Example of ring(s) with precisely 2 maximal ideals?

Question

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.

Answer 1

The product of two fields will always have this property. For example $\mathbb{Q}\times\mathbb Q$, or $\mathbb{Z}/(2) \times \mathbb{Z}/(3) \cong \mathbb{Z}/(6)$.

Answer 2

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.

Answer 3

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.

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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.