# How does one find the Zariski closure of a set?

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!

• 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? Jul 22, 2015 at 19:11
• That's exactly the approach that i've tried @Mohan, but I could not find all such polynomials... Jul 22, 2015 at 19:13
• Can you find some? Jul 22, 2015 at 19:14
• Sure, $y^2-x^3$ has the desired property, for example. Is it possible that $I(S)=<y^2-x^3>$? Jul 22, 2015 at 19:15
• That is correct. Try to prove that. Jul 22, 2015 at 19:17

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 $f=0$.