Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

On Wikipedia, in the description of the conormal exact sequence, it is described as arising from a closed immersion, which corresponds in the affine case to a surjection of algebras. However, in Eisenbud (CA with a View Towards AG), we have proposition 16.3 (page 389) as follows:

If $\pi:S\to T$ is an epimorphism of $R$-algebras, with kernel $I$, then there is an exact sequence of $T$-modules $$ I/I^2 \overset d\to T\otimes_R\Omega_{S/R} \overset {D\pi}\to\Omega_{T/R}\to0 $$ where the right-hand map is given by $D\pi:c\otimes db\mapsto cdb$ and the left-hand map takes the class of $f$ to $1\otimes df$.

An epimorphism of algebras is certainly not in general a surjection, and yet Eisenbud later writes $S/I= T$, making it quite clear that this map is intended to be a surjection. I've checked Eisenbud's published list of errata, and this is not there. Is this a typo or am I missing something?

share|cite|improve this question
This abuse of language is quite common. But it should be clear from the context that no epimorphisms of commutative rings are considered, they are only relevant in a small part of algebraic geometry (namely monomorphisms of schemes). Anyway, this leads us to the following question: Let $S \to T$ be an epimorphism of commutative $R$-algebras. Is $T \otimes_R \Omega_{S/R} \to \Omega_{T/R}$ an epimorphism of $T$-modules? Remark that for localizations it is even an isomorphism. – Martin Brandenburg Feb 4 '13 at 11:54
up vote 1 down vote accepted

I believe the hypotheses are that the map is surjective, as this is also the way the result you mention is presented in Hartshorne's "Algebraic Geometry".

This choice of language probably stems from the fact that the notions of mono- and epimorphisms are often introduced in the context of $R$-modules as a shorthand for injective/surjective linear map. Thus one way to interpret the sentence you mention is "Let $\pi: S \rightarrow T$ be a homomorphism of $R$-algebras that is an epimorphism as a map of $R$-modules".

share|cite|improve this answer
I hadn't even considered that possibility, but indeed Eisenbud defines on pg 15 that an epimorphism of modules is a map which is a surjection of underlying sets, and makes no other definition of the term throughout the text. Thanks a lot! – Xander Flood Feb 4 '13 at 4:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.