## 108.37 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

$k[\mathbf{Q}] \longrightarrow k, \quad x_\alpha \longmapsto 1.$

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

$(x_\alpha - 1)(x_\beta - 1) = x_{\alpha + \beta } - x_\alpha - x_\beta + 1 = (x_{\alpha + \beta } - 1) - (x_\alpha - 1) - (x_\beta - 1)$

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

$0 = pz_{\alpha /p} = z_{\alpha /p} + \ldots + z_{\alpha /p} = z_\alpha$

and we see that $J/J^2 = 0$. If the characteristic of $k$ is zero, then

$J/J^2 = \mathbf{Q} \otimes _{\mathbf{Z}} k \cong k$

(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

$\xymatrix{ k \ar[r] \ar@{..>}[rd] & A \\ k[\mathbf{Q}] \ar[u] \ar[r]^\varphi & A' \ar[u] }$

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.

Lemma 108.37.1. There exists a formally smooth ring map which is not flat.

Proof. See discussion above. $\square$

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