Suppose we have a field $F$ with norm function $|| \cdot ||.$

Let $\mathcal{F}$ be the set of Cauchy sequences in $F$ (with respect to the given norm), made into a ring in the usual way by addition and multiplication of corresponding terms in the sequence.

That is, we define $$\left\{\alpha_n\right\}_{n\ge 0} + \left\{\beta_n\right\}_{n\ge 0} = \left\{\alpha_n+\beta_n\right\}_{n\ge 0}$$ and $$\left\{\alpha_n\right\}_{n\ge 0} \cdot \left\{\beta_n\right\}_{n\ge 0} = \left\{\alpha_n\cdot\beta_n\right\}_{n\ge 0}$$ for any $\left\{\alpha_n\right\}_{n\ge 0}, \left\{\beta_n\right\}_{n\ge 0} \in \mathcal{F}.$

Finally, take $I$ to be the set of sequences $\left\{\alpha_n\right\}_{n\ge 0} \in \mathcal{F}$ satisfying $$\lim_{n\to \infty} ||\alpha_n|| = 0.$$

One can show that $I$ is an ideal.

Is there a direct way to prove that $I$ is a maximal ideal?

By direct method, I mean a proof that does not appeal to the fact that $\mathcal{F}/I$ is a field.


Let $e = (1,1,1,...)$.

Then $e$ is the multiplicative identity of $\mathcal{F}$.

Suppose $x \in \mathcal{F}$, $x \notin I$.

To show that $I$ is a maximal ideal of $\mathcal{F}$, it suffices to show $(I,x) = (e)$.

Since $x \notin I$, at most finitely many terms of $x$ are zero.

Thus, suppose $x_n \ne 0$ for $n > N$.

Let $f$ be the element of $\mathcal{F}$ such that $f_k = 1$ for $0 \le k \le N$ and $f_k = 0$ for $k > N$. Clearly $f \in I$.

Let $y = x - xf + f$.

Then $y \in (I,x)$, $\;y \notin I$.

By choice of $f$, all terms of $y$ are nonzero.

Let $1/y$ denote the sequence $(1/y_0,1/y_1,1/y_2,...)$.

Since $y$ is a Cauchy sequence with all terms nonzero, and $y \notin I$, the terms of $y$ are bounded away from zero, hence the products $y_my_n$ are also bounded away from zero.


\begin{align*} &\bullet\;\; \left\Vert{1/y_m - 1/y_n}\right\Vert = \left\Vert\frac{y_n - y_m}{y_my_n}\right\Vert\\[6pt] &\bullet\;\; \text{The denominators $y_my_n$ are bounded away from zero}\\[6pt] &\bullet\;\; y \text{ is a Cauchy sequence}\\ \end{align*}

it follows that $1/y$ is a Cauchy sequence, so $1/y \in \mathcal{F}$.

Since $y \in (I,x)$ and $1/y \in \mathcal{F}$, it follows that $y(1/y) \in (I,x)$.

Thus $e \in (I,x)$, so $(I,x) = (e)$.

Therefore $I$ is a maximal ideal of $\mathcal{F}$, as was to be shown.


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