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Let $R$ be a ring with unit and let $M(R)$ be a collection of all non invertible elements in $R$.

If $M(R)$ is a ideal in $R$, prove that this is the only and maximal ideal

My thoughts:

suppose there is a bigger ideal choose some invertible element in the bigger ideal so the bigger ideal is the whole ring because there are two elements such that $u\cdot v=1$.

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Error 404
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    You have proved the "maximal" part, but what about the "only" part? – DHMO Apr 13 '17 at 18:21
  • I don't really understand why I proved the maximal part, what's the problem if the ideal is the whole ring? – Error 404 Apr 13 '17 at 18:22
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    @Error404 a maximal ideal is an ideal such that if another ideal is a superset of that ideal, then it must be the whole ring. – DHMO Apr 13 '17 at 18:24
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    If $I$ is a maximal ideal then it doesn't contain any invertible element, so it is $\subset$ ... – reuns Apr 13 '17 at 18:26
  • @Error404 A maximal ideal $I$ is one such that, if it is properly contained in any other ideal $J$, then we must have that $J = R$. One way this is often shown is by supposing $I \subset J$ proper, and proving that $1 \in J$. An ideal that contains $1$ must be the full ring, $R$, because (by definition of an ideal) we have that for any $r \in R$ and $j \in J$, we must have $rj \in J$. But let $r \in R$ be arbitrary and pick $j = 1$ to show that $rj = r \in J$, so that $J = R$. I recently wrote up a similar maximal ideal example on MSE here... – Benjamin Dickman Apr 13 '17 at 18:40

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