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So basically I want to prove the following.

Assume $R$ is communtative, I want to prove $R[x_1,....,x_n] \cong R[x_{\pi(1)},.......,x_{\pi(n)}]$ where $\pi$ is a fixed permutation map.

So I want to consider a similar easier, problem. Prove $R[x,y] \cong R[y,x]$

Even then I'm stuck.

I cant seem to construct a bijective, homomorphic map between these 2 Rings.

Any help or insight regarding this problem is deeply appreciated.

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    What is the most general form of an element of $R[x,y]$? Now, if you think about it, the map is just a swap between $y$ and $x$, so what should the map take the general form to? – 2017-02-01
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    What exactly is your definition of $R[y,x]$? For some definitions, it may literally be the same ring as $R[x,y]$... – 2017-02-01

2 Answers 2

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Let be $R[x,y]=(R[x])(y)=R_1[y]$ ,then $f(x,y)\in R[x,y]$ can be written by form $f(x,y)=\sum_{j=0} ^{n} a_j(x)y^j; a_j(x)=\sum_{i=0} ^{m}a_{ij}x^i; a_{ij} \in R $, so $f(x,y)=\sum_{i=0} ^{m}\sum_{j=0} ^{n} a_{ij}x^iy^j $. In similar way we have $f(y,x)=\sum_{i=0} ^{m}\sum_{j=0} ^{n} b_{ij}y^jx^i=\sum_{i=0} ^{m}\sum_{j=0} ^{n} b_{ij}x^iy^j$ and we prove that the written is unique $\sum_{i=0} ^{m}\sum_{j=0} ^{n} a_{ij}x^iy^j=\sum_{i=0} ^{m}\sum_{j=0} ^{n} b_{ij}x^iy^j \Leftrightarrow a_{ij}=b_{ij} $. Therefore $R[x,y]=R[y,x]$

In general $R[x_1,..,x_n]=(R[x_1,..,x_{n-1})[x_n]; n>1$ in similar way we have $R[x_1,..,x_n]=R[x_{\sigma(1)},..,x_{\sigma(n)}]$

Second way:

Hint: we know in $R[x]$ that $ax=xa \forall a\in R$, then $x\in C(R[x])$

by using then hint in general case ($R[x_1,..,x_n]$) we note that $x_1,..,x_n \in C(R[x_1,..,x_n])$ then the arrangement of elements $x_i $ to $R$ has no effect

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In the case of two variables convince yourself that $R[x,y] \cong R[a,b]$, by simply relabeling the indeterminates. For the same reason we have that $R[y,x] \cong R[a,b]$.