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I have four sets that I've come up with, which I think fail to be algebraic. However, I don't know how to prove this. They are:

  1. The graph of $\mathbb{C} \rightarrow \mathbb{C} : z \mapsto e^z$
  2. {$(z, w) \in \mathbb{A}_\mathbb{C}^2 \mid |z|^2 + |w|^2 = 1$}
  3. {$(\cos t, \sin t, t) \in \mathbb{A}_\mathbb{R}^3 \mid t \in \mathbb{R}$}
  4. {$(x, y) \in \mathbb{A}_\mathbb{R}^2 \mid |x| \leq 1, |y| \leq 1$}

My basic reasoning so far is that for (1) has an infinite polynomial expansion; (2) since we're working over $\mathbb{C}$, things break down when our point is not strictly real nor strictly imaginary; (3) a spiral just seems really difficult to be a solution set; (4) I don't see how a box can be a solution set to polynomial equations.

If anyone can help me fill out my reasoning, or correct me where I'm wrong, that would be great.

user93779
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    For 3 and 4, assume $f$ is a polynomial vanishing on your specified set. Then by restricting to each line through the origin (i.e., set $y=\lambda x$ for some $\lambda$, or $x=0$), you can get a bunch of polynomials of one variable, each of which has infinitely many roots, so must be $0$. Thus you can show that $f$ vanishes everywhere. Possibly a similar method can be adapted to the others. – mdp Sep 10 '13 at 21:52
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    Smells Like Fulton Spirit... – Matemáticos Chibchas Sep 11 '13 at 01:24

1 Answers1

3

Let $f\in\mathbb R[X,Y,U,V]$ be a polynomial with $f(x,y,u,v)=0$ whenever $e^{x+iy}=u+iv$. Then $f(0,Y,1,0)$ is a polynomial $\in\mathbb R[Y]$ with infinitely many roots, hence the zero polynomial. Therefore, any algebraic set $\subseteq \mathbb R^4\cong\mathbb C^2$ that contains $\{(z,e^z)\}$ necessarily also contains $(ti,1)$ for any $t\in\mathbb R$.

The second set is algebraic in $\mathbb A_{\mathbb R}^4$. But it is not over the complex: If $f\in\mathbb C[X,Y]$ vanishes on the set, then $f(X,0)\in\mathbb C[X]$ vanishes on the $S^1\subset \mathbb C$, hence is zero, hence vanishes on too many points - again.

For 3 you can do almost the same as for 1.

For 4 you can show that any polynomial vanishing on the set also vanishes on every line through the origin, hence is the zero poynomial.

Hagen von Eitzen
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