# 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 $2$.

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

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.
• 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