Proof:

We know that, if is a finite abelian group of order then considered as a -module, is annihilated by , the -primary component of is isomorphic to the direct product of its Sylow subgroups.

Here, is a finite abelian group of order . As a -module, is the direct sum of its Sylow -subgroups, say where is the finite set of primes dividing . Tensor product commute with finite direct sums, so as -homomorphism.

We show if , then . Since and are relatively prime (in this case) in the PID , then there are s.t. . Let be a simple tensor in . Then, . Hence, .

Now we show . Here, every simple tensor in can be written as . Then any tensor, . Now define by . Here, if divides then . Hence is well-defined. It is easy to see that is a -balanced map, so induces a group homomorphism and is the necessary natural inclusion and .

Let then .

It is easy to see that is injective. Also, is surjective, since for each , . Thus is an isomorphism, i.e. .

Hence, .

Proof:

We know that for any ring , is a field if and only if is a maximal ideal of . To use it, we show that is maximal if and only if is irreducible.

Let be maximal. We show is irreducible. If possible, let where the degree of and is at least one, for some . Here, and so the ideals and are proper and also contains , which is a contradiction since is maximal. So the polynomial is irreducible.

Let be irreducible. We show is maximal. If possible, let for an ideal of . Since is a field so is a PID, so every ideal of is principal. Then for a polynomial is principal. Then for a polynomial , .

Since , so for there is some s.t. . Since is irreducible, so either or is of degree zero. If then is a unit in . Here, i.e. contains a unit, so . If then , for , constant. Then i.e. i.e. . Hence, . Which suggests that is maximal.

Now, let is a field, so the ideal is maximal. We have that if is maximal then is irreducible.

Let is irreducible then is maximal so is a field. ]]>