# 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.

• Try to see what fails in your argument taking $R$ the ring of integers. The problem is that your inverse is an inverse modulo $M$. – Crostul Nov 25 '15 at 22:43
• All you've proven is that there is some $y$ such that $(x+M)(y+M)\in M$, i.e. such that $xy\in M$. – rogerl Nov 25 '15 at 22:43
• @rogerl: No. He's shown that there is some $y$ such that $xy-1\in M$. – tomasz Nov 25 '15 at 22:47
• Let $R$ be the integers, (5) is maximal since $\mathbb{Z}_5$ is a field. So $4 + 5\mathbb{Z}$ is a unit in $\mathbb{Z}_5$ since it is not congruent to $0$. $4\cdot 4 \equiv1$ (mod 5) so that is it's inverse. But you claim this means $4$ is a unit in the orginal ring. $4$ is definitely not a unit in $\mathbb{Z}$. It is the opposite. If $x\in U(R)$, then $x\in U(R/I)$. – CPM Nov 25 '15 at 22:51
• @Kevin My mistake, that's correct, $xy\in 1+M$. But that still doesn't show $xy=1$. – rogerl Nov 25 '15 at 23:04

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 $R$.
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.