3
$\begingroup$

I am learning about left (right) semi-hereditary rings, and got stuck with the following questions. Just to recall that being left semi-hereditary means that every finitely generated submodule of a projective left module is projective.

Let $S$ and $T$ be two unital non-commutative rings which are left semi-hereditary.

(1) Is it true that the opposite ring $S^{op}$ is left semi-hereditary. I expect the answer to be no, but my knowledge of non-commutative ring theory is not enough to get a counterexample.

(2) Is it true that $S\otimes_{\mathbb Z} T$ is left semi-hereditary?

$\endgroup$
4
$\begingroup$

Concerning (1), the answer is no. An example, attributed to Chase, of a left semihereditary ring that is not right semihereditary is given in T. Y. Lam, Lectures on modules and rings, Springer GTM 189 (1999) as Example 2.34.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

The answer to (2) is "no" even for hereditary rings. For example, if $S=T$ is the algebra of upper triangular $2\times 2$ matrices (or, more generally, pretty much any finite dimensional hereditary algebras) over the field $\mathbb{F}_p$ ($p$ prime), then $S\otimes_\mathbb{Z}T$ is not hereditary (and since it's a finite dimensional algebra over a field, not semihereditary).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you please say a few words, why hereditary and semihereditary are the same over finite dimensional algebras ? $\endgroup$ – Todd Leason Dec 22 '16 at 22:04
  • 1
    $\begingroup$ @ToddLeason A ring is left hereditary if every left ideal is projective, and left semihereditary if every finitely generated left ideal is projective. For a finite dimensional algebra, or more generally any left noetherian ring, all ideals are finitely generated, and so left hereditary is the same as left semihereditary. $\endgroup$ – Jeremy Rickard Dec 23 '16 at 9:39

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.