Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was hoping to ask a small follow up to the question I asked here.

Suppose $V$ is an algebraic variety over arbitrary field $k$. (For this situation, I'll take the definition $\dim\ V=\deg_k(k(x))$, where $(x)=(x_1,\dots,x_n)\in V$ is a generic point, and by $\deg$ I mean the transcendence degree.) As usual, $V(f_1,\dots,f_s)$ is the set of zeroes of the homogeneous forms $f_1,\dots,f_s$ in the affine space.

Now say you take $U$ to be an algebraic set $x_1=\cdots=x_p=0$, (so $U$ is the algebraic set with associated ideal $(x_1,\dots,x_p)$) that is the algebraic set of coordinates in $\mathbb{A}^n$ where the first $p$ coordinates are $0$, and where $p<\dim\ V$. Is it now the case that $U\cap V\neq\{0\}$? Many thanks.

share|cite|improve this question
I don't understand your question. In particular, what is the "trivial zero" in this context? – Mariano Suárez-Alvarez Feb 12 '12 at 2:10
@MarianoSuárez-Alvarez Sorry, I think my wording was unnecessarily poor. I've removed it now. – Vika Feb 12 '12 at 3:17

Answer: No.

Consider $k=\mathbb{R}$ and $f(x_1,x_2,x_3):=x_1^2-x_2^2-x_3^2$. Then $V:=V((f))\subset\mathbb{A}^3_\mathbb{R}$ is a double-cone, where the singuarity is placed at $(0,0,0)$. In particular $\dim (V)=2$.

The plane $U$ defined by $x_1=0$ intersects $V$ precisely in $(0,0,0)$.

share|cite|improve this answer

I suppose that you want to know if $U \cap V$ can be bigger than a point, or a finite set of points...

In general, as Hagen said, it is false but it is true when $k$ is algebraically closed. In fact, in this case, you have an intersection theorem which says that if $V$ and $W$ are two irreducible affine sets in $k^n$, with $\dim V = r$ and $\dim W=s$ then each irreducible component of $V \cap W$ has dimension $\geq r + s - n$. This means that in your case, if $V \cap U$ is nonempty, its dimension is necessarily $\geq 1$, and contains then more than a finite set of points.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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