Renaming the variables in a polynomial ring does not change its isomorphism type

Let R be a commutative ring and x_1, x_2, … x_n be symbols. Prove that R[x_1,\ldots,x_n] \cong R[x_{\sigma(1)}, \ldots, x_{\sigma(n)}], where \sigma is any permutation in S_n.

We will approach this problem from a slightly different perspective; we will see that polynomial rings are really monoid rings. Given a ring R and a monoid M, the monoid ring R[M] is defined just like the group ring R[G]. (Recall that a monoid is a set with (1) an associative binary operator and (2) an identity element.)

Lemma 1: Let R be a commutative ring. Then the polynomial ring R[x] is isomorphic to the monoid ring R[\mathbb{N}], where \mathbb{N} is considered a monoid under addition. Proof: Define \varphi : R[x] \rightarrow R[\mathbb{N}] as follows: \varphi(\sum_{k=0}^n r_kx^k) = \sum_{k=0}^n r_k \cdot k. It is easy to see that this mapping is an isomorphism. \square

Lemma 2: Let R be a commutative ring and let M,N be monoids. Then R[M \times N] \cong (R[M])[N]. Proof: Define \varphi : R[M \times N] \rightarrow (R[M])[N] by \varphi(\sum_{i,j} r_{i,j}(m_i,n_i)) = \sum_{i,j} (r_{i,j}m_i)n_j, where i and j index M and N, respectively. It is clear that this mapping is surjective. It is also clear that its kernel is trivial, so that \varphi is injective. That \varphi is a ring homomorphism is also clear. \square

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 R[\mathbb{N}^k] for the same k. 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.

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