# Where do I use the fact that $F$ is algebraically closed in this proof?

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?

• Since $F[X_1,\dots,X_n]$ is not a p.i.d., there is no such thing as a GCD. – Thomas Andrews Mar 13 '13 at 0:53
• What is $\gcd(X_1,X_2)$? If you think it is $1$, is the ideal $\left<X_1,X_2\right>$ equal to $\left<1\right>$? – Thomas Andrews Mar 13 '13 at 0:57
• Im a derp. ${}{}$ – Kanye West Mar 13 '13 at 1:00
• @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. – Ragib Zaman Mar 13 '13 at 1:30
• @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 – Thomas Andrews Mar 13 '13 at 2:25

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.
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...