Let be an integral domain, and let be its field of fractions.
- Considering as an -module in the usual way, prove that .
- Let be a submodule. Prove that is torsion for all .
- Exhibit an integral domain and an -submodule such that for all .
Let be a simple tensor. Note that . In particular, we have in .
Suppose be nonzero; then the are nonzero, and certainly the are nonzero. Now in , and evidently . So every element of is torsion, and thus is torsion as an -module.
Now consider , and let . We claim that if , then there exist such that . To see this, choose some and consider . Since is a domain, divides the right hand side of this equation; say . Note that every term on the left hand side is divisible by some other than . Hence every term in is divisible by some other than . In particular, , and .
Now consider the elements of as “column vectors”- that is, write as . Note that this does not give a unique representation of the elements of . However, if two column vectors and represent the same element in , then there exists a third column vector such that , and moreover the entries of are in .
Now write the elements of as (square) matrices. Let us consider the determinant of such a matrix , as an element of , reduced mod . This is an alternating bilinear map on the set of matrices over ; we claim that this is also well-defined up to our nonunique identification of matrices over with elements of . To that end, suppose matrices and represent the same element of ; then we have a matrix such that . Consider the determinant of both sides mod . Using the combinatorial formula for computing determinants, we have . Note that each is divisible by some , and so goes to zero in the quotient . So in fact ; thus the map is a well-defined alternating bilinear map. Since is nonzero, this map is nontrivial. Thus for all .