1
$\begingroup$

This is a question in Passman's "A Course in Ring Theory", and I'm struggling to prove it. I have an idea, which involves adapting some of his proofs for results on hereditary rings, but I'm not sure about it. His Lemma 6.2 states that if $\{V_\alpha:\alpha\in I\}$ is a family of right hereditary $R$ modules, and if $V=\bigoplus_{\alpha\in I}V_\alpha$ with $W$ a submodule of $V$, then $W\cong \bigoplus_{\alpha\in I}V_\alpha'$, with $V_\alpha '$ a submodule of $V_\alpha$. Using this lemma we can prove, if $R$ is hereditary, that every submodule of a free $R$ module is isomorphic to a direct sum of right ideals of $R$ (This is Theorem 6.3).

Now if I can show that Lemma 6.2 holds for a collection of semihereditary rather than hereditary rings, then I believe I can adapt the proof of Theorem 6.3 (which uses Lemma 6.2) to show that a finitely generated submodule of a free $R$ module must be isomorphic to a finitely generated right ideal of $R$, which will give us what we need.

Is there any problem with my proposed argument? My main problem is that I am unsure whether my proposed alteration to Lemma 6.2 is correct. If you know of a more elegant solution, please let me know as I'd be interested to see it.

References:

Donald Passman. A Course in Ring Theory. Wadsworth and Brooks/Cole, 1st edition, 1991.

$\endgroup$
  • 1
    $\begingroup$ so much jargon, ew $\endgroup$ – terrace Jun 10 '17 at 23:08
  • 1
    $\begingroup$ @terrace I know- I think that is half the reason why I'm struggling to work through Passman's book at a reasonable pace, but it was recommended to me by the resident expert on ring theory at my university, so I persevere. $\endgroup$ – K.Power Jun 10 '17 at 23:10
  • $\begingroup$ Good luck :) $ \, $ $\endgroup$ – terrace Jun 10 '17 at 23:11
1
$\begingroup$

So the argument is more or less done if you can show that if $R$ is a (left) semihereditary ring, then every f.g. submodule of a free $R$-module is a direct sum of finitely many finitely generated (left) ideals.

First, reduce to the case where the free module is actually finitely generated. Let $F$ be a free $R$-module, with basis $\{x_i\}_{i\in I}$, possibly infinite. Let $A=\langle a_1,\dots,a_m\rangle$ be a f.g. submodule of $F$. Each $a_i$ has finite support, that is, it has only finitely many nonzero coefficients when written in terms of the basis $\{x_i\}$. Let $X$ be the set of $x_i$ which are used with nonzero coefficient when writing the $a_1,\dots,a_m$. Then $A\subseteq\langle X\rangle$, and $X$ is free and finitely generated. So without loss of generality, we can assume $F$ is finitely generated, say with basis $\{x_1,\dots,x_n\}$.

Now induct on $n$ to see that $A$ is a direct sum of f.g. ideals. If $n=1$, this is clear. Otherwise, let $B=Rx_1\oplus\cdots\oplus Rx_{n-1}$. Then every $a\in A$ has a unique expression $a=b+rx_n$ for some $b\in B$. Define a projection map onto the last coordinate based on this expression $$ \pi\colon A\to R:a\mapsto r. $$ Then $\operatorname{im}(\pi)$ is an ideal in $R$, and it is f.g. since $A$ is f.g. Explicitly, if $a\in A$, then $a=c_1a_1+\cdots c_ma_m$, for some $c_i\in R$. But since $a_i=b_i+r_ix_n$ for unique $b_i,r_i$, we see $a\in B+(c_1r_1+\cdots+c_mr_m)x_n$, so $\pi(a)=c_1r_1+\cdots+c_mr_m$.

Since $R$ is semihereditary, $\operatorname{im}(\pi)$ is projective. There is then an exact sequence $$ 0\to B\cap A\hookrightarrow A\to\operatorname{im}(\pi)\to 0, $$ which splits as $A\simeq (B\cap A)\oplus\operatorname{im}(\pi)$. By induction, since $B\cap A$ is a f.g. submodule of a free module of lesser dimension, it is a direct sum of finitely many finitely generated ideals, and thus so is $A$.

Each of these finitely generated ideals is projective, hence $A$ is projective as well as a direct sum of projectives.

$\endgroup$

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.