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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 $cl(X)=\mathbb{A}^{2}$.

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Yes, it does. Do you have any ideas for proving it? –  Qiaochu Yuan Jun 15 '12 at 21:56
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1 Answer

up vote 10 down vote accepted

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.

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but wouldn't the fact that $f$ vanishes on $\mathbb{R} \times [-1,1]$ implies $f=0$ is based on the Euclidean product topology and not the Zariski topology? Can you please ellaborate more? –  user10 Jun 15 '12 at 23:52
    
@user10 Yes, it does. We are using a fact outside algebraic geometry here. –  Alex Becker Jun 16 '12 at 1:12
    
thanks, actually to apply the result that $cl(X)=V(f)$ for some irreducible $f$ we need to know that $X$ (or equivalently cl(X) is irreducible, why is this? can we simply say because $X$ is not algebraic? –  user10 Jun 17 '12 at 0:57
    
@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. –  Alex Becker Jun 17 '12 at 1:58
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