2) $\{(x,y): x^2+y^2 < 1 \} \subset \mathbb{A}^2(\mathbb{R})$

Any help with these? In general, how does one usually attacks this kind of problem?

Thank you!

  • 3
    $\begingroup$ Let us try the first. $(a,b)\in\mathbb{A}^2(\mathbb{Q})$ is in the Zariski closure of your set means for $ every$ polynomial $f(x,y)\in\mathbb{Q}[x,y]$ such that $f(n^2,n^3)=0$ for all $n\in\mathbb{N}$, we must have $f(a,b)=0$. Can you find all such polynomials first? $\endgroup$
    – Mohan
    Jul 22, 2015 at 19:11
  • $\begingroup$ That's exactly the approach that i've tried @Mohan, but I could not find all such polynomials... $\endgroup$
    – u1571372
    Jul 22, 2015 at 19:13
  • $\begingroup$ Can you find some? $\endgroup$
    – Mohan
    Jul 22, 2015 at 19:14
  • $\begingroup$ Sure, $y^2-x^3$ has the desired property, for example. Is it possible that $I(S)=<y^2-x^3>$? $\endgroup$
    – u1571372
    Jul 22, 2015 at 19:15
  • $\begingroup$ That is correct. Try to prove that. $\endgroup$
    – Mohan
    Jul 22, 2015 at 19:17

2 Answers 2


Since the first part has been dealt with in the comments:

Assume $f(X,Y)=\sum_{i,j\ge 0}a_{i,j}X^iY^j\in\mathbb R[X,Y]$ is a polynomial with $f(x,y)=0$ whenever $x^2+y^2<1$. Let $\alpha\in\mathbb R$ and consider the polynomial $g(T)=f(T,\alpha T)\in\mathbb R[T]$. Then $g(t)=0$ whenever $t^2(1+\alpha^2)<1$. As there are infinitely many such $t$, $g$ must be the zero polynomial. We conclude that each of its coefficients $$\sum_{i+j=k}a_{i,j}\alpha^j $$ is zero - no matter what $\alpha\in\mathbb R$ we pick. Hence each of the polynomials $\sum_{i+j=k}a_{i,j}X^j$ has infinitely many roots, hence is the zero polynomials, hence all $a_{i,j}$ are zero. We conclude $f=0$.

  • $\begingroup$ Nice trick, thank you for your answer. So, concerning my question "how does one usually attacks this kind of problem?", there is not a general way, just certain tricks that work for some cases? $\endgroup$
    – u1571372
    Jul 26, 2015 at 17:27
  • $\begingroup$ You can simply compute iterated derivatives of f and evaluate them at 0. $\endgroup$ Mar 17, 2018 at 4:27

It So question is “find Zariski closure for infinite set of 1-dimension objects (points) in two-dimension space”, and the answer is f=0. Isn’t that obvious? To some extent. We need projection to 1st dimension (x, and to find “closure” there, it will be f(x, y) = g(y) space. Because non-trivial polynomial of (x) can have only finite number of roots. Then second projection f(x,y) = h(x). And intersection of these spaces is f(x,y) = 0.

  • 6
    $\begingroup$ "Isn' t that obvious?" is among the worst things you can say to someone you are trying to help. $\endgroup$ Mar 16, 2018 at 17:45
  • $\begingroup$ I am sorry. Yes I am trying to help. Thank you! $\endgroup$ Mar 17, 2018 at 3:43

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