The Stacks project

Lemma 37.7.4. Let $Z \to X$ be a formally unramified morphism of schemes. Then the universal first order thickening $Z'$ is formally unramified over $X$.

Proof. There are two proofs. The first is to show that $\Omega _{Z'/X} = 0$ by working affine locally and applying Algebra, Lemma 10.149.5. Then Lemma 37.6.7 implies what we want. The second is a direct argument as follows.

Let $T \subset T'$ be a first order thickening. Let

\[ \xymatrix{ Z' \ar[d] & T \ar[l]^ c \ar[d] \\ X & T' \ar[l] \ar[lu]^{a, b} } \]

be a commutative diagram. Consider two morphisms $a, b : T' \to Z'$ fitting into the diagram. Set $T_0 = c^{-1}(Z) \subset T$ and $T'_ a = a^{-1}(Z)$ (scheme theoretically). Since $Z'$ is a first order thickening of $Z$, we see that $T'$ is a first order thickening of $T'_ a$. Moreover, since $c = a|_ T$ we see that $T_0 = T \cap T'_ a$ (scheme theoretically). As $T'$ is a first order thickening of $T$ it follows that $T'_ a$ is a first order thickening of $T_0$. Now $a|_{T'_ a}$ and $b|_{T'_ a}$ are morphisms of $T'_ a$ into $Z'$ over $X$ which agree on $T_0$ as morphisms into $Z$. Hence by the universal property of $Z'$ we conclude that $a|_{T'_ a} = b|_{T'_ a}$. Thus $a$ and $b$ are morphism from the first order thickening $T'$ of $T'_ a$ whose restrictions to $T'_ a$ agree as morphisms into $Z$. Thus using the universal property of $Z'$ once more we conclude that $a = b$. In other words, the defining property of a formally unramified morphism holds for $Z' \to X$ as desired. $\square$

Comments (0)

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Your email address will not be published.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 04F6. Beware of the difference between the letter 'O' and the digit '0'.