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Show that a torsion-free divisible module over a commutative integral domain is injective. ($M$ is torsion-free if $rx=0$, $0\not=x\in M \implies r=0$.)

I tried to use Baer's criterion for injectivity to prove that $M$ is injective but I don't see how $R$ being a commutative integral domain gives me additional information. The first two conditions, namely $M$ being torsion-free and divisible tell me about the injectivity of the maps $\varphi_x(r)=rx$ and $\varphi_r(x) = rx$ for $r\in R-\{0\}$ and also the surjectivity of $\varphi_r$ but I don't know how to proceed from this point.

darij grinberg
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user375190
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3 Answers3

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Following Baer's criterion, let $I\subset R$ be an ideal and $f:I\to M$ be a homomorphism. We want to extend $f$ to a homomorphism $\bar{f}:R\to M$. To define a homomorphism $\bar{f}(r)$, you just have to pick an element $m\in M$ to be $\bar{f}(1)$ and then define $\bar{f}(r)=rm$ for each $r\in R$. For $\bar{f}$ to be an extension of $f$, you need to have $im=f(i)$ for each $i\in I$.

So you want to find $m\in M$ such that $im=f(i)$ for all $i\in I$. The condition that $M$ is divisible suggests a way to find such an $m$: just pick some nonzero $i\in I$, and then divisiblity of $I$ guarantees there exists $m\in M$ such $im=f(i)$. (This step requires $I\neq 0$; I'll let you figure out how to handle the case $I=0$ separately.)

Now of course, you only know that $im=f(i)$ for the particular $i\in I$ you chose. You still need to check that $jm=f(j)$ for all other $j\in I$. You might think it's not reasonable to expect this to be true--maybe we picked the wrong $m$. However, since $M$ is torsion-free, there is only one possible $m$ such that $im=f(i)$ (since $im=f(i)=im'$ implies $m=m'$). So if there is any $m$ that works, it must be the one we chose!

So see if you can prove that $jm=f(j)$ for all $j\in I$, using the fact that $M$ is torsion-free. The details are hidden below.

Note that $ijm=j(im)=jf(i)=f(ij)=if(j)$. Since $i\neq 0$ and $M$ is torsion-free, we can cancel $i$ and conclude that $jm=f(j)$.

Eric Wofsey
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  • Thank you. Where did we use the condition that $R$ is a commutative integral domain? Isn't that condition redundant? – user375190 Dec 29 '16 at 18:03
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    @user375190 The cancellation performed in the spoiler box may not be possible, depending on what definition of "torsion-free" you have for general rings. If you have an integral domain, this is not a problem. Perhaps a more general version of your problem is possible, with an appropriate definition. – rschwieb Dec 29 '16 at 19:16
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There is an extremely direct proof by contradiction. Let $E(M)$ be the injective envelope of $M$, and suppose $x\in E(M)\setminus M$.

  1. There exists an $r\in R$ such that $xr\in M\setminus\{0\}$ (since $E(M)$ is an essential extension.)

  2. There exists a $y\in M$ such that $yr=xr$ (since $M$ is divisible.)

  3. $x=y$ (since $M$ is torsion-free.)

  4. This is a contradiction since $x\notin M$. Therefore $M=E(M)$. QED

rschwieb
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I don't think Step 3 in @rschwieb's proof is legal. This is because the element $x$ is not known to be in $M$. So the cancellation as shown doesn't make sense, I think. However one can show that the injective hull of a torsion-free module is torsion-free and then repeat Step 3 in the injective hull instead of $M$.

Here is an alternate proof: any torsion free divisible module $M$ over a domain $A$ is naturally a module over the fraction field $K$ of $A$ by the universal property of localization. But all modules over a field are injective.

Now use the following result which follows from restriction-extension of scalars adjunction: if $A \to B$ is a flat ring map and $M$ is an injective $B$-module, then $M$ is an injective $A$-module.

Rankeya
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