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This question is motivated by my most spectacular answer on MO (:

Let $A$ be a module over $\mathbb Z$. $A$ is said to be torsion-free if $na=0$ implies $n=0$ or $a=0$ for any $n\in \mathbb Z, a\in A$. $A$ is torsionless if the map $\phi: A \to A^{**}$ is injective (here ${}^*$ means $\text{Hom}_{\mathbb Z}(A,\mathbb Z)$).

If $A$ is finite, then torsion-free and torsionless are equivalent. In general, it is not hard to see that being torsionless implies torsion-free. On the other hand, $A=\mathbb Q$ is torsion-free but not torsionless since $A^*=0$. But the question and answers quoted above (which shows that for $A=\mathbb Z[x]$, $\phi$ is an isomorphism) raised the following:

Question: If $A$ is a finite $\mathbb Z[x_1,...,x_d]$-module, are being torsion-free and torsionless equivalent?

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2 Answers 2

up vote 5 down vote accepted

If by finite you mean finitely presented, then the answer is no. For instance, Let $A = \mathbb{Z}[x]/(2x-1) = \mathbb{Z}[\frac12]$. Then, like $\mathbb{Q}$, $\text{Hom}(A,\mathbb{Z}) = 0$.

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I forgot to say: Thanks! –  Hailong Dao Jan 18 '10 at 1:24
    
Sure, you're welcome. –  Greg Kuperberg Jan 18 '10 at 2:46

I'm sorry, if you take $A = \mathbb{Z}[x]/(2x-1) = \mathbb{Z}[\frac{1}{2}]$ why is ${\rm Hom}_{\mathbb{Z}}(A,\mathbb{Z})=0$? Isn't the map $\frac{1}{2} \mapsto 1$ (and of course $1 \mapsto 2$) an element of such a dual?

Sorry, I can see that the post is pretty old but it's the first time I visit this website.

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2  
The notation refers to the ring of rationals with power-of-two denominators, not the group $\frac12 \mathbb{Z}$. The map you describe is not defined on elements like $x^2$ (which becomes $\frac14$ under the identification). –  S. Carnahan Feb 7 '11 at 3:46

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