##R[x,y]## and ##R[y,x]## are Ring Isomorphic

[Bashyboy]

Homework Statement

Let $R$ be a commutative ring,. Show that $R[x,y]$ is isomorphic to $R[y,x]$.

Homework Equations

The Attempt at a Solution

Let $f : R[x,y] \to R[y,x]$ be defined by $\sum_{i,j=1}^{n,m} a_{ij} x^i y^j \mapsto \sum_{i,j=1}^{n,m} a_{ij} y^i x^j$. Verifying that $f$ is additive is rather trivial, however I am having a little trouble verifying that $f$ is multiplicative. Given $\sum_{i,j=1}^{n,m} a_{ij} x^i y^j$ and $\sum_{i,j=1}^{n,m} b_{ij} x^i y^j$, what is the general form of the product? Multiplying double sums hasn't been introduced yet in my book, as far as I can tell.

 

[fresh_42]

It can probably be done by the formula, which you can easily find out by a simple multiplication. It would be boring in any case. But given additivity, why not use if for polynomials $ax^ny^m$?

 

[Bashyboy]

Okay. Let me see if I follow you. Since $\sum_{i,j=1}^{n,m} a_{ij} x^i y^j$ can be expanded as
$$(a_{11} xy + ... + a_{1m} xy^m) + (a_{21} x^2 y + ... + a_{2m} x^2 y^m) + ... + (a_{n1} x^n y + ... + a_{nm} x^n y^m),$$

we see that the product of it with $\sum_{i,j=1}^{n,m} b_{ij} x^i y^j$ is

$$\sum_{i,j=1}^{n,m} a_{ij} x^i y^j \cdot \sum_{i,j=1}^{n,m} b_{ij} x^i y^j$$

$$= \bigg[(a_{11} xy + ... + a_{1m} xy^m) + (a_{21} x^2 y + ... + a_{2m} x^2 y^m) + ... + (a_{n1} x^n y + ... + a_{nm} x^n y^m) \bigg] \cdot \sum_{i,j=1}^{n,m} b_{ij} x^i y^j$$

$$= (a_{11} xy \cdot \sum_{i,j=1}^{n,m} b_{ij} x^i y^j+ ... + a_{1m} xy^m \cdot \sum_{i,j=1}^{n,m} b_{ij} x^i y^j) + (a_{21} x^2 y \cdot \sum_{i,j=1}^{n,m} b_{ij} x^i y^j+ ... + a_{2m} x^2 y^m \cdot \sum_{i,j=1}^{n,m} b_{ij} x^i y^j) + ... + (a_{n1} x^n y \cdot \sum_{i,j=1}^{n,m} b_{ij} x^i y^j + ... + a_{nm} x^n y^m \cdot \sum_{i,j=1}^{n,m} b_{ij} x^i y^j )$$

From there I can break down each $a_{pq} x^p y^q \cdot \sum_{i,j=1}^{n,m} b_{ij} x^i y^j$ down in a similar manner, which is extraordinarily tedious, and then it suffices to show that ##f(a x^p y^q \cdot b x^n y^m) = f(a x^p y^q) f(b x^n y^m)? Does that sound right?

 

[fresh_42]

I'm not sure whether your indices are all correct, especially why you omit the constant terms. However, it isn't needed. We can simply write
$$
\left( \sum_{i,j} a_{ij} x^iy^j \right) \left( \sum_{k,l} b_{kl}x^ky^l \right) = \sum_{p,q} \left( \sum_{i+k=p}\sum_{j+l=q}a_{ij}b_{k,l}\right) x^p y^q
$$
This form has the advantage that we don't need to bother constant terms or degrees. The only question is, what does this mean, if we swap $x$ and $y$ on the left and on the right? Here you need additivity and a result for products $c_{pq}x^py^q$.