[Math] Every element outside the maximal ideal of a local ring is a unit

abstract-algebraidealsmaximal-and-prime-ideals

A homework question from my algebra class asks:

Show that in a local ring $R$ with maximal ideal $M$, every element outside $M$ is a unit.

My argument is that since $M$ is maximal $R /M $ is a field and so for any $ x \in R \backslash M $, $ x + M $ has a multiplicative inverse, which implies $ x $ is a unit.

I don't see where we need the fact that $R$ is a local ring.

Best Answer

I hope you know the following theorem:

Let $R$ be a commutative ring with $1$, $a \in R$ a non-unit. Then there exists a maximal ideal $M$ of $R$ such that $a \in M$.

This is a standard consequence of Zorn's lemma. In particular this implies that the set of units of $R$ coincides with the complement of the union of maximal ideals of $R$.

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