I believe that the statement of exercise 2.22 in Rotman's Introduction to Homological Algebra is wrong. It states that if $R$ is an integral domain, $M$ is an $R$-module, and $\textrm{Hom}_{R}(M,R/I) = 0$ for all non-zero ideals $I$ of $R$, then $\textrm{Hom}_R(M,R) = 0$.

If $R$ is a field, then the only non-zero ideal is $(1)$ and $\textrm{Hom}_{R}(M,R/(1))$ is trivially zero, but $\textrm{Hom}_R(M,R)$ need not be.

Am I right that the exercise is mistaken? I did not find a correction in the errata.

Edit: Thank you metalspringpro for your answer. The exercise is correct when we exclude the possibility that $R$ is a field. To see this one can use the following statement: If the intersection of all non-zero ideals of an integral domain R is non-zero, show that R is a field.


1 Answer 1


Yes, but it's a rather small mistake; the case you point out is the only one where there is an issue. To fix it, change the claim to

If $R$ is an integral domain and not a field, $M$ is an $R$-module, and $\operatorname{Hom}_R(M,R/I)=0$ for all non-zero ideals of $R$, then $\operatorname{Hom}_R(M,R)=0$.

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    $\begingroup$ Maybe it's just me being stupid, but I don't see how this resolves the issue. The property $\operatorname{Hom}_R(M,R/I)=0$ is always true for $I=(1)$, so what difference would it make to remove $I=(1)$ from consideration? As you wrote, when $R$ is a field, the hypothesis “$\operatorname{Hom}_R(M,R/I)=0$ for all nonzero proper ideals $I$” is vacuously true (since there are no such ideals), but it's still true, so then again a true hypothesis leads to a false conclusion, just as the original true hypothesis “$\operatorname{Hom}_R(M,R/I)=0$ for all nonzero ideals $I$” did. $\endgroup$ Sep 13, 2021 at 5:26
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    $\begingroup$ @HansLundmark That's a good point, it's now fixed. The point is that you just have to exclude the case where R is a field. I tried to do it in a cute way, but you are correct this doesn't quite work. $\endgroup$ Sep 13, 2021 at 5:38

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