The Unapologetic Mathematician

Mathematics for the interested outsider

Ideals and quotients

Now that we know what a homomorphism of rings is, we can do the same thing we did for groups: form quotients. Remember that a ring is already an abelian group if we just consider the additive structure. We’re going to use that and see how the multiplication restricts us.

A subring S of a ring R is (as you might guess) a subset of the ring that’s still a ring, using the same operations. It’s got the sum of any two of its elements, the products, the negatives… In particular it has to be a subgroup of the additive group structure. Even better, since this is an abelian group any subgroup is normal, so we don’t have to worry about that holding us back.

Since (forgetting all but the additive structure) the ring is a group and the subring is a subgroup, we can slice R into cosets of S. I’ll write a general one as r+S, since the group is additive. Recall that this is the collection of elements of R we get by adding r to every element of S. Now, since S is a normal subgroup of R (additive structure only) there’s clearly an abelian group structure on the set of cosets. This is just how we made the quotient group before. But we also want a multiplication on the set of cosets.

Well, the obvious way to multiply cosets r+S and r'+S is to use rr'+S. But as usual we have to be careful that our choice of representative doesn’t throw us off. What if we chose r+s and r'+s' instead of r and r', where s and s' are in S? Then we get the coset rr'+sr'+rs'+ss'+S. If this is going to be the same as rr'+S, we’d better have a few things cancel.

For one thing, pick s'=0, which is in every subring. Now to have rr'+S=rr'+sr'+S we have to require that sr' land back in S for any s in the subring S and any r' at all in the ring R. That is, S has to not only contain all the products of its elements, but their products sr' with any other elements of R. If we pick s=0 we can see that rs' also has to be in S, so we have to have those products as well to make multiplication of cosets make sense.

This special kind of subring we call an “ideal”. To restate: an ideal I of a ring R is a subgroup of the additive structure of R which contains all products on either side with any element of R. Some authors further distinguish “left ideals” and “right ideals”, which contain products with an arbitrary element of R on the left side or the right side, respectively. Our ideal is then both a left and a right ideal. Of course, for a commutative ring all ideals are two-sided like this.

At first glance it rather seems like an ideal has to contain an awful lot of the ring, and often it does. Still, enough is left over to be interesting, as we’ll see.

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April 5, 2007 - Posted by | Ring theory

4 Comments »

  1. [...] way requires examining R[x]/(PR[x]), where P is a prime ideal in R. We haven’t defined such quotient rings, but we’ll freely link/steal the two results we need from the blogosphere. Since P is a prime [...]

    Pingback by units of a polynomial ring, take two « Information Flow | November 25, 2009 | Reply

  2. [...] so does each , whether each does or not. Thus we see that the integrable simple functions form an ideal of the algebra of all simple [...]

    Pingback by Basic Properties of Integrable Simple Functions « The Unapologetic Mathematician | May 25, 2010 | Reply

  3. [...] setting r(x) = rx takes each element of R to a endomorphism in End(R). Even better, every (left) ideal I of R is also a (left) R-module. This is because the restriction of each map r(x) = rx to x in I [...]

    Pingback by Relearning algebra « Information Flow | August 15, 2010 | Reply

  4. [...] a subalgebra of . Indeed, it’s actually an “ideal” in pretty much the same sense as for rings. That is, if and then . And we can check [...]

    Pingback by Ideals of Lie Algebras « The Unapologetic Mathematician | August 13, 2012 | Reply


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