Let $R$ be an arbitrary ring (not necessarily commutative or have a $1$), and $S=R[x]$. Then fix an $r\in R$, and define the substitution map $\phi=\phi_r: S \to R$, mapping $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ to $a_0+a_1r+a_2r^2+\cdots+a_nr^n$.
Then I need to show that if we take two arbitrary elements $f,g\in R$, and if $R$ has a multiplicative identity, $$\phi(f\cdot g)=\phi(f)\cdot\phi(g) \iff r \in Z(R).$$
The right to left implication is simple fairly simple. However I'm having trouble with the left to right implication. This is what I've tried,
Define $f(x)=\sum\limits_{k=0}^{\alpha}a_kx^k$ and $g(x)=\sum\limits_{k=0}^{\beta}b_kx^k$. Then we follow through with evaluating the LHS and RHS of the equality.
$$\phi(f\cdot g)=\sum_{n=0}^{\alpha+\beta}\sum_{k=0}^{n}a_kb_{n-k}r^n$$
$$\phi(f)\cdot\phi(g)=\sum_{k=0}^{\alpha}\sum_{n=0}^{\beta}a_kr^kb_nr^n=\sum_{k=0}^{\alpha}\sum_{n=k}^{\beta+k}a_kr^kb_{n-k}r^{n-k}=\sum_{n=0}^{\alpha+\beta}\sum_{k=0}^{n}a_kr^kb_{n-k}r^{n-k}$$
On the interchange of summation I'm not quite sure whether it's correct or not, however, assuming it is correct. By assumption these two sums are equal, so their inner parts must be equal. Therefore,
$$a_kb_{n-k}r^n=a_kr^kb_{n-k}r^{n-k}$$
From here it would be easy to prove if I could cancel things out from both sides, but the assumptions didn't include that $R$ was a division ring, so it may or may not have multiplicative inverses. As such, from this point on I'm stuck on what to do next or whether this is the right approach in the first place. Additionally, nowhere did I have to assume that $R$ has a multiplicative identity, even in the first implication, so I'm confused as to where that comes into play.
Thanks in advance.
The full question is displayed below, if this is more understandable, it is part (c).
Best Answer
We have to show $r$ commutes with all the elements of $R$. So let $a\in R$. You can take $f(x)=x$ and $g(x)=a$. Note that by definition of multiplication in the polynomial ring we have $fg=xa=ax$. Hence $\phi(fg)=ar$. On the other hand we also have $\phi(f)\phi(g)=ra$. Since by assumption $\phi(fg)=\phi(f)\phi(g)$ it follows that $ar=ra$.