Are irreducible components of a flat family flat?

Let $f:X\rightarrow Y$ be a flat morphism of schemes of finite type over a field $k$, and assume $Y$ is irreducible. Let $X_1, \dots, X_n$ be the scheme-theoretic irreducible components of $X$ (i.e., including embedded components).

• Is it true that each $X_i$ is flat over $Y$?
• If there are counterexamples to flatness of the $X_i$, is it true at least that each of them has equidimensional fibers?
• If you take a plane and a line intersecting at the origin, isn't that a flat map of finite type over your ground field with components of different dimensions? Nov 22, 2012 at 23:22
• By each component having equidimensional fibers I mean that $dim (X_i)_y$ is independent of $y$. Of course these dimensions will depend on $i$.
– quim
Nov 23, 2012 at 9:27

No to the first question. Let $Y$ be a nodal cubic curve and let $X$ be its connected two-sheeted covering space. Each of the two components of $X$ is the normalization of $Y$.

• Sometimes the nomenclature is weird... That a connected covering of something have two components! Nov 22, 2012 at 14:25
• @Mariano: if the two components intersect, what is the problem in calling the covering "connected"? Nov 22, 2012 at 14:46
• The weirdness is the conflict between topology language and algebraic geometry language. In topology the connected components of a space are sometimes referred to as simply "components". In alg geo the irreducible components of a scheme are sometimes referred to as simply "components". But connectedness plays a role in alg geo, too. Nov 22, 2012 at 15:33
• @Tom Goodwillie: Thank you! I have been able to track down this example to EGA IV, Remarques 6.5.5(ii) and 12.1.2(i). On the other hand, the immediately preceding proposition 12.1.1.5 shows that my second question has positive answer
– quim
Nov 29, 2012 at 21:13