1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you explain why my construction doesn't work? I think I'm having a bit of trouble with the language, because what you say is obvious but I have problems reasoning it out mathematically. $\endgroup$ – user119264 Mar 12 '18 at 17:35
  • $\begingroup$ I edited. What are your problems exactly? $\endgroup$ – Levent Mar 12 '18 at 17:38
  • $\begingroup$ @user119264 In case $k$ is algebraically complete, think in terms of dimension: a single polynomial can at best give you a subset of codimension 1, whereas the graph has codimension $m$. $\endgroup$ – Daniel Schepler Mar 12 '18 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.