Classifying Algebra Extensions over a fixed extension? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T13:50:48Z http://mathoverflow.net/feeds/question/9190 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9190/classifying-algebra-extensions-over-a-fixed-extension Classifying Algebra Extensions over a fixed extension? Chris Schommer-Pries 2009-12-17T17:56:49Z 2009-12-17T19:16:44Z <p>There are lots of "Ext groups" in homological algebra which measure extensions of various things. I'm sure there must be a homological algebra machine for computing the following, and I'm hoping that someone out there knows about it. </p> <p>I'm interested in the following situation. Let R and S be commutative rings and fix a ring homomorphism $f:R \to S$. Also fix a commutative S-algebra A. I'm interested in understanding/classifying those R-algebras B, together with a (surjective?) ring homomorphism $g: B \to A$ which intertwines the algebra structures in the following sense:</p> <p>$g(rb) = f(r) g(b)$</p> <p>for all $r \in R$, and $b \in B$. Is there a homological algebra way to do this? </p> <p>A particular example that I am interested in is when we have the equality $B \otimes_R S = A$, but I am also interested in other cases as well. </p> http://mathoverflow.net/questions/9190/classifying-algebra-extensions-over-a-fixed-extension/9194#9194 Answer by Alberto García-Raboso for Classifying Algebra Extensions over a fixed extension? Alberto García-Raboso 2009-12-17T19:04:39Z 2009-12-17T19:16:44Z <p>I think your problem is not constrained enough to have an interesting answer. Notice that your intertwining condition can be rephrased by saying that $g: B \to A$ is a homomorphism of $R$-algebras, where $A$ is given the structure of $R$-algebra given by $f$. In these terms, what you are looking for is the comma category $\mathcal{C} = (\mathbf{Alg}_R \downarrow A)$, whose objects are precisely the pairs $(B, g)$ as above, and whose morphisms <code>$\operatorname{Hom}_{\mathcal{C}}((B, g), (B', g'))$</code> are the $R$-algebra homomorphisms $h: B \to B'$ such that $g = g' \circ h$. I am not sure it is possible to capture this beast with a cohomology group of any sort.</p> <p>What you can do is restrict the class of objects that you are looking at. For example, you can classify square-zero extensions: fixing an $A$-module $I$, you can look at $R$-algebras $B$ such that you have a short exact sequence $0 \to I \to B \to A \to 0$; the name square-zero comes from the fact that $I$ is an ideal of $B$ with $I^2 = 0$. You can read about them in the first chapter of Sernesi's <em>Deformations of Algebraic Schemes</em>.</p>