Let be a commutative ring and , , … be symbols. Prove that , where is any permutation in .
We will approach this problem from a slightly different perspective; we will see that polynomial rings are really monoid rings. Given a ring and a monoid , the monoid ring is defined just like the group ring . (Recall that a monoid is a set with (1) an associative binary operator and (2) an identity element.)
Lemma 1: Let be a commutative ring. Then the polynomial ring is isomorphic to the monoid ring , where is considered a monoid under addition. Proof: Define as follows: . It is easy to see that this mapping is an isomorphism.
Lemma 2: Let be a commutative ring and let be monoids. Then . Proof: Define by , where and index and , respectively. It is clear that this mapping is surjective. It is also clear that its kernel is trivial, so that is injective. That is a ring homomorphism is also clear.
Now it is clear that renaming the variables in a polynomial ring (with finitely many variables) does not change its isomorphism type, as in both cases the ring is isomorphic to for the same . We could extend this to infinitely many variables by defining the direct sum of monoids in analogous fashion to direct sums of groups- that is, the subset of a direct products in which each element has only finitely many nonidentity entries.