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Let $R$ be a commutative ring. Before I proved that every submodule of a free $R$-module is free over a P.I.D.

Now I'm trying to prove the reciprocal, if every submodule of a free $R$-module is free, is $R$ a PID?

I couldn't find my second question here, only the first one.

Another question regarding this if it's true is how much can we "stretch" the conditions, this is, could be concluded that $R$ is a PID knowing that some submodules of a free $R$-module verify some condition? "Some condition" may sounds vague, but I couldn't think of any that will do.

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3 Answers 3

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Yes, it is.

If $a\in R$, $a\ne 0$, then $aR$ is free (as a submodule of the free $R$-module $R$), so $a$ is a non-zero divisor. In order to prove this let $ax$ be an element of an $R$-basis of $aR$. If $ba=0$ for some $b\in R$ then $b(ax)=0$ and therefore $b=0$. (For a characterization of free ideals see here.) This shows that $R$ is necessarily an integral domain.
Now let $I$ be a non-zero ideal of $R$. Since $I$ is free (as a submodule of the free $R$-module $R$) it must have a basis with a single element (as you already know), so $I$ is principal.

(As you may noticed, I've only used the property for the free $R$-module $R$.)

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  • $\begingroup$ is $\{a\}$ a basis for the free module $aR$? Why when considering an arbitrary basis element does it have the form $ax$? Wouldn't an arbitrary basis element just be an arbitrary element of $R$? $\endgroup$
    – user637978
    Commented Jul 23, 2019 at 16:27
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    $\begingroup$ So the proof is just that you take a finite basis for an arbitray ideal, and show that it must have a single element, because any 2 elements of a ring are linearly dependent over that ring. Thanks man!! $\endgroup$
    – user637978
    Commented Jul 23, 2019 at 16:30
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I prove that $R$ is an integral domain. Let $a\in R\setminus\{0\}$. The exercise is equivalent to prove that the multiplication-by-$a$ morphism $m_a:R\rightarrow R$, $x\mapsto ax$ is injective.

By assumption $aR$ is free since is a submodule of the free $R$-module $R$. In particular there exists some isomorphism of $R$-modules $\phi:\oplus_{I}R \overset{\simeq}{\rightarrow} aR$, for some nonempty set $I$.

As Bahador explained this implies that $I$ consists of one element.

Thus there is an isomorphism $\phi: R \rightarrow aR\subset R$ of (cyclic) $R$-modules. Let $b=\phi(1)$ then $\phi=m_b$ and $bR = Im\, (m_b)= Im\,(\phi) = aR$.

In particular $b$ is not zero-divisor since $m_b$ is injective. Generators of a principal ideal differ by a unit if one of them is non-zero divisor so that there is some unit $u$ of $R$ such that $a=bu$.

In particular $m_a= m_u \circ m_b = m_u\circ \phi$ is injective.

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We know that $R$ is a free $R$-module with basis $X=\{1\}$. Now let $I$ be an ideal of $R$.(the submodules of a ring are exactly its ideals.) By hypothesis $I$ has a basis $X$. We claim that $X$ has exactly one element. If $a$ and $b$ are two distinct elements of $X$, then $ab-ba=0$, since $R$ is commutative, a contradiction to the fact that $X$ is linearly independent. Therefore $X$ has only one nonzero element, i.e. $I$ is a principal ideal.

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