# Is ideal an “anti-field”?

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.

• The key word in both theorems is maximal. – lhf Jul 31 '15 at 11:18
• I can't see it that way: a field has a unit element, and an ideal usually doesn't. – Bernard Jul 31 '15 at 11:29
• The zero ideal has a multiplicative inverse. I think you also need to work with maximal ideals defined as being proper ideals, otherwise $R$ is an ideal of itself. – Mark Bennet Jul 31 '15 at 11:45

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.

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 $0$.
• This is good, I was trying to think of an example of this when we cross-posted: the ideal $I$ contains no element with an inverse in $R$, but when you consider $I$ as a ring in its own right, it is a field and every element has an inverse. – Morgan Rodgers Jul 31 '15 at 12:36
• @A.Magnus An ideal in general (even a maximal ideal) might not have a multiplicative identity (definitely won't contain the identity of $R$); consider the ideal $I$ of $\mathbb{Z}_{12}$ generated by $2$: it is maximal, but contains the elements ${0,2,4,6,8,10}$, so it does not only not contain the multiplicative identity of $\mathbb{Z}_{12}$, but also does not contain any element that acts as an identity when $I$ is considered a ring on its own. – Morgan Rodgers Jul 31 '15 at 14:34
Though it is in general true that no element of a maximal ideal has an inverse: if an ideal contains a unit $u$, then it must be the whole ring (because $u(u^{-1}x)$ must be in the ideal for every $x$), I don't know that this is a valuable interpretation.