Find all maximal ideals in a ring formed by all infinite sequences of rational numbers that are $0$ after a point

Question

For all sequence $(a_1,a_2,\cdots,a_{n-1},a_n=0,a_{n+1}=0,\cdots)$ for some $n>0$. I saw some hints: 1) take the projection map $\pi_i:A→\mathbb{Q}$ that sends $a=(a_1,a_2,\cdots,a_{n-1},a_n=0,a_{n+1}=0,\cdots)$ to $a_i$.

2. We conclude that for each $i$, ker($\pi_i$) is a maximal ideal. ok.

Are there any others?

Answer 1

I guess that the ring $R$ you're considering is the set of eventually zero sequences with componentwise addition and multiplication.

For $r=(a_1,a_2,\dots,a_n,\dotsc)\in R$, define $Z(r)=\{n\in\mathbb{N}:a_n=0\}$ (the zero set of $r$).

^{I usually start natural numbers from $0$. Since you start from $1$, I'll follow this convention.}

Note that $$ Z(r+s)\supseteq Z(r)\cap Z(s),\qquad Z(rs)=Z(r)\cup Z(s) $$ For an ideal $I$ of $R$, define $Z(I)=\bigcap\{Z(r):r\in I\}$.

Exercise 1. $Z(I)=\emptyset$ if and only if $I=R$.

Let $X$ be any subset of $\mathbb{N}$. Define $\mathcal{I}(X)=\{r\in R:Z(r)\supseteq X\}$.

Exercise 2. Prove that $\mathcal{I}(X)$ is an ideal of $R$.

If $r,s\in \mathcal{I}(X)$, then $$Z(r+s)\supseteq Z(r)\cap Z(s)\supseteq X.$$ If $r\in\mathcal{I}(X)$ and $s\in R$, then $$Z(rs)=Z(r)\cup Z(s)\supseteq X.$$ Therefore $\mathcal{I}(X)$ is an ideal of $R$.

Exercise 3. If $\mathcal{I}(X)=R$, then $X=\emptyset$.

Exercise 4. If $X\subsetneq Y$, then $\mathcal{I}(X)\supsetneq\mathcal{I}(Y)$.

Let $I$ be an ideal of $R$. Suppose $r\in I$; then $Z(r)\supseteq Z(I)$, which implies $r\in\mathcal{I}(Z(I))$; thus $I\subseteq\mathcal{I}(Z(I))$.

In particular, if $I$ is maximal, then either $I=\mathcal{I}(Z(I))$ or $\mathcal{I}(Z(I))=R$. The latter case implies $Z(I)=\emptyset$ and therefore $I=R$ (exercise 3).

Thus $I=\mathcal{I}(Z(I))$ and this implies $Z(I)$ just contains one element (exercise 4).

Answer 2

Hint: If $I$ is maximal, and unequal to $\ker \pi_i$, then $I$ contains some element $e_i$ whose $i^{th}$ coordinate is nonzero. If $I$ is unequal to any of $\ker\pi_i$, you can use these elements $e_i$ to show that $I$ is all of $A$, contradiction.