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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?

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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.

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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).

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  • $\begingroup$ Can you please say a few words, why hereditary and semihereditary are the same over finite dimensional algebras ? $\endgroup$
    – Todd Leason
    Commented Dec 22, 2016 at 22:04
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    $\begingroup$ @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
    Commented Dec 23, 2016 at 9:39

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