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# showing a set is affine

2018-03-12 20:22:19

Let $F_1,\ldots, F_m \in k[x_1,\ldots, x_n]$. Let $\psi: \mathbb{A}^n\to \mathbb{A}^m$ be defined by

$$\psi(p) = (F_1(p),\ldots,F_m(p)).$$

Show: The $\textit{graph}$ of $\psi$

$$\Gamma_\psi := \{(p,q)\in \mathbb{A}^{m+n}\, | \,\psi(p) = q\}$$

is an affine algebraic set.

$\textbf{Solution}$

The problem boils down to finding a polynomial that attains its zeroes points at the points. One construction of a polynomial $g\in k[x_1,\ldots, x_n,y_1,\ldots,y_m]$ with this property is given as:

$$g(p,q) = (y_1- F_1(p) + (y_2- F_2(p)+\cdots + \big(y_m- F_m(p)\big),$$

another would be

$$g(p,q) = \sum_{i=1}^m\big(y_i- F_i(p)\big)^i.$$

It is clear that $g(p,q) = 0$ if and only if $y_i = F_i(p)$ for all $i=1,\ldots, m$

To me this doesn't seem to really prove anything. Are my steps correct?

Neither of your polynomials gives the desired set. Luckily, an affine algebraic set does not have to be defined by a single polynomial but a set of polynomials. You can use $S=\{y_1-F_1 • Neither of your polynomials gives the desired set. Luckily, an affine algebraic set does not have to be defined by a single polynomial but a set of polynomials. You can use$S=\{y_1-F_1,y_2-F_2,...,y_m-F_m\}$so$V(S)=\Gamma_\psi$. Edit : To see why your polynomials do not work, consider the example where$F_1=x$and$F_2=x-1$. Then the graph of$\psi$is a line. The first$g$you define is$y_1-x+y_2-x+1$, the zero set of$g$is a plane. The second$g$you define is$(y_1-x)\cdot (y_2-x+1)^2$, the zero set is two planes, the union$V(y_1-x)\cup V(y_2-x+1)$. Note that the zero set of polynomials you define always contain the graph$\Gamma_\psi\$ but in general they are not equal.

2018-03-12 21:21:10