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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.

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  • $\begingroup$ Try to see what fails in your argument taking $R$ the ring of integers. The problem is that your inverse is an inverse modulo $M$. $\endgroup$ – Crostul Nov 25 '15 at 22:43
  • $\begingroup$ All you've proven is that there is some $y$ such that $(x+M)(y+M)\in M$, i.e. such that $xy\in M$. $\endgroup$ – rogerl Nov 25 '15 at 22:43
  • $\begingroup$ @rogerl: No. He's shown that there is some $y$ such that $xy-1\in M$. $\endgroup$ – tomasz Nov 25 '15 at 22:47
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    $\begingroup$ 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 $x\in U(R/I)$. $\endgroup$ – CPM Nov 25 '15 at 22:51
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    $\begingroup$ @Kevin My mistake, that's correct, $xy\in 1+M$. But that still doesn't show $xy=1$. $\endgroup$ – rogerl Nov 25 '15 at 23:04
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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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  • $\begingroup$ Good answer, haha ;) $\endgroup$ – CPM Nov 25 '15 at 22:47
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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.

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