The Stacks project

Lemma 10.138.2. Let $R \to S$ be a formally smooth ring map. Let $R \to R'$ be any ring map. Then the base change $S' = R' \otimes _ R S$ is formally smooth over $R'$.

Proof. Let a solid diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rrd] & R' \otimes _ R S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[u] \ar[r] & R' \ar[r] \ar[u] & A \ar[u] } \]

as in Definition 10.138.1 be given. By assumption the longer dotted arrow exists. By the universal property of tensor product we obtain the shorter dotted arrow. $\square$

Comments (0)

There are also:

  • 5 comment(s) on Section 10.138: Formally smooth maps

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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00TJ. Beware of the difference between the letter 'O' and the digit '0'.