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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 $\mathbb{A}^2$.

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Any thoughts of yours so far? Giving us information on what you've tried will help us giving you a better answer. If you haven't tried anything yet, then you should probably do so prior to asking here; this way you will learn much more. – Nils Matthes Feb 5 '13 at 14:48

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.

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Now that this is merged, I can note that the procedure above the line in my answer will solve part i) with a slight modification to adjust for more variables in a general field. – Arthur Feb 6 '13 at 12:45

An idea: for $\,f\in I(Z)\,$ , look at

$$g(x_1,...,x_{n-1}):=f(x_1,...,x_{n-1},c)\in\Bbb C[x_1,...,x_{n-1}]$$

with $\,c\in\Bbb C\,$ s.t. for some $\,a_1,...,a_{n-1}\,\,,\,(a_1,...,a_{n-1},c)\in Z\,$

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

Begin by writing $Z=V(I)$ for $I \subset \mathbb C[x,y]$. Take generators of $I$ evaluate at $c$ and consider these as elements of $\mathbb C[x,y]$ and determine what their zero sets can be to find $Y$. The next part should follow.

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Let $Z=Z(f_1,\ldots,f_r)$ with certain polynomials $f_i\in\mathbb C[x_1,\ldots,x_n]$. Then, we have polynomials $g_i := f_i(x_1,\ldots,x_{n-1},c)\in\mathbb C[x_1,\ldots,x_{n-1}]$. We claim that $Y=Z(g_1,\ldots,g_r)$. Indeed, $$\begin{align*} Y &= \{ (b_1,\ldots,b_{n-1}) \mid (b_1,\ldots,b_{n-1},c)\in Z \} \\ &= \{ (b_1,\ldots,b_{n-1}) \mid \forall i: f_i(b_1,\ldots,b_{n-1},c)=0 \} \\ &= \{ (b_1,\ldots,b_{n-1}) \mid \forall i: g_i(b_1,\ldots,b_{n-1})=0 \} \\ &= Z(g_1,\ldots,g_r). \end{align*}$$

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