Lemma 110.41.1. There exists a formally smooth ring map which is not flat.
110.41 A formally smooth non-flat ring map
Let k be a field. Consider the k-algebra k[\mathbf{Q}]. This is the k-algebra with basis x_\alpha , \alpha \in \mathbf{Q} and multiplication determined by x_\alpha x_\beta = x_{\alpha + \beta }. (In particular x_0 = 1.) Consider the k-algebra homomorphism
It is surjective with kernel J generated by the elements x_\alpha - 1. Let us compute J/J^2. Note that multiplication by x_\alpha on J/J^2 is the identity map. Denote z_\alpha the class of x_\alpha - 1 modulo J^2. These classes generate J/J^2. Since
we see that z_{\alpha + \beta } = z_\alpha + z_\beta in J/J^2. A general element of J/J^2 is of the form \sum \lambda _\alpha z_\alpha with \lambda _\alpha \in k (only finitely many nonzero). Note that if the characteristic of k is p > 0 then
and we see that J/J^2 = 0. If the characteristic of k is zero, then
(details omitted) is not zero.
We claim that k[\mathbf{Q}] \to k is a formally smooth ring map if the characteristic of k is positive. Namely, suppose given a solid commutative diagram
with A' \to A a surjection whose kernel I has square zero. To show that k[\mathbf{Q}] \to k is formally smooth we have to prove that \varphi factors through k. Since \varphi (x_\alpha - 1) maps to zero in A we see that \varphi induces a map \overline{\varphi } : J/J^2 \to I whose vanishing is the obstruction to the desired factorization. Since J/J^2 = 0 if the characteristic is p > 0 we get the result we want, i.e., k[\mathbf{Q}] \to k is formally smooth in this case. Finally, this ring map is not flat, for example as the nonzerodivisor x_2 - 1 is mapped to zero.
Proof. See discussion above. \square
Comments (0)