7
$\begingroup$

Let $X=\{(x,\sin(x)): x \in \mathbb{A}^{2}\}$. I want to find the closure (with respect Zariski topology) of $X \subseteq \mathbb{A}^{2}$.

OK I've already shown that $X$ is not a closed set. Now consider $cl(X)$ this is a closed subset of $\mathbb{A}^{2}$ so its dimension is $0,1$ or $2$, it is not $0$ because it is not a point. So either $cl(X)$ has dimension $1$ or $2$. I suspect the answer is $2$. If the dimension is $1$ then $X=V(f)$ for some $f \in k[x,y]$ with $f$ irreducible. This implies then that $f(a,\sin(a))=0$ for every $a \in \mathbb{A}^{1}$.

Question: does this implies that $f$ is the zero polynomial?

From this it would follow that the dimension is $2$ so $cl(X)=\mathbb{A}^{2}$.

  • 0
    Yes, it does. Do you h$a$ve $a$ny ide$a$s for proving it?2012-06-15

1 Answers 1

10

The Zariski closure is the whole set. Suppose $f(x,\sin(x))=0$ for all $x$. Define $g_y(x)=f(x,y)$ for $y\in [-1,1]$ and note that $g_y$ has infinitely many zeroes, thus must be the zero polynomial. Thus $f$ vanishes on the strip $\mathbb R\times [-1,1]$, so must be the zero polynomial by any number of properties (for example, the fact that polynomials are analytic).

Edit: To clarify, since polynomials are analytic, it suffices to show that $f$ is $0$ on an open set (in the usual metric topology, NOT the Zariski topology) in order to conclude that $f$ is $0$ everywhere. Otherwise, we have some $n$ such that the first $n-1$ derivatives vanish on the set but the $n^{th}$ does not, so is nonzero and of constant sign on some open set, so integrating gives us $f$ is nonzero somewhere on this set. Since $\mathbb R\times [-1,1]$ contains the unit open ball around $(0,0)$, we are done.

  • 0
    @user10 In general, yes. But since we are trying to show $cl(X)$ is the entire set, we just need to show that it is not contained in $V(f)$ for any $f\neq 0$. Note that nowhere did I assume $f$ is irreducible.2012-06-17