Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.18
Solution:
We first prove a lemma.
Lemma: The map $\varphi : R[[x]] \rightarrow R$ given by $\sum r_ix^i \mapsto r_0$ is a surjective ring homomorphism and $\mathsf{ker}\ \varphi = (x)$.
Proof: $\varphi$ is clearly a surjective homomorphism. Now suppose $\sum r_ix^i \in \mathsf{ker}\ \varphi$. Then $r_0 = 0$, and $$\sum r_ix^i = x \sum r_{i+1}x^i \in (x).$$ If $\alpha = x\sum a_ix^i$, then $\varphi(\alpha) = 0$, so that $\alpha \in \mathsf{ker}\ \varphi$. $\square$
By the First Isomorphism Theorem for rings, $R[[x]]/(x) \cong R$. The problem at hand then follows from Propositions 12 and 13 in the text.
