# Is dimension an open condition?

Let's consider a family of varieties defined by two equations in $\operatorname{Spec}(\mathbb{C}[x_1,\dots,x_n])$.

First, Let $\Delta_i, i=1,2$ be a finite set of monomials in $\mathbb{C}[x_1,\dots,x_n]$, for simplicity, I use $X^v$ to denote an element of $\Delta_i$. Then $$F_i=\{\sum_{X^v \in \Delta_i}c_vX^v \mid c_v \in \mathbb{C}\}$$ is a family of polynomials parameterized by $\mathbb{C}^{\#(\Delta_i)}$, where $\#(\Delta_i)$ is the cardinality of $\Delta_i$.

Let $V_{f,g}$ be a variety in $\operatorname{Spec}(\mathbb{C}[x_1,\dots,x_n])$ defined by $f=g=0$ with $f \in F_1, g \in F_2$ respectively. Then $V_{f,g}$ is parameterized by $\mathbb{C}^{\#(\Delta_1)} \times \mathbb{C}^{\#(\Delta_2)}$.

Is the family of codimesion $2$ intersection $V_{f,g}$ parameterized by an open set (might be empty) of $\mathbb{C}^{\#(\Delta_1)} \times \mathbb{C}^{\#(\Delta_2)}$?

I think the point is to find an "algebraic" characterization of dimension. However, it will be good even if such set is not open, but contains a nonempty open set (or the set itself is empty). Geometrically, it seems quite plausible this is the case.

Edit I think the result might be true in $\operatorname{Proj}(\mathbb{C}[x_0,\dots,x_n])$ by the following argument:

First, the parameterized space is really $\mathbb{P}^{\#(\Delta_1)-1} \times \mathbb{P}^{\#(\Delta_2)-1}$ after ignoring the case $f=g=0$, and module the constant.

Let $a \in \mathbb{P}^{\#(\Delta_1)-1} \times \mathbb{P}^{\#(\Delta_2)-1}$, and $V_a$ be the intersection parameterized by $a$. Then we have a "variety" $S$ (I am a little worry about $S$ to be a variety, but since the following construction is "algebraic", it should be a variety )

$$S = \{(a, V_a) \mid a \in \mathbb{P}^{\#(\Delta_1)-1} \times \mathbb{P}^{\#(\Delta_2)-1}\} \subset \mathbb{P}^{\#(\Delta_1)-1} \times \mathbb{P}^{\#(\Delta_2)-1} \times \mathbb{P}^n.$$

It should be a projective variety, with a natural projection

$$S \to \mathbb{P}^{\#(\Delta_1)-1} \times \mathbb{P}^{\#(\Delta_2)-1}, \quad (a, V_a) \mapsto a.$$ This projection is proper, then by the upper semicontinuity of the fibre, we know there is an open set of $\mathbb{P}^{\#(\Delta_1)-1} \times \mathbb{P}^{\#(\Delta_2)-1}$ such that the fibres are of codimension $2$.

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To downvoter: I think one line explanation would be much helpful for a downvote. –  Li Yutong Jul 16 '13 at 10:19

## 1 Answer

Probably this was downvoted because it is follows from standard material in most algebraic geometry textbooks. It might be more appropriate for Math Stack Exchange.

At any rate, with the question as formulated, this locus is not an open subset. For instance, consider a copy of $\mathbb{C}$ with coordinate $t$ inside your parameter space, where $f$ and $g$ are $f=x_1(1-tx_3)$ and $g=x_2(1-tx_3)$. For $t\neq 0$, then $V_{f,g}$ contains the codimension one hypersurface $\{(x_1,x_2,x_3) : x_3= 1/t\}$. But for $t=0$, $V_{f,g}$ has codimension $2$. Thus the intersection of this copy of $\mathbb{C}$ with your locus is the Zariski closed subset $\{t : t=0\}$, rather than a Zariski open subset. This type of example is one reason that algebraic geometers frequently work with proper / projective varieties rather than affine varieties.

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Thank you so much!! From your last sentence, it seems that in projective varieties, this result holds, and could you kindly explain why? –  Li Yutong Jul 16 '13 at 23:50
I have edited my original question with a possible proof of projective case. Do you think the argument sounds plausible? I am a little worry about $S$ in the construction to be a variety. –  Li Yutong Jul 17 '13 at 5:36
Yes, $S$ is a variety. It is defined by homogeneous equations in your product of projective spaces. –  Dan Petersen Jul 17 '13 at 6:49