# Proving that $\varphi :R[X]\rightarrow F$ is a ring homomorphism

Let $R$ be a commutative ring and let $F=F(R,R)$ be the set of functions $f:R\rightarrow R$. Let $I$ denote the identity function $I(r)=r$ in $F$. Prove that the map $\varphi:R[X]\rightarrow F$ given by $$\varphi (a_nX^n+\cdots+a_1X+a_0)=a_nI^n+\cdots+a_1I+a_0$$ is a ring homomorphism.

I easily showed that $\varphi$ is a group homomorphism from $(R[X],+)$ to $(F,+)$. Now I'm stuck on how to show $\varphi(xy)=\varphi(x)\varphi(y)$ for all $x,y\in R[X]$. How should I tackle this proof?

I want to show that $$\varphi[(a_nX^n+\cdots+a_1X+a_0)(b_mX^m+\cdots+b_1X+b_0)]=\varphi(a_nX^n+\cdots+a_1X+a_0)\varphi(b_mX^m+\cdots+b_1X+b_0)$$ but I don't know whether I should combine the terms through multiplication because it's hard to determine what happens inside the dots $\cdots$. Should I expand or is there a better way of doing this?

Hint: The easiest way to do this is using the universal property of a polynomial ring, but the question would be trivial if you were allowed to do that so I'm assuming you haven't seen that yet. In which case the formula you want to use is: $$(a_nx^n + \cdots + a_1x + a_0)(b_mx^m + \cdots + b_1x + b_0) = \sum_{k = 0}^{n + m}\left[\sum_{i = 0}^ka_ib_{k - i}\right]x^k$$ where we set $a_i = 0$ if $i > n$ and $b_i = 0$ if $i > m$.
Note that this is just what you get when you expand the product and collect powers of $x$, so this formula holds in any ring (not just a polynomial ring) and in place of $x$ could be any element of the ring. In particular, this formula holds in $F$ when you substitute $I$ in for $x$.
• Could I have a hint on how I can show that $\sum_{k = 0}^{n + m}\left[\sum_{i = 0}^ka_ib_{k - i}\right]I^k=\left(\sum_{i=0}^n a_iI^i\right)\left(\sum_{i=0}^m b_iI^i\right)$? – Patrick Shambayati Apr 6 '15 at 7:22
• Just distribute and collect terms. Every term will be of the form $a_iI^i\cdot b_jI^j = a_ib_jI^{i + j}$ for some $i$ and $j$. So if you're collecting all the $I^k$ terms look for all $i, j$ such that $i + j = k$. – Jim Apr 6 '15 at 7:38