**Background:**

Let $\mathbb{F}$ be an algebraically closed field. Let $X \subset \mathbb{F}^n$ be an affine variety. Let $\pi(X)$ be the projection of $X$ to the first $m < n$ coordinates, $$ \pi(X) = \{(x_1,\ldots,x_m): x \in X\}, $$ and for a point $a \in \mathbb{F}^m$ let $\phi(a,X)$ be the fiber of $X$ over $a$, $$ \phi(a,X) = \{x \in X: x_1=a_1,\ldots,x_m=a_m\}. $$ It is known that $\dim(\pi(X)) + \dim(\phi(a,X)) \ge \dim(X)$ for all points $a$, and that equality holds for all $a \in U$ where $U \subset \mathbb{F}^m$ is a Zariski open set (so, dimension equality holds for "typical" fibers).

**Question:**

Are "a-typical" fibers, where the dimension equality doesn't hold, have lower degree than $X$?

That is, for all fibers we have that $\deg(\phi(a,X)) \le \deg(X)$ since they are the intersection of $X$ with the degree $1$ variety given by $x_1=a_1,\ldots,x_m=a_m$. Can it be that when $\dim(\pi(X))+\dim(\phi(a,X))>\dim(X)$ it implies that $\deg(\phi(a,X))<\deg(X)$?

**Example:**

Let $X$ be defined by $x_1 x_3 + x_2 x_4=0$. Then $\dim(X)=3,\deg(X)=2$. The projection of $X$ to the first $2$ coordinates $x_1,x_2$ has dimension $2$. Fibers over $(a_1,a_2)$ if $(a_1,a_2) \ne (0,0)$ have as expected dimension $3-2=1$. However, the fiber over $(0,0)$ has dimension $2$ (which is $>1$) but degree $1$ (which is $<2$).