The Stacks project

Lemma 90.3.3. Let $f: B \to A$ be a surjective ring map in $\mathcal{C}_\Lambda $. Then $f$ can be factored as a composition of small extensions.

Proof. Let $I$ be the kernel of $f$. The maximal ideal $\mathfrak {m}_ B$ is nilpotent since $B$ is Artinian, say $\mathfrak {m}_ B^ n = 0$. Hence we get a factorization

\[ B = B/I\mathfrak {m}_ B^{n-1} \to B/I\mathfrak {m}_ B^{n-2} \to \ldots \to B/I \cong A \]

of $f$ into a composition of surjective maps whose kernels are annihilated by the maximal ideal. Thus it suffices to prove the lemma when $f$ itself is such a map, i.e. when $I$ is annihilated by $\mathfrak {m}_ B$. In this case $I$ is a $k$-vector space, which has finite dimension, see Algebra, Lemma 10.53.6. Take a basis $x_1, \ldots , x_ n$ of $I$ as a $k$-vector space to get a factorization

\[ B \to B/(x_1) \to \ldots \to B/(x_1, \ldots , x_ n) \cong A \]

of $f$ into a composition of small extensions. $\square$


Comments (0)

There are also:

  • 4 comment(s) on Section 90.3: The base category

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Your email address will not be published.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 06GE. Beware of the difference between the letter 'O' and the digit '0'.