Show that the ring isomorphism $\phi :R[x]\to S[x]$ is well-defined. Could someone please verify whether my answer is too simple or needs more detail or something is wrong?

Let $\phi_{1}:R\to S$ be a ring isomorphism. Show that the ring isomorphism $\phi_{2} :R[x]\to S[x]$ defined by $\phi_{2}(a_{0}+a_{1}x+\cdots +a_{n}x^{n})=\phi_{1}(a_{0})+\phi_{1}(a_{1})x+\cdots + \phi_{1}(a_{n})x^{n}$ is well-defined.

Let $f(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n},g(x)=b_{0}+b_{1}x+\cdots +b_{m}x^{m}\in R[x]$. If $f(x)=g(x)$, by polynomial equality, $a_{0}=b_{0}$, ..., $a_{n}=b_{m}$. Then $n = m$ (?). Then $\phi_{2}(f)=\phi_{1}(a_{0})+\phi_{1}(a_{1})x+\cdots + \phi_{1}(a_{n})x^{n}=\phi_{1}(b_{0})+\phi_{1}(b_{1})x+\cdots + \phi_{1}(b_{m})x^{m}=\phi_{2}(g)$.
 A: This is a pretty weird problem.  When you are asked to show something is "well-defined", that means it is defined by using representatives of some equivalence classes, and you want to show the resulting definition does not depend on the choice of representatives.  But here, there do not appear to be any such equivalence classes.  At worst, you might say that the expression $$a_{0}+a_{1}x+\cdots +a_{n}x^{n}$$ for a polynomial is not unique, since you could add extra terms to the end with coefficients $0$ (so, in the context of what you wrote, the problem is that you might have $m\neq n$).  So you would need to check that adding extra terms with coefficient $0$ to the end doesn't change the definition of $\phi_2$.  (And this is completely trivial to check, since $\phi_1(0)=0$.)  But actually, the usual approach would not be to make this definition with this ambiguity and then ask you to prove that it's well-defined, but instead to stipulate that $a_n\neq 0$ (unless $n=0$), to make the expression above unique.  In that case, there is literally nothing to check to prove that $\phi_2$ is well-defined; you don't even need to do the work that you did.
My guess would be that whoever wrote the problem actually intended to ask you to prove that this map is an isomorphism, not that that it is well-defined.
