Lemma 10.148.2. Let $R \to S$ be a formally unramified map. Let $R \to R'$ be any ring map. Then the base change $S' = R' \otimes _ R S$ is formally unramified 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 10.148.1 be given. By assumption there exists at most one longer dotted arrow. By the universal property of tensor product we conclude there is at most one shorter dotted arrow. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)