Let $X, Y$ be irreducible projective varieties and $Y$ be smooth. Let $f:X \to Y$ be a flat projective morphism. Assume that a special fiber of $f$ is non-reduced i.e., there exists an irreducible component of the fiber which is of multiplicity greater than $1$. When can we say that the generic fiber or $X$ is non-reduced?
EDIT Is there any criterion on $f$ or on $X, Y$ under which the generic fiber or $X$ is non-reduced provided it satisfies the above condition?
I think there is some confusion here. Either on your part or on mine. I don't think being non-reduced is equivalent to having a non-reduced component. A scheme may have a fat point, but be irreducible and generically reduced. Similarly, I don't quite understand what you want. Do you want the generic fiber be non-reduced or do you want $X$? Those are different questions. Or I am missing something. I would say that the generic fiber being non-reduced implies that $X$ is, but not the other way around.
My first reaction was that there is no such criterion, because it is quite common to have a multiple fiber, even for smooth surface flat over a smooth curve.
Then I realized that there might be a condition that implies this: If $f$ admits a section that intersects the multiple component of your special fiber, then you have a fighting chance.
Here is the idea: Let's say for safety that we are in characteristic zero. If $X$ were smooth, and say $D$ is a section and $F$ is a fiber, then $f$ is smooth at a general point of $D$ and hence $D\cdot F=1$, so $D$ cannot intersect a multiple component in any fiber.
Of course, you don't want to assume that $X$ is smooth, because you want to conclude that it isn't. However, there might be some ways to weaken this. So here is a criterion that does what you ask for, but I am not claiming that this is easy to achieve or that there isn't anything simpler and for sure I do not expect this to work for you. (But if it does, that would be great). Also, it has some limitations.
Claim Assume that everything is defined over an algebraically closed field of characteristic zero and $\dim Y=1$. If $f$ admits a section that intersects the multiple component of the special fiber and contains a general point of $X$, then the general fiber of $f$ is everywhere non-reduced and hence so is $X$.
Proof Suppose $X$ (and hence the general fiber) is not everywhere non-reduced, that is, it is generically reduced. Then $X$ is generically smooth and hence $f$ is generically smooth. By the assumption the section, say $Y'$, intersects this locus and within that locus it intersects each fiber transversally in a single smooth point. By the assumptions we may consider a resolution of singularities of $X$, say $\pi:X'\to X$ and the proper transform of $Y'$ will be a section of the composition $f\circ\pi$. The preimage of the multiple component of the special fiber will still be (a union of) multiple component(s) of a fiber of the composition and the proper transform of $Y'$ will still intersect this. (I am saying preimage, not proper transform!!). But then we get a contradiction because the intersection number of the section and the fiber should be $1$. Therefore the original assumption was false and hence the general fiber of $f$ and $X$ are both everywhere non-reduced.
I have the feeling that this could be improved to include the case $\dim Y>1$ and not having to make the assumption on characteristic (that is, using resolutions), but I don't see an easy way to do that. The point is to ensure that if a section intersects the general fiber at a single smooth point transversally, then it cannot intersect a multiple component of another fiber. If $X$ has a reasonable intersection theory, then this should be OK, but what if it is just horribly singular but still reduced.....?
Uh, and just to note that the condition can be indeed satisfied, observe that if $X=Y\times Z$ where $Z$ is everywhere non-reduced, then all the assumptions hold.
