Lemma 10.138.14. Let $R \to S$ be a smooth ring map. Then there exists a subring $R_0 \subset R$ of finite type over $\mathbf{Z}$ and a smooth ring map $R_0 \to S_0$ such that $S \cong R \otimes _{R_0} S_0$.

**Proof.**
We are going to use that smooth is equivalent to finite presentation and formally smooth, see Proposition 10.138.13. Write $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and denote $I = (f_1, \ldots , f_ m)$. Choose a right inverse $\sigma : S \to R[x_1, \ldots , x_ n]/I^2$ to the projection to $S$ as in Lemma 10.138.5. Choose $h_ i \in R[x_1, \ldots , x_ n]$ such that $\sigma (x_ i \bmod I) = h_ i \bmod I^2$. The fact that $\sigma $ is an $R$-algebra homomorphism $R[x_1, \ldots , x_ n]/I \to R[x_1, \ldots , x_ n]/I^2$ is equivalent to the condition that

for certain $a_{kl} \in R[x_1, \ldots , x_ n]$. Let $R_0 \subset R$ be the subring generated over $\mathbf{Z}$ by all the coefficients of the polynomials $f_ j, h_ i, a_{kl}$. Set $S_0 = R_0[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$, with $I_0 = (f_1, \ldots , f_ m)$. Let $\sigma _0 : S_0 \to R_0[x_1, \ldots , x_ n]/I_0^2$ defined by the rule $x_ i \mapsto h_ i \bmod I_0^2$; this works since the $a_{lk}$ are defined over $R_0$ and satisfy the same relations. Thus by Lemma 10.138.5 the ring $S_0$ is formally smooth over $R_0$. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)