TITLE: Maximal ideal of Cauchy nullsequences
QUESTION [1 upvotes]: Let $(K, \left|\phantom{x}\right|)$ be a valued field. Let $\mathscr{C}$ be the commutative ring with unity given by Cauchy sequences of elements in $K$, with $+$ and $\cdot$ defined term by term. Finally, let $\mathscr{M}$ be the subset of $\mathscr{C}$ given by Cauchy sequences converging to $0$. I've proved that $\mathscr{M}$ is an ideal of $\mathscr{C}$. Now I want to prove that is maximal. To do this, I think it suffices to show that $\mathscr{C}-\mathscr{M}\subseteq\mathscr{C}^{\star}$, i.e. that every $\{a_n\}$ in $\mathscr{C}$, not in $\mathscr{M}$, is invertible. I've been able, given such a sequence $\{a_n\}$, to find a sequence $\{b_n\}$ in $\mathscr{C}$, such that $\{a_n\}\cdot\{b_n\}=\{1\}+\zeta$, where $\zeta$ is a sequence in $\mathscr{M }$. How can I conclude, from this, that $\{a_n\}$ is invertible?

REPLY [2 votes]: Let $\zeta$ be a sequence in $\mathscr{C}$. Denote by $\bar{\zeta}$ the element of $K$ to which $\zeta$ converges.
The homomorphism $$ \varphi \colon \mathscr{C} \to K $$ that sends $\zeta$ to $\bar{\zeta}$ is surjective. It is well known that the kernel of a surjective homomorphism is a maximal ideal. The kernel of $\varphi$ is $\mathscr{M}$, hence $\mathscr{M}$ is a maximal ideal in $\mathscr{C}$.
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