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I'm wondering where did this complicated proof that $J$ is a maximal ideal $\iff$ $A/J$ is a field. Is there an easy to look case where we can clearly see that when we take the quotient of the ring with its maximal ideal we must have a field? What about the inverse?

Also, where did this appear in history? Often the necessity of such proof came from somewhere else.

user26857
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Guerlando OCs
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    Using the correspondence theorem for rings, we know that taking the quotient ring is the same as looking at all ideals containing the kernel. We also know that a ring is a field if and only if it has two ideals. Thus a quotient is a field if and only if the kernel is contained in only two ideals. But since every nontrivial ideal is contained in itself and the whole ring, this is the same as saying the kernel is maximal. To me this is the reason that it is obvious that a quotient of a ring with a maximal ideal is a field. Not sure where this comes from in history. I hope this helps. – jgon Nov 07 '15 at 19:29
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    There is an easy to look case: $A=\mathbb{Z}$ and $J=p\mathbb{Z}$ with a prime $p$. – Dietrich Burde Nov 07 '15 at 19:30
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    I guess you can find some posts about the proof of this result on this site. For example, here and here. – Martin Sleziak Nov 08 '15 at 10:53

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