I am trying to show that the zero set of $y - e^x$ is not an affine algebraic variety in $\mathbb{A}^2$. My idea has been to show that any polynomial vanishing on the zeros of $y - e^x$ must vanish on the whole plane, since then the nullstellensatz will imply that this zero set cannot be an affine algebraic variety. But I am having difficulty establishing this vanishing condition. Can someone help?
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I'm assuming you're working over an algebraically closed field field? – Brian Fitzpatrick Apr 21 '13 at 22:03
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1He mentions Nullstellensatz and $e^x$, so he almost surely works over $\mathbb{C}$. – xyzzyz Apr 21 '13 at 22:06
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Yes. All of this is done over $\mathbb{C}$. – James Miller Apr 21 '13 at 22:09
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Hint: Suppose that $V = \{ y - e^x = 0 \}$ is a closed subset. Intersect it with line given by $L = \{y = 1\}$ to obtain $V \cap L = \{ (2 k \pi i, 1): k \in \mathbb{Z}\}$. The set $V \cap L$ is closed as intersection of two closed sets. Deduce a contradiction.
xyzzyz
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For any polynomial $P(Y)=a_nY^n+\ldots +a_0$ with $a_n\ne 0$, we can estimate that it isn't even close to zeroif we are too far away from the origin: too large absolute value: If $|P(y)|<1$ with $|y|>1$, then $$|a_ny|\le|a_{n-1}+a_{n-2}y^{-1}+\ldots+a_0y^{n-1}|+|y|^{n-1}\le1+\sum_{k<n}|a_k|,$$ hence $$ |y|\le \frac{1+\sum_{k<n}|a_k|}{|a_n|}.$$ For a nonzero bivariate polynomial $P(X,Y)$, considered as polynomial in $Y$ with coefficients in $\mathbb C[X]$, the leading coefficient is $\le1$ in absolute value only for bounded $x$, hence for big $x$, we can estimate $y$ with $P(x,y)=0$ by a polynomial expression in $x$. Since the exponential grows faster than any polynomial, this is absurd.
Hagen von Eitzen
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