Example of ring(s) with precisely 2 maximal ideals? I am looking for examples of ring(s) with precisely 2 maximal ideals. Anyone know of any, or how I should look for them? Just need examples of rings with this property.
 A: The product of two fields will always have this property. For example $\mathbb{Q}\times\mathbb Q$, or $\mathbb{Z}/(2) \times \mathbb{Z}/(3) \cong \mathbb{Z}/(6)$.
A: Alex's answer works. But in general given a ring $R$, find an ideal $I$ that is contained in exactly two maximal ideals and then consider the quotient $R/I$. You know by one of the isomorphism theorems that the maximal ideals of $R/I$ are simply the maximal ideals of $R$ containing the ideal $I$. More concretely, if you know two maximal ideals, you could just take their product!
Live example: Let $R=k[x]$, $m_1=(x)$ and $m_2=(x-1)$. Then let $I=m_1m_2=(x(x-1))$. Then the ring $k[x]/(x(x-1))$ has exactly two maximal ideals.
A: 1. See my answer to this question, which has more. You can generalize it to the ring that has any number of finite maximal ideals. 

2. Gilmer 's book Multiplicative Ideal Theory Th(22.8) has:
let $V_1, \cdots, V_n$ be a finite collection of valuation rings on a field $K$ s.t. $V_i\nsubseteq V_j$. Let $V=\cap V_i$. Then $V$ is a prufer domain Which has exactly $n$ maximal ideals.
