Let $A \subseteq B \subseteq C$ be commutative rings.

A known result, which can be found in De Meyer and Ingraham's book, says that separability of $A \subseteq C$ implies separability of $B \subseteq C$.

When separability of $A \subseteq C$ implies separability of $A \subseteq B$? By 'when' I mean what additional condition guarantees that implication.

Another, somewhat similar, question: Let $A \subseteq B$, $A \subseteq C$, $B \twoheadrightarrow C$ be commutative rings.

When separability of $A \subseteq C$ implies separability of $A \subseteq B$?

Thank you very much for any help!

Edit: (1) By separable I mean the definition as in the above mentioned book on page 40 (this page is available in the preview), namely: $A \subseteq B$ is separable if $B$ is a projective $B \otimes_A B$-module. The definition of separability also appears here. (2) There is an answer to a special case of my first question, due to Adjamagbo's transfer theorem. In that special case, 'when'= $B \subseteq C$ is flat, or 'when'= $B$ is normal.


I do not know what definition of separability you are using. $A\subset C$ is separable means every element of $C$ is separable over $A$ in the usual definition. Thus, for any subring $A\subset B\subset C$, clearly $B$ is separable over $A$, since every element of $B$ is an element of $C$ and thus separable.

  • $\begingroup$ I really apologize for not being clear enough. My rings are not assumed to be fields. The definition of separability can be found in the book I mentioned above, and also in people.brandeis.edu/~buchsbau/miscpapers/024.pdf and in ams.org/journals/tran/1960-097-03/S0002-9947-1960-0121392-6/… $\endgroup$ – user237522 Jul 19 '17 at 12:26
  • $\begingroup$ If we deal with three fields, then the second question also has a positive answer: By assumption $B \twoheadrightarrow C$ (surjective) and trivially it is injective ($B$ is a field), so $B$ and $C$ are isomorphic, hence separability of $A \subseteq C$ clearly implies separability of $A \subseteq B$. $\endgroup$ – user237522 Jul 19 '17 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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