# Why is the finitely generated submodule of a free module over a semihereditary ring projective?

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.

• so much jargon, ew – terrace Jun 10 '17 at 23:08
• @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. – K.Power Jun 10 '17 at 23:10
• Good luck :) $\,$ – terrace Jun 10 '17 at 23:11

## 1 Answer

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.