Is the set $\{z\in\mathbb{C}:|z|=1\}$ and algebraic set? Intuitively, I think the answer is no because it is not possible to use a polynomial to split an arbitrary complex number into its real and imaginary components.
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You are right. If it were an algebraic set (the zero set of a polynomial), then since we're in $\mathbb{C}$, it would be the zero set of a set of polynomials in one variable. Now a polynomial in one variable (if it's non-zero) always has finite zeros. This means that any algebraic set in $\mathbb{C}$ that is not all $\mathbb{C}$ must be finite. Reciprocally, any finite set is easily seen to be an algebraic set. So no, your set cannot be algebraic.
EDIT On the other hand, if we're looking at $\mathbb{C}$ like $\mathbb{R}^2$, then it IS the zero set of the polynomial $x^2+y^2-1$. So it depends on if we're looking at complex polynomials in one variable or real polynomials in two variables.
rfauffar
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Unless he meant over R, in which case it is. – Alex Youcis Aug 14 '13 at 23:24