Flat map – fibre dimension

Flat map has a nice property : the fibre dimension is very well behaved. Assume we have a flat map . If we take a point on , its image, then the dimension of the fibre over ( going through ) is exactly as we expect : the difference between the local dimension at and that at :

Idea of proof :

-The proof use a lot of steps involving simplifying the situation. First we can base change to make a closed point, is a local affine scheme ( Spectrum of a local ring ) , by base change to . This is in effect focus all attention to the local property at . Note that the local information at is preserved because we infact take the direct limit of the inverse of a local basis at , which must as not refined as a local basis at . This can be check using using affine open sets at and .

- Next we can, by base change ( cut out the sheaf of nilradical), get rid of all the nilpotents in , make it into a reduced scheme.

. This does not change dimensions at all. Now we can proceed by induction : we can choose an element in that is not a zero divisor. Then the subscheme defined by is one dimension less. The pull back of the subscheme ( corresponds to the base change by the map that injects the subscheme into ) is defined on by the inverse image ideal of , which is non-zero-divisor at by flatness, hence define a subschem passing through which is 1 dimension less, and this agrees with the hypothesis.

Since we need the dimension to decrease by exactly one after being cut out by a non zero divisor, we need the schemes to be of finite type over an a.c. field .

The above result is a local property. If we need to make claims about the relative dimension between the components of with their images are all the same, we need to make an assumption that all the fibres have the same dimensions. The two things are infact equivalent.