Fibre dimension Let $f:X\rightarrow Y$ be a surjective morphism of algebraic schemes (over an algebraically closed field $F$ of characteristic $0$). I think in good cases ($X$ and $Y$ irreducible) we have that every for every $y$ in $Y$ the dimension of the fibre $f^{-1}(y)$ is at least $\dim X-\dim Y$. What happens if we drop the irreducibility assumption? Are the counterexamples to this lower bound for the ddimension of $f^{-1}(y)$?
 A: Yes, if you do not require $Y$ to be irreducible then this bound can be violated.
The cheapest example is to take $f: X \rightarrow Y$ surjective of whatever relative dimension you like, bigger than $0$, let $X'$ and $Y'$ be the disjoint union of each of them with a point, and define $f'$ to equal $f$ on $X$ and to map the new point on $X$ to the new point on $Y$.
If you require connectedness you can do something similar, now attaching say a line transversely to $X$ at some point $x$ and another line to $Y$ at $f(x)$, and extending $f$ by mapping the line to the line isomorphically.

With a little more fiddling, we can also cook up an example where $Y$ is irreducible, as follows.
Let $Y=\mathbf A^1$, let $Z \subset \mathbf A^2$ be defined by $xy=1$, and let $q:Z \rightarrow Y$ be the projection onto say the first factor. This map is an isomorphism onto the open set $Y \setminus \{0\}$.
Now let $W=Z \times \mathbf A^1 \subset \mathbf A^3$. Then the projection map $W \rightarrow Y$ has 1-dimensional fibres over every point $y \neq 0$, while the fibre over $0$ is empty.
Finally let $L$ be the axis $\{(t,0,0) \mid t \in k \} \subset \mathbf A^3$, and let $X=W \cup L \subset \mathbf A^3$. Then the projection $X \rightarrow Y$ is surjective, with 1-dimensional fibres over every $y \neq 0$, and 0-dimensional fibre over $0 \in Y$.
