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I'd like some hints for the problem:

Show that the following set is not algebraic:

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


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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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I am having trouble completing the argument here. So, for each fixed $\theta\in [0, 2\pi)$, we have $f(\cos(\theta), \sin(\theta), z)=0$ as a polynomial of $z$. How can we conclude that $f(x, y, z)=0$? Can't we have something like $f(x, y, z)=(x^2+y^2-1)z$? – Prism Jul 26 '15 at 14:34
Hmmm, I think it is somewhat easier to argue as follows: Let $L\subset\mathbb{A}^{3}$ be the line given by $x=1$ and $y=0$, i.e. $L=\{(1, 0, z): z\in\mathbb{R}\}$. If $A$ were algebraic set, then $L\cap A = \{(1, 0, 2\pi k): k\in\mathbb{Z}\}$ would be an infinite proper closed subset of $L\cong \mathbb{A}^{1}$ in the Zariski topology, contradiction. – Prism Jul 26 '15 at 14:50
@Prism: That just seems like a more complicated way of saying my argument, since the reason what you reached is a contradiction is that a polynomial in $\mathbb{R}[z]$ with infinitely many zeros must be the zero polynomial, i.e. it must vanish everywhere on the real line. That's what my answer is hinting at. – Zev Chonoles Jul 26 '15 at 16:40
I think you are right. But I still don't know how to complete your argument. So you start with $f(x, y, z)\in V(A)$, and then $f(\cos(\theta), \sin(\theta), z)=0$ (as a polynomial in $z$) for each fixed $\theta\in [0, 2\pi)$. Can you explain why this forces $f=0$ (as a polynomial in $x$, $y$, $z$)? – Prism Jul 26 '15 at 16:47
@Prism: It doesn't force that, it forces $f(\cos(\theta),\sin(\theta),z)=0$ for all $\theta\in[0,2\pi)$ and all $z\in \mathbb{C}$, i.e. $f$ must vanish on the entire unit cylinder. The point is that $A$ is Zariski-closed if and only if $V(I(A))=A$, but if every element of $I(A)$ vanishes on the entire unit cylinder, then $V(I(A))$ must contain the unit cylinder. – Zev Chonoles Jul 26 '15 at 16:54

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