Maximal ideals in the ring of eventually constant sequences of real numbers For homework I am studying the ring $R$ of eventually constant sequences of real numbers (with multiplication and addition defined componentwise).
What are the maximal ideals of $R$?
By looking at surjective homomorphisms it is easy to find maximal ideals, but I haven't been able to give a thorough characterization. Below are the maximal ideals I have identified.
Let $s$ be some sequence in $R$. Denote by $\bar{s}$ the real number to which $s$ converges and denote by $s_i$ the $i$-th element in $s$.
For each $i \in \mathbb{N}$ define the homomorphism
$$\varphi_i \colon R \to \mathbb{R}$$
which maps $s$ to $s_i$. $\varphi_i$ is surjective so its kernel is a maximal ideal. The kernel of $\varphi_i$ is the ideal of sequences with $i$-th component zero.
Also define the homomorphism
$$\varphi \colon R \to \mathbb{R}$$
which maps $s$ to $\bar{s}$. $\varphi$ is also surjective and its kernel is the ideal of sequences that converge to 0. Mapping $s$ to $\bar{s} - r$ for $r\in \mathbb{R}$ seems to yield more maximal ideals.
Are there any others and how can I show there aren't other ideals once I've found all the ideals?
Thank You
 A: I think you've found them all.  Let's call $\mathfrak m_i$ the ideal of $s$ with $s_i = 0$ and $\mathfrak m_\infty$ the $s$ with $\overline s = 0$.  What you want to do is assume you have a maximal $\mathfrak m$ such that $\mathfrak m \neq \mathfrak m_i$ for any finite $i$ and then show that this implies $\mathfrak m = \mathfrak m_\infty$.
Let $e_i$ be the sequence with a $1$ in the $i^\text{th}$ position and zero's elsewhere.  If $\mathfrak m \neq \mathfrak m_i$ then there's an element in $\mathfrak m$ that has an $a \neq 0$ in the $i^\text{th}$ position.  Multiply that element by $\frac1a e_i$ to get that $e_i$ is in $\mathfrak m$ for all $i$.
Now assume $\mathfrak m$ contains an $s$ such that $\overline s \neq 0$.  Use the $e_i$ to show that this would imply that $1 \in \mathfrak m$.
Edit: As rschwieb suggests, you can also observe that an element is a unit iff it's nonzero at every spot in the sequence, so one can also show that there's a unit in $\mathfrak m$ instead of jumping straight to showing that $1 \in \mathfrak m$.
