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December 25th, 2011, 01:23 PM   #1
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Free modules .. does this solution work?

Hi all, could someone take a look at my claim and solution below and see if it makes sense? Thanks!

Claim: Let R be a commutative ring with , and let be a surjective homomorphism of free -modules. Then .

Proof:
Take to be any linearly independent elements of . Then there exists in such that . But since are linearly independent, we have that if , then . Also, there exists some such that . So, as is a ring homomorphism, we have the following:

but in order for this to be true, again for , since are linearly independent. But this means that must be linearly independent in , and so .
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December 30th, 2011, 07:21 AM   #2
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Re: Free modules .. does this solution work?

Your approach works, provided, that your last implication is known: "If there are n linear independent elements in then ." But unless this is a theorem in a book or your course, the last assertion seems similarly hard to prove as the original problem.

An alternative approach would be: Recall that every commutative ring with has at least one maximal ideal, say M. Then check for all positive integers i. Then you get a surjective map and these are vector spaces of dimiesion m and n respectively over the field . Thus the problem is reduced to the case where R is a field, i.e., the case well-known from linear algebra.
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