Maximal ideals for the commutative ring formed by all infinite sequences of rational numbers that are constant after a point I am being asked to find all maximal ideals in a ring $A$, where $A$ is the commutative ring formed by all infinite sequences of rational numbers that are constant after a point, i.e. all $(a_1, a_2, a_3, ...)$ such that $a_n=a_{n+1}=\cdots$.
I've seen a questions related to this but A in that case is formed by real numbers, so I think this case may be different.
I don't know how to begin this question...
Thank you so much for your answers!
 A: First let's find some maximal ideals. For each coordinate, $n\in\newcommand{\NN}{\mathbb{N}}\NN$, we have a homomorphism $\pi_n : A\to \newcommand{\QQ}{\mathbb{Q}}\QQ$ given by projecting onto the $n$th coordinate. This is a surjective homomorphism to a field, so its kernel is a maximal ideal. Call it $\newcommand{\mm}{\mathfrak{m}}\mm_n$. This is the set of sequences with $n$th coordinate 0. There is also the homomorphism $\lim : A\to \QQ$, which takes the limit of the sequence. Again this is a surjective homomorphism to a field, so $\mm_\infty := \ker\lim$, the set of all sequences which are eventually 0, is a maximal ideal as well. 
Now we want to show that these are the only possible maximal ideals. Let $e_i$ be the sequence that is 0 at indices other than $i$ and $1$ at $i$. Then $e_ie_j = 0$ when $i\ne j$, so if $\mm$ is maximal, and $e_i\not \in \mm$ for some $i$, $e_j\in \mm$ for all other $j$, and since $(1-e_i)e_i=0$, $(1-e_i)\in\mm$ as well, and you can check that these generate $\mm_i$. Otherwise, $e_i\in \mm$ for all $i$, and you can check that these elements generate $\mm_\infty$. Hence these are all the maximal ideals.
Edit
Editing in response to ZFR's comment below to attempt to clarify the second paragraph.
To be precise, either $e_i$ is not in $\mm$ for some $i$, in which case $\mm$ contains the elements $e_j$ for $j\ne i$ and $1-e_i$, and these elements $\{e_1,\ldots,e_{i-1},1-e_i,e_{i+1},\ldots\}$ generate $\mm_i$. Otherwise $e_i$ is in $\mm$ for all $i$, and the set of elements $\{e_1,\ldots,e_n,\ldots\}$ generates $\mm_\infty$.
