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I have to find the closure, in the Zariski topology, of the curve $C = \{(x,y): y = e^x, x \in \mathbb{C}\}$. I suppose the answer is $\mathbb{C}^2$ and I guess I have to use the Taylor expansion of the function $e^x$ or the implicit function theorem to prove it, but I have no idea.

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up vote 4 down vote accepted

Suppose that a polynomial $p(x, y)$ vanishes on $C$. Then $p$ has the property that for all $a$, $p(a, e^a) = 0$. Because $e$, and hence $e^2, e^3, \ldots$ are transcendental, we have that for each natural number, $p(n, y)$ is the zero polynomial in $y$. It follows that $p$ is the zero polynomial. (You could also argue directly from the algebraic independence of $x$ and $e^x$ over $\mathbb{C}$, which is much easier to prove than the transcendence of $e$.)

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How could not I see that? Thank you. – Diego Silvera Dec 19 '12 at 21:40

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