Torsion-free precover

Question

Consider R is an integral domain and K is its filed of fractions. Definition:A linear map $f:T\rightarrow M$ where T is torsion-free module is called a torsion-free precover of M if for every torsion-free module G and linear map $g:G\rightarrow M$,there is a linear map $h:G\rightarrow T$ such that $fh=g$.

the mistake is in proof of Lemma 4.1.4. G may not be direct sum of K,since we can consider $\prod K$. So how to prove Every module has a torsion-free precover?

Thank you in advance!

score 1 · Accepted Answer · answered Sep 10 '17 at 03:50

1

There is no error. A torsion-free divisible $R$-module $G$ is the same thing as a $K$-module, also known as a $K$-vector space. The fact that such a $G$ is a direct sum of modules isomorphic to $K$ is just the statement that every $K$-vector space has a basis.

answered Sep 10 '17 at 03:50

Eric Wofsey

330,363

But $\prod K$ is not direct sum of copies of K$ – Jian Sep 10 '17 at 03:53
Sure it is. It's a vector space, so it has a basis. – Eric Wofsey Sep 10 '17 at 03:55
so how to explain $\prod K$is not free K module – Jian Sep 10 '17 at 03:56
It is a free $K$-module. – Eric Wofsey Sep 10 '17 at 03:57
I don't know what your statement of "Baer's theorem" is, but it only applies to certain rings, and fields are not included among them. – Eric Wofsey Sep 10 '17 at 04:10
https://math.stackexchange.com/questions/320444/why-isnt-an-infinite-direct-product-of-copies-of-bbb-z-a-free-module – Jian Sep 10 '17 at 04:30
yeah ，I think you are right – Jian Sep 10 '17 at 04:32
thank you，I understand – Jian Sep 10 '17 at 04:34

Torsion-free precover

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