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Let $R$ be a conmutative local ring, $I$ its maximal ideal. I want to prove that $x\notin I$ implies $x$ unit.

So far I have: Let $x\notin I$, I consider $x+I\in A/I$, which is a field (because $I$ is maximal). So there exists $y+I\in A/I$ such that $(x+I)(y+I)=1+I\Longrightarrow xy+I=1+I$. How do I deduce that $y$ is the inverse of $x$ (which I think it is true). All I have is $xy-1\in I$

user203327
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3 Answers3

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Suppose $x$ is not a unit. Then by Zorns lemma there exists a maximal ideal in $R$ containing $x$. But $R$ is a local ring, so has a unique maximal ideal, which is a contradiction.

Marc
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    Just a side note: This is fine if the ring is commutative (which may well suffice for the OP) but the reasoning doesn't work for noncommutative rings (and the claim in the post is true for noncommutative local rings.) you can have nonunits that do not lie in any proper ideal. – rschwieb Jan 13 '15 at 11:53
  • @rschwieb, if $R$ is noncommutative then can't we say that the left ideal $(x)=Rx$ is a proper ideal of $R$ containing $x$? Do you mean by "ideal", being both left & right ideal? – Ninja Dec 30 '16 at 16:06
  • @Ninja Yes, "ideal" means "two-sided ideal" by default. I am saying that it is possible for $Rx=R$ and $xR\neq R$ at the same time. Such an $x$ would be a nonunit such that $(x)=R$. – rschwieb Dec 30 '16 at 16:45
  • @rschwieb Thank you, do you have an example in your mind? I am though the real quaternions $\mathbb{H}$ but could not find an ideal satisfying $\mathbb{H}x = \mathbb{H}$ and $x\mathbb{H} \neq \mathbb{H}$ – Ninja Dec 30 '16 at 17:27
  • @Ninja There are examples appearing many places on the site, one of which I'm sure is this example at DaRT. $\mathbb H$ is Dedekind finite, so it will not be an example. You need a much more wild type of ring to get it to happen. – rschwieb Dec 30 '16 at 17:42
  • @rschwieb, thank you very much, I am checking the examples. – Ninja Dec 31 '16 at 00:51
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Here's a different idea.

Suppose $x \not \in I$. Then since $R$ is local, $x$ must generate $R$ (unless $x=0$). Hence there is some $z \in R$ such that $xz=1$.

Fredrik Meyer
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    I think the whole thinking part of the question is contained in that $x$ must generate $R$ since $R$ is local. – Marc Jan 13 '15 at 10:17
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if $x$ is not a unit, then $(x) \neq R$ and since $I$ is the unique maximal Ideal you must have $(x) \subseteq I$, which is impossible.

Blah
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