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## The ideal generated by the variable is maximal iff the coefficient ring is a field

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.