Flatness under reduction Suppose that $f : X \to Y$ is a flat morphism of schemes.  Is $f_\text{red} : X_\text{red} \to Y_\text{red}$ necessarily flat?  Are there any hypotheses that would guarantee this?
 A: No !
The result is not true: it may happen that  $f : X \to Y$ is a flat morphism but that  $f_\text{red} : X_\text{red} \to Y_\text{red}$ is not flat.
A counterexample
Consider $\mathbb A^3$ with coordinates $x,y,z$ and for $X\mathbb \subset \mathbb A^3$ take the curve defined by the equations $4y^3+27z^2=0,x^3+yx+z=0$  .
For $Y$ take the  subscheme of $\mathbb  A^2$ with coordinates $y,z$ defined by the equation $4y^3+27z^2=0$.
Note carefully that $Y$ is already reduced but that $X$ is not.
And finally for $f:X\to Y$ take  the projection $(x,y,z)\mapsto (y,z)$.
Now, $f$ is flat because it is finite and all its fibers have length $3$ .
However $f_\text {red}:X_\text {red}\to Y_\text {red}=Y$ is not flat because its  fibers have length $2$ above all fibers except the fiber over $(0,0)\in Y$ which is of length $3$.
Attribution
According to Gerd Fischer in his fine book Complex Analytic Geometry (page 151) this example is due to the late Adrien Douady, the master of counterexamples.
However I could not find the original source.
This is not surprising: I witnessed several episodes where Douady stunned the audience consisting of some of the best complex analysts in the world (in Oberwolfach for example) with counterexamples which he found on the spot with lightning velocity.
I guess that, unfortunately, most of these counterexamples went unrecorded.
Appreciation
This is a very nice question, apparently not addressed in books about algebraic geometry, not even apparently (but I may have overlooked something) in EGA nor Stacks Project.
