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I am comparing theorems on normal subgroup and ideal from Fraleigh's, and come to this strange intuition. I hope my conclusion does not screw up, I hope I won't get ridiculed:

Theorem 15.18: $M$is a maximal normal subgroup of $G$ if and only if $G/M$ is simple.

To me this theorem makes sense because in $G/M$, the $M$ has been "collapsed" into either $0$ or $e$ (borrowing from Fraleigh's term.) In other words, the maximal normal subgroup $M$ has been "modded out" of $G$ such that all that is left is a simple group. Having said that, let's us now go to the second theorem:

Theorem 27.9: (Analogue of Theorem 15.18) Let $R$ be a commutative ring with unity. Then $M$ is a maximal ideal of $R$ if and only if $R/M$ is a field.

Since $R$ becomes a field only after it is "modded out" of the ideal $M$, may I thus conclude that ideal can intuitively be seen as an "anti-field," meaning that each and every element of an ideal does not have multiplicative inverse, whereas each and every element of field has multiplicative inverse?

Thank you for your time and effort.

POST SCRIPT: I found another theorem of similar flavor:

An ideal $I$ of $R$ is prime if and only if $R/I$ is an integral domain.

In similar vein, may I conclude that each element of prime ideal has zero divisor? This 4-year-old posting here strikes me as relevant to my conclusion. Thanks again.

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A.Magnus
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1 Answers1

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Here's an example to consider. Take a field $F$ and form $R=F\times F$. The ideal $I=F\times \{0\}$ satisfies simultaneously that $F\cong I\cong R/I$ (ring isomorphisms.) Should $I$ be an anti-field? It is not really clear that is a fruitful viewpoint...

It is certainly an interesting perspective on things. In my experience, though, I don't see that it illuminates any corners.

For one thing, the idea breaks down in that every ideal is "anti-field" in the sense that it can't contain the ring's identity. The quotient by an arbitrary ideal is usually not a field, of course. So the maximality condition has not really been incorporated into this idea of paying attention to inverses.

The idea about inverses falls even further apart if you consider what the noncommutative version of this observation is: $I$ is a maximal two-sided ideal iff $R/I$ is a simple ring (has only trivial two-sided ideals.) Commutative simple rings are fields, but in general the elements of simple rings need not have inverses.

I think what you're picking up on is mainly the pairing of maximality with simplicity via the quotient. The smaller a quotient becomes, the larger the thing being modded out is. That much is true.

PS for your PS

No, a prime ideal need not contain a nonzero zero divisor. This doesn't even work for $\Bbb Z$, where $(2)$ is prime, but contains no zero divisors other than $0$.

rschwieb
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  • @rschwieb : Thanks to you & Morgan Rogers. I am interested to understand more about your 3rd. paragraph: "For one thing,..." May I summarize the paragraph into these: (1) The observation that every element of ideal does not have inverse is not true because the ideal's identity is a counterexample; (2) The Theorem 27.9 is true only if the ideal is maximal. (Note that what I am looking for is a "working definition" of an ideal to get me going, which maybe an eye-sore to black-belts like you guys. I understand this working definition has to be re-calibrated as I advance further. Thanks again!) – A.Magnus Jul 31 '15 at 13:55
  • @A.Magnus added one more paragraph about the noncommutative ring situation. – rschwieb Jul 31 '15 at 14:00
  • @rschwieb : May I ask you to pitch in on my post script question? Sorry for being stubborn. :-) Thanks again. – A.Magnus Jul 31 '15 at 14:08
  • @MorganRodgers : Thank you! – A.Magnus Jul 31 '15 at 15:02
  • +1 for addressing the intuition behind the OPs question. – Ethan Bolker Jul 31 '15 at 17:24