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 5Let 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…)