Lemma 38.23.2. Let $A$ be a ring. Let $u : M \to N$ be a surjective map of $A$-modules. If $M$ is projective as an $A$-module, then there exists an ideal $I \subset A$ such that for any ring map $\varphi : A \to B$ the following are equivalent
$u \otimes 1 : M \otimes _ A B \to N \otimes _ A B$ is an isomorphism, and
$\varphi (I) = 0$.
Proof. As $M$ is projective we can find a projective $A$-module $C$ such that $F = M \oplus C$ is a free $A$-module. By replacing $u$ by $u \oplus 1 : F = M \oplus C \to N \oplus C$ we see that we may assume $M$ is free. In this case let $I$ be the ideal of $A$ generated by coefficients of all the elements of $\mathop{\mathrm{Ker}}(u)$ with respect to some (fixed) basis of $M$. The reason this works is that, since $u$ is surjective and $\otimes _ A B$ is right exact, $\mathop{\mathrm{Ker}}(u \otimes 1)$ is the image of $\mathop{\mathrm{Ker}}(u) \otimes _ A B$ in $M \otimes _ A B$. $\square$
Your email address will not be published.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)
There are also: