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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.

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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

2 Answers 2

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 $(0,0,0)$.

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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.

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