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

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.

Solution 1:

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 $R$.

Solution 2:

You need to use the fact that every non-unit is contained in a maximal ideal. To prove it is an easy application of Zorn's lemma, but is probably a theorem in your book. Let $x$ be an element outside of $M$. If it is not a unit, it is contained in a maximal ideal. Since $R$ is local, there is only one maximal ideal, $M$. This is a contradiction, so $x$ must be a unit.