Proving algebraic sets i) Let $Z$ be an algebraic set in $\mathbb{A}^n$. Fix $c\in \mathbb{C}$. Show that $$Y=\{b=(b_1,\dots,b_{n-1})\in \mathbb{A}^{n-1}|(b_1,\dots,b_{n-1},c)\in Z\}$$ is an algebraic set in $\mathbb{A}^{n-1}$.
ii) Deduce that if $Z$ is an algebraic set in $\mathbb{A}^2$ and $c\in \mathbb{C}$ then $Y=\{a\in \mathbb{C}|(a,c)\in Z\}$ is either finite or all of $\mathbb{A}^1$. Deduce that $\{(z,w)\in \mathbb{A}^2 :|z|^2 +|w|^2 =1\}$ is not an algebraic set in $\mathbb{A}^2$. 


This answer was merged from another question so it only covers part ii).

$Z$ is algebraic, and hence the simultaneous solution to a set of polynimials in two variables. If we swap one variable in all the polynomials with the number $c$, you will get a set of polynomials in one variable, with zero set being your $Y$. $Y$ is therefore an algebraic set, and closed in $\Bbb A^1$, therefore either finite or the whole affine line.

Assume for contradiction that $Y = \{ ( z,w) \in \mathbb{A}^2 : |z|^2 + |w|^2 = 1 \}$ is algebraic. Set $w = 0$ (this is our $c$). The $Y$ we get from this is the unit circle in the complex plane. That is an infinite set, but certainly not all of $\Bbb C$. Thus $Y$ is neither finite nor all of $\Bbb A^1 = \Bbb C$, and therefore $Z$ cannot be algebraic to begin with.
