why we need "torsion-free"

Asked Jun 25 '20 at 09:46

Active Jun 25 '20 at 09:46

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This is from P 134 of Rotman's Homological Algebra book.

If R is a PID, then every torsion-free R-module is flat.

Proof. If R is a PID, then every finitely generated R-module M is a direct sum of cyclic modules. If M is torsion-free, then it is a direct sum of copies of R.

I'm not sure why we need M to be torsion-free to have each cyclic module isomorphic to R. Any help would be appreciated! Thank you

asked Jun 25 '20 at 09:46

scsnm

1,283

$$a\left(\frac{R}{aR}\right)=0$$ – Brian Moehring Jun 25 '20 at 09:55
Because an integral domain has no torsion. – Bernard Jun 25 '20 at 10:23
Hi Brian and Bernard! I am still not sure ... So suppose M is not torsion-free, then there is a nonzero r and nonzero m s.t. rm=0. Could one get a contradiction from this? Sorry, I am new to this. – scsnm Jun 25 '20 at 10:26
hi Brian! I got what the expression of yours means. But I still can't see the connection with torsion-free. Thx! – scsnm Jun 25 '20 at 10:31

why we need "torsion-free"

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