Hello I'm reading the proof for Theorem 7 here and I'm struggling to grasp at which point it's necessary that the ring be commutative. I tried looking elsewhere and through other posts here but am still at loss. Help is appreciated.
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1At no point does the linked document suggest that the result only holds for commutative rings. Clearly this is part of a course about commutative rings, and therefore the result is stated in that context. – Captain Lama Feb 25 '22 at 13:04
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In non-commutative case we have three distinct notions of ideals: left ideals, right ideals and two-sided ideals. They all coincide in commutative rings, but not necessarily in non-commutative case.
With that we have:
Krull's theorem. For every unital ring $R$ and a proper left/right/two-sided ideal $I\subseteq R$ there is a proper maximal left/right/two-sided ideal $M\subseteq R$ containing $I$.
The standard proof is by applying the Zorn's lemma to the collection $$\{J\subseteq R\ |\ J\text{ is a proper ideal containing }I\}$$ and indeed it doesn't depend on commutativity at all. The main observation is that a left/right/two-sided ideal $J\subseteq R$ is proper if and only if $1\in J$. And therefore union of a chain of proper ideals is again proper.
Note that the theorem is not true for non-unital rings, even when commutative. Indeed, given any abelian group $A$ we can turn it into a non-unital commutative ring by setting $g\cdot h:=0$. Then ideals correspond to subgroups, and we all know that abelian groups don't have to have maximal subgroups (e.g. $\mathbb{Q}$). And so the multiplicative unity plays very important role here, unlike commutativity.
Anyway, for unital rings the assumption is superfluous. I suppose that the author (of the paper you read) simply deals with commutative rings only.
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