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I'd like some hints for the problem:

Show that the following set is not algebraic:

$ \{ (\cos(t),\sin(t),t) \in \mathbb{A}^3 : t \in \mathbb{R} \} $

Thanks.

1 Answers 1

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Let $A=\{ (\cos(t),\sin(t),t) \in \mathbb{A}^3 : t \in \mathbb{R} \}$. Suppose that $f(x,y,z)\in I(A)$, i.e. $f$ vanishes on $A$. Because $f(\cos(\theta+2\pi k),\sin(\theta+2\pi k),\theta+2\pi k)=f(\cos(\theta),\sin(\theta),\theta+2\pi k)=0$ for all $k\in\mathbb{Z}$, we have that for each $\theta\in[0,2\pi)$, the polynomial $f(\cos(\theta),\sin(\theta),z)\in\mathbb{R}[z]$ has infinitely many zeros. What does that imply?

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    Thanks a lot Zev! This makes more sense now. Just to paraphrase: We start with a function $f\in I(A)$ that is supposed to vanish on $A$, and we find that $f$ actually vanishes on the unit cylinder! So the Zariski closure of $A$, which is $V(I(A))$, is strictly larger than $A$, and so $A$ is not Zariski-closed.2015-07-26