This has to do with Remark 8.14 (b) on page 75 of Analysis I by Amann and Escher.
I am not defining all the notation used because I'm trying to keep the length of this post manageable.
Excerpt from text:
This excerpt provides some additional context as well as including Remark 8.14 (b).
Discussion:
I am a bit confused regarding the proof of Remark 8.14 (b). It seems that the function defined in Remark 8.14 (b) is automatically a ring homomorphism because of how it's defined. (I'm referring to the fact that $(p + q)(x) = p(x) + q(x)$ and similarly for the multiplication.)
What's especially confusing is justifying why the homomorphism sends $1_{R[X]}$ to $1_{R^R}$. It is my understanding (which seems to be corroborated on page 72) that $1_{R[X]}$ is the formal power series with $1_R$ in the first position and $0_R$ in all other positions. This may be written as $X^0$. The text says that $X^0$ is unity in $R [[ X ]]$ (the formal power series ring) so as far as I know $X^0$ is also unity in $R[X]$ (the polynomial ring).
If $X^0$ is unity in $R[X]$, then its "summation form" would be $p = \sum_k p_k X^k = X^0$, right? So the corresponding polynomial function $\underline p \in R^R$ would be $p(x) = x^0 = 1_R$, right? If that is true, then the function $R[X] \to R^R, p \mapsto \underline p$ takes $1_{R[X]}$ to the constant function in $R^R$ that sends everything to $1_R$.
That's not the answer I'm supposed to get. I'm supposed to show that $1_{R[X]}$ gets sent to $\text{id}_R \in R^R$. If you look at Remark 8.14 (a) (see excerpt) then you see that $X$ gets sent to $\text{id}_R$, not $X^0$. That makes sense to me, but the problem is that $X$ is not unity in $R[X]$; $X^0$ is.
Presumably I've misunderstood something here. I appreciate any clarification.
I think Qi Zhu and Brian Moehring have helped me realize that I'm getting bamboozled by the sheer number of rings that are floating around. There are at least the following rings:
Focusing on the second ring, it is the ring that is arises due to the operations on $R^R$ induced from the operations on $R$. This is discussed on page 27 of the text. In particular, given $f, g \in R^R$, the induced multiplicative operation is given by $(f \cdot g)(x) := f(x) \cdot g(x)$ So the new function $f \cdot g \in R^R$ is the rule that sends $x \mapsto f(x) \cdot g(x) \in R$, for all $x \in R$.
It is therefore clear that $1_{R^R}$ should be the constant function that sends everything to $1_R$. In my question I correctly deduced that the homomorphism described sends $1$ to $1$, but I was just confused as to what the latter $1$ really meant. I was somehow imagining that $\text{id}_R$ is the identity element of the ring $R^R$, but of course function composition has nothing to do with how the ring $R$ is defined.