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Let $E$ be a ring.

And let $N$ be the set of non-units. Assume that it is a additive subgroup of $E$(so $E$ is local).

I need to prove $N$ is two-sided ideal and the commutative case is trivial.

For the non-commutative case I have that $N$ is mult. closed and multiplying a (two-sided)unit with a non-unit yields a non-unit. But it gets difficult for me when you look at elements with a one-sided inverse.

Assume $b$ only has a right inverse and let $u\in N$. Then, why, for all $a$ with only a left inverse:

$$bua\neq1$$

Since this is required if $N$ is a ideal. Any hints? Where can I use the additive subgroup assumption?

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    $\begingroup$ The assumption it is an additive subgroup makes it necessary only to check that $a$ non unit implies $ax$ non unit, for all $x$. $\endgroup$ – egreg Nov 17 '13 at 16:55
  • $\begingroup$ That is very clear. And whenever $x$ has a two-sided inverse or no inverse at all I can prove it. But the problem lies with the elements with a one-sided inverse. $\endgroup$ – bbnkttp Nov 17 '13 at 20:05
  • $\begingroup$ I'm sure I did this several years ago. But now I don't remember. I guess it's easier showing that $N$ is the Jacobson radical $\endgroup$ – egreg Nov 17 '13 at 20:10
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First, here are a pair of lemmas that might help you get over your immediate difficulty. I left them mostly but not completely proven:

Lemma 1: In a local ring $R$, the only idempotents are $\{0,1\}$.

Proof: If $e^2=e$ and $e\notin \{0,1\}$, then $eR\oplus (1-e)R=R$ is a decomposition of $R$ into two nonzero pieces. But if $R$ has a unique maximal right ideal, it would need to contain both $eR$ and $(1-e)R$...

Lemma 2: In a local ring $R$, one-sided units are two-sided.

Proof: Suppose $ab=1$. Then $ba$ is an idempotent. By lemma 1...

Where can I use the additive subgroup assumption? Well, after showing that $nr\in N$ and $rn\in N$ for any $n\in N$ and $r\in R$, you would invoke the additive subgroup hypothesis immediately and conclude $N$ is an ideal.

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