$V(f)\cup V(g) = V((f)(g))=V((fg))=V(fg)$?

Question

$f,g$ are all in $k[x_1,...,x_n]. k$ is a field. I find the title confusing: for example $f=(x_1-a_1)...(x_n-a_n), g=(x_1-b_1)...(x_n-b_n)$. Then $V(f)\cup V(g)$ is a two-point set. But $V(fg)$ clearly has more than two points e.g. $(a_1,b_2,a_3,...,a_n)$. What is wrong here?

Sorry if this question sounds silly.

@AlexWertheim in my example, from the forms of $f,g$, $V(f)$ is just a single point, and so is $V(g)$. Right? — Jun Xu, Jun 19 '21 at 04:46
No. Why do you think $V(f)$ is a single point? What is $V(f)$? — Alex Wertheim, Jun 19 '21 at 04:49
@AlexWertheim Thanks... I was thinking about $V(x_1-a_1, x_2-a_2,...)$. How silly. — Jun Xu, Jun 19 '21 at 04:54
No problem - happens to us all. You can write an answer to your own question, which will help future users who have the same confusion. — Alex Wertheim, Jun 19 '21 at 05:03

score 1 · Accepted Answer · answered Jun 22 '21 at 03:10

1

So the catch is $V(x_1-a_1, ...,x_n-a_n)\neq V((x_1-a_1)\cdot(x_2-a_2)\cdot ...\cdot (x_n-a_n))$.

answered Jun 22 '21 at 03:10

Jun Xu

449

In order for this to be useful for other people, please consider expanding the answer. For example, one may write out explicitly what is $V((x_1-a_1)\cdot(x_2-a_2)\cdot \cdots \cdot (x_n-a_n))$. – Arctic Char Jun 22 '21 at 03:31

$V(f)\cup V(g) = V((f)(g))=V((fg))=V(fg)$?

1 Answers1