12

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!

u1571372
  • 3,205
  • 3
    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? – Mohan Jul 22 '15 at 19:11
  • That's exactly the approach that i've tried @Mohan, but I could not find all such polynomials... – u1571372 Jul 22 '15 at 19:13
  • Can you find some? – Mohan Jul 22 '15 at 19:14
  • Sure, $y^2-x^3$ has the desired property, for example. Is it possible that $I(S)=<y^2-x^3>$? – u1571372 Jul 22 '15 at 19:15
  • That is correct. Try to prove that. – Mohan Jul 22 '15 at 19:17
  • Hmm, what about this: using the monomial order $y>x$, write $f=(y^2-x^3)q(x,y)+h(x)y+c$. By hip., $0=f(n^2,n^3)=h(n^2)n^3+c$ for all natural number $n$. Putting $n=0$, we get $c=0$. Therefore, $0=h(n^2)n^3 \forall n \geq 1$. But $h$ is a polynomial in one variable - thus, if it has infinite zeros, it must be the zero polynomial. One concludes that $f=(y^2-x^3)q(x,y) \in <y^2-x^3>$. – u1571372 Jul 22 '15 at 19:25
  • Ups, the above argument does not work: I forgot to add a polynomial in $x$ in the remainder and I don't think that I can adapt the same idea. Any help? – u1571372 Jul 22 '15 at 19:40
  • This is fine, but no monomial order is necessary. Use division algorithm (which is what you have used) for monic polynomials (again what you did with respect to $y$). Finally, you do not have to use 0 (some people think of natural numbers as numbers positive integers. – Mohan Jul 22 '15 at 19:40
  • For $h(x)y+g(x)$, how does the two terms grow for large $n$ when you put $x=n^2, y=n^3$? – Mohan Jul 22 '15 at 19:44
  • Of course, silly me. By the division algorithm, write $f=(y^2-x^3)q(x,y)+a(x)y+b(x)+c$. By hip., $0=a(n^2)n^3+b(n^2)+c$. Now, the idea is to notice that this $a(n^2)n^3$ and $b(n^2)$ grow (in absolute value) with $n$ and there is no possible cancelation in the highest power of $a(n^2)n^3$ with the highest power of $b(n^2)$ as the first highest power must be an odd integer and the second one must be an even integer. So, after expanding $a(n^2)n^3+b(n^2)$ one gets a polynomial of the form $k n^d + \cdots + c$ ($d \geq 1$) which is not zero for large $n$ unless the original polynomials $a,b,c=0$. – u1571372 Jul 22 '15 at 19:51
  • Does this seem correct to you, @Mohan? – u1571372 Jul 22 '15 at 19:52
  • @Mohan, do you have any similar hint for the 2) question? – u1571372 Jul 26 '15 at 00:50
  • First try to find the Zariski closure of the set $x^2<c$, $c>0$ in $\mathbb{A}^2(\mathbb{R})$. – Mohan Jul 26 '15 at 02:06

2 Answers2

4

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 $f=0$.

Hagen von Eitzen
  • 374,180
  • 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? – u1571372 Jul 26 '15 at 17:27
  • You can simply compute iterated derivatives of f and evaluate them at 0. – Mariano Suárez-Álvarez Mar 17 '18 at 04:27
-2

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.

Fuad Efendi
  • 11