3
$\begingroup$

Suppose $F$ is a free module of rank $n$ over a PID, with $N$ a submodule. Is $N$ always a direct summand of $F$?

I think the answer is yes, $N$ is also free of rank $m\leq n$ since we are working over a PID, and then $N\simeq R^m$, thus $F/N\simeq R^{n-m}$ which is a free $R$-module, so $N$ is a direct summand of $F$.

I'm just a little unsure since I can't find a reference for this online, and it seems like it should be a commonly documented fact if it is indeed true.

$\endgroup$
1
  • $\begingroup$ What examples did you consider? Does this work for submodules of $\mathbb Z$?! $\endgroup$
    – Mariano Suárez-Álvarez
    Feb 13, 2016 at 1:46

2 Answers 2

Reset to default
5
$\begingroup$

No. Let $R$ be a principal ideal domain that is not a field, and let $p\in R$ be a prime element. Then the ideal $I=pR$ cannot be a summand of $R$. Indeed, if there were an $R$-module isomorphism $R\cong I\oplus M$ for some $R$-module $M$, then we would have an injection $R/I\cong M\hookrightarrow R$, which is impossible as the source is a non-zero torsion $R$-module.

Your error is in concluding that the quotient of two free modules is free. This is not so, as the above example shows.

$\endgroup$
2
  • $\begingroup$ Thanks for pointing out my mistake. $\endgroup$
    – Valorum
    Jan 27, 2014 at 3:38
  • $\begingroup$ The simplest case is that of $\Bbb Z/2\Bbb Z$ which, being finite, is certainly not $\Bbb Z$-free $\endgroup$
    – Andrea Mori
    Mar 16, 2014 at 10:39
-2
$\begingroup$

Suppose $F$ is a free module of rank $n$ over a PID, with $N$ a submodule. Is $N$ always a direct summand of $F$?

Yes this is true. First, $N$ is a free module, as it is a submodule of a free module over a PID. Now let $\{x_1, x_2, \ldots, x_n\}$ be a basis for $F$, and let $\{y_1, y_2, \ldots, y_k\}$ be a basis $N$. Since we are working over a PID, we know $k \le n$. Then we can define an $R$-linear map $\phi:F\rightarrow N$ where $$\phi(x_i) = y_i$$ for $1 \le i \le k$, and $$\phi(x_j) = 0$$ for all $k \le j \le n$. Then $\phi$ is surjective. Then consider the short exact sequence $$0 \rightarrow \ker(\phi) \xrightarrow{\iota} F \xrightarrow{\phi} N \rightarrow 0,$$ where $\iota:\ker(\phi)\rightarrow F$ is the inclusion map. Then since $N$ is free and hence projective, the above exact sequence splits. Then $$F \cong \ker(\phi)\oplus N,$$ hence $N$ appears as a direct summand.

$\endgroup$
3
  • 2
    $\begingroup$ This shows $N$ is isomorphic to a direct summand of $F$, but the submodule $N$ itself may not be an (internal) direct summand of $F$. $\endgroup$
    – Eric Wofsey
    Feb 13, 2016 at 1:22
  • $\begingroup$ I never knew "direct summand" required it to be internal. Is this a common definition? For example, in Dummit and Foote, section 10.5, proposition 30, they also use the term "direct summand," but are not necessarily requiring it to be internal, as $P$ is not necessarily a subset of the free module $F$ they construct. But rather $P$ is isomorphic to a subset in $F$. $\endgroup$
    – Holbaron
    Feb 13, 2016 at 1:31
  • $\begingroup$ When you talk about a submodule of a module being a direct summand, this always means an internal direct summand. When you talk about some unrelated module, then obviously it can only be meant in an external sense, although even then it is common to say only that it is "isomorphic to a direct summand". $\endgroup$
    – Eric Wofsey
    Feb 13, 2016 at 1:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .