5
$\begingroup$

I have to do the following. Let $F$ be an algebraically closed field. $I\in F[X_1,...,X_n]$ an ideal. Denote by $S(I)$ the subset in $F^n$ consisting of all $n$-tuples $(a_1,...,a_n)\in F^n$ such that $f(a_1,...,a_n)=0$ for all $f\in I$. A subset $S\subset F^n$ is called closed if there exists an ideal $I$ (of $F[X_1,...,X_n]$ such that $S=S(I)$. Prove that the union of any two closed sets is closed.

Here is my attempt:

First: I claim that if $S$ is closed then I can find an $I$ such that it is generated by one element and $S(I)=S$.

Proof: By definition of closed we know there is a $J$ such that $S=S(J)$. Since our ring is Noetherian we know that $J$ is f.g. so $J=\langle f_1,..,f_k\rangle$. Let $h=\gcd(f_i)$. I claim that $I=\langle h\rangle $ works. If $(a_1,..,a_n)\in S$, then $f_i(a_1,..,a_n)=0$. Since $h=\sum \alpha_i f_i$ (Bezout), we just plug in $(a_1,...,a_n)$ at both sides and obtain that $h(a_1,...,a_n)=0$, so indeed $(a_1,...,a_n)\in S(I)$. For the reverse inclusion if $(a_1,..,a_n)\in S(I)$, then $h(a_1,...,a_n)=0$. As $h\mid f_i$, we have that $f_i(a_1,...,a_n)=0$, so $(a_1,...,a_n)\in S(J)$.

Now using this fact, I will solve the question:

Let $S_1$ and $S_2$ be closed with $I_1=\langle f\rangle$ and $I_2=\langle g\rangle$ such that $S_1=S(I_1)$, and $S_2=S(I_2)$. Then let $h=$lcm$ (f,g)$. Then we have that $S(\langle h\rangle)$ works. If $(a_1,...,a_n)\in S_1\cup S_2$, then either $f(a_1,...,a_n)$ or $g(a_1,...a_n)=0$ (wlog say the first case happens). Then since $f\mid h$ then $(a_1,...,a_n)\in S(\langle h\rangle)$. For the other inclusion just note that if $h(a_1,...,a_n)=0$ then eithe $f(a_1,...,a_n)=0$ or $g(a_1,...,a_n)=0$ (because otherwise we would have $fg(a_1,....,a_n)\neq 0$ and since $h\mid fg$ we would get a contradiction here) so we have that either $(a_1,...,a_n)\in S_1$ or $(a_1,...,a_n)\in S_2$.

Question: Where do I need the algebraic closure here?

$\endgroup$
  • 1
    $\begingroup$ Since $F[X_1,\dots,X_n]$ is not a p.i.d., there is no such thing as a GCD. $\endgroup$ – Thomas Andrews Mar 13 '13 at 0:53
  • $\begingroup$ What is $\gcd(X_1,X_2)$? If you think it is $1$, is the ideal $\left<X_1,X_2\right>$ equal to $\left<1\right>$? $\endgroup$ – Thomas Andrews Mar 13 '13 at 0:57
  • $\begingroup$ Im a derp. ${}{}$ $\endgroup$ – Kanye West Mar 13 '13 at 1:00
  • 2
    $\begingroup$ @ThomasAndrews Every UFD is a GCD domain (a domain in which every pair of elements has a gcd). In fact, being a noetherian GCD domain is equivalent to being a UFD. If R is a UFD then so is R[X]. So even though $F[X_1,\cdots, X_n]$ is not a PID, it is still a GCD domain. $\endgroup$ – Ragib Zaman Mar 13 '13 at 1:30
  • 1
    $\begingroup$ @KanyeWest We're both derps. There is a GCD, but the idael generated by two elements is not the ideal of the GCD - Bezout doesn't apply in any UFD, only p.i.d.s $\endgroup$ – Thomas Andrews Mar 13 '13 at 2:25
7
$\begingroup$

Your claim is not correct, there are many algebraic sets which can not be defined by a single equation. In your proof you use properties of the polynomial ring which only hold for one variable.

The correct proof is much easier, just show $V(I \cdot J) = V(I) \cup V(J)$, where $V(I)$ denotes the vanishing set of the ideal $I$ (which you denote by $S(I)$, but this notation is not common). And in fact this holds over every field.

$\endgroup$
1
$\begingroup$

I think you don't use it... But I also think that you cannot use Bezout Theorem in $k[X_1,..,X_n]$.

$K[X_1,..,X_n]$ is not a PID, and in your proof you basically "prove" and use that it is...

$\endgroup$

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.