**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 $<2$).



Degree of fibers with too large dimension
This is an interesting idea, but I think there are some issues with it. First of all, it seems to me that if the image $\pi(X)$ has degree $>1$, then all fibers will have strictly smaller degree than $\mathrm{deg}X$. So one might ask, if it is true that the fibers with excess dimension have smaller degree than those fibers whose dimension is the expected one. However, this actually fails. 


**Example**: Let $X\subseteq \mathbb A^5$ be defined by $x\_1=x\_3x\_4,x\_2=x\_3x\_5$, and $x\_4x\_5=1$ and project to the first three coordinates.


For points on $\pi(X)$ with $x\_3\neq 0$ the fiber consists of the single point $\bigg(x\_1,x\_2,x\_3,\dfrac{x\_1}{x\_3}, \dfrac{x\_2}{x\_3}\bigg)$. If $x\_3=0$, then the first two equations of $X$ imply that $x\_1=x\_2=0$ as well, so the only point on $\pi(X)$ with $x\_3=0$ is $(0,0,0)$. The fiber over that is the conic $\{(0,0,0,t,u) | tu=1 \}$. So, the general fiber has degree $1$, while the (only) higher dimensional fiber has degree $2$. 


This degree is still smaller than the degree of $X$, but perhaps suggest that the defect is not a consequence of the excess dimension. In fact, notice that $\pi(X)\subseteq \mathbb A^3$ is contained in the quadric cone defined by $x\_1x\_2=x\_3^2$, and my guess is that a proper degree estimate would have to take the degree of the image into account.


*Addendum* to answer Shachar's question in the comments to the original question.  

Here is why I think that the degree of $\pi(X)$ matters:  

1) Your definition of degree seems to agree with the following: Take the projective closure of $X$ and take the degree of that in $\mathbb P^n$.  

2) To determine the degree of $X$ you need a linear space of appropriate dimension in general position. That may be described as the intersection of an appropriate number of hyperplanes in the target space of the projection which would give $\mathrm{deg} \pi(X)$ number of points.  

3) The intersection of the hyperplanes that project onto these ones would give the same number of fibers. To get down to finitely many points, just take the appropriate remaining number of hyperplanes to intersect.  

4) So now we have $\deg X$ number of points on $\deg\pi(X)$ number of fibers, therefore the expected number of points on a single fiber is $\deg X/\deg\pi(X)$.  

5) The reason I'd prefer to work projectively is that then the intersections would always give the right number of points counted with multiplicities.  

6) This is not a rigorous argument, but it seems convincing to me. Cheers!


