How do I prove that $R[x] \cong R[y]$? I understand that $R[n]$ represents the set of all polynomials in $n$ with coefficients from the commutative ring $R$, but I don't know how to even start this problem.

Any help would be great, thank you!


I will outline an sketch for how you might start thinking about this problem. First, recall that two sets are isomorphic if there exists an isomorphism between the two sets. Consider a polynomial in $R[x]$. How could you map this to $R[y]$? Then show that this mapping is well-defined, injective, surjective, and operation-preserving. Then, you have your explicit isomorphism, so you can say that $R[x]\cong R[y]$.

  • $\begingroup$ Is it just $f(x)=y$? $\endgroup$
    – James Done
    Mar 25 '19 at 4:40
  • $\begingroup$ That map is neither injective, nor surjective, nor operation-preserving. (Also, you do not generally denote mapping with an equal sign.) Try $f(x)\mapsto f(y)$. You can do the property verification on your own. $\endgroup$
    – user656966
    Mar 25 '19 at 6:57

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