In this post, we shall accomplish the goal stated earlier: we shall show that a formally smooth morphism of noetherian rings (which is essentially of finite type) is flat. We shall even get an equivalence: flatness together with smoothness on the fibers will be both necessary and sufficient to ensure that a given such morphism is formally smooth.
In order to do this, we shall use a refinement of the criterion in the first post for when a quotient of a formally smooth algebra is formally smooth. We shall need a bit of local algebra to do this, but the reward will be a very convenient Jacobian criterion, which will then enable us to prove (using the results from last time on lifting flatness from the fibers) the final characterization of smoothness.
3. The Jacobian criterion
So now we want a characterization of when a morphism is smooth. Let us motivate this with an analogy from standard differential topology. Consider real-valued functions . Now, if are such that their gradients form a matrix of rank , then we can define a manifold near zero which is the common zero set of all the . We are going to give a relative version of this in the algebraic setting.
Recall that a map of rings is essentially of finite presentation if is the localization of a finitely presented -algebra.
Proposition 5 Let be a local homomorphism of local rings such that is essentially of finite presentation. Suppose for some finitely generated ideal , where is a prime ideal in the polynomial ring.Then is generated as a -module by polynomials whose Jacobian matrix has maximal rank in if and only if is formally smooth over . In this case, is even freely generated by the . (more…)