# Prove that $R[x] \cong R[y]$ [closed]

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 $$R[x]\cong R[y]$$.
• Is it just $f(x)=y$? Mar 25 '19 at 4:40
• 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.