A subgroup of a group is a subset of the group which is a group in
its own right, using the operation it inherits from its parent group.
Likewise, a * subring* of a ring is a subset of
the ring which is a ring in its own right, using the addition and
multiplication it inherits from its parent ring.

* Definition.* Let R be a ring. A * subring* is a subset such that:

(a) S is closed under addition: If , then .

(b) The zero element of R is in S: .

(c) S is closed under additive inverses: If , then .

(d) S is closed under multiplication: If , then .

It turns out to be useful to consider certain other kinds of
"subobjects" of rings: * Ideals*. I'll
use ideals to construct * quotient rings*, which
just as I used normal subgroups to construct quotient groups.

* Definition.* Let R be a ring. An * ideal* S of R is a subset such that:

(a) S is closed under addition: If , then .

(b) The zero element of R is in S: .

(c) S is closed under additive inverses: If , then .

(d) If and , then and . In other words, S is closed under multiplication (on either side) by arbitrary ring elements.

What's the difference between a subring and an ideal? A subring must
be closed under multiplication of elements *in the subring*.
An ideal must be closed under multiplication of an element in the
ideal by *any* element in the ring.

Since the ideal definition requires *more* multiplicative
closure than the subring definition, every ideal is a subring. The
converse is false, as I'll show by example below.

In the course of attempting to prove Fermat's Last Theorem,
mathematicians were led to introduce rings in which *
unique factorization* failed --- that is, it might be possible to
factor a ring element into primes in more than one way. They were led
to introduce *ideal numbers* (essentially what are now called
*ideals*) in an attempt to restore unique factorization.

What I've defined above is usually called a *
two-sided ideal*. If I only require that for and , I get * left ideals*. Likewise,
if I only require that for and , I get *
right ideals*.

From now on, if I just say "ideal", I will mean a two-sided ideal.

If R is commutative, then , so you only need to check that one of , , is in S. In the commutative case, there's no difference between left ideals, right ideals, and two-sided ideals.

* Lemma.* Let R be a ring. Then R and are ideals.

* Proof.* R is a group under addition, and as
such I've already proved that R (the whole group) and (the set consisting of the identity) are subgroups of
R. Thus, they are both closed under addition, contain 0, and are
closed under taking additive inverses. I only have to verify the
fourth ideal axiom in each case.

For R, if and , then , because R is closed under multiplication (being the whole ring!). Therefore, R is an ideal.

For , take --- what other choice do you have? --- and . Then

Therefore, is an ideal.

* Definition.* Let R be a ring. *
A proper ideal* is an ideal other than R; a *
nontrivial ideal* is an ideal other than .

* Example.* (* The integers as a
subset of the reals*) Show that is a subring of , but not an ideal.

is a subring of : It contains 0, is closed under taking additive inverses, and is closed under addition and multiplication. With regard to multiplication, note that the product of two integers is an integer.

However, is *not* an ideal in . For example, and
, but .

* Example.* (* An ideal in the
ring of integers*) Show that the subset is an ideal in for .

We already know that is a subgroup of under addition. So I just need to check closure under multiplication.

Let and let , where . Then

Therefore, is an ideal.

* Example.* (* An ideal in a
product ring*) In the ring , consider the subset

Show that I is a subring, but not an ideal.

It's easy to check that I is a subring of . First, I contains the additive identity .

Next, a typical element of I has the form . The additive inverse is

If you add two elements of I, you get an element of I:

(Of course, you'll reduce mod 4, but the two components remain the same.)

Finally, if you multiply two elements of I, you get an element of I:

However, I is not an ideal; for example, , but

In other words, I is closed under multiplication of elements
*inside* I, but not closed under multiplication by an element
from *outside* I.

* Definition.* Let R be a commutative ring, and
let . The * principal ideal
generated by* a is

For example, in the ring of polynomials with real coefficients , this is the principal ideal generated by :

It's the set consisting of all multiples of . For example, here are some elements of :

We'd better check that the principal ideal really is an ideal!

* Lemma.* Let R be a commutative ring, and let
. Then is a
two-sided ideal in R.

* Proof.* First, .

If , then .

Finally, if , then .

Thus, is an additive subgroup of R.

If and , then

Therefore, is a two-sided ideal.

* Definition.* Let , ..., be ideals in a ring R.
The * ideal sum* is

* Definition.* Let I and J be ideals in a ring R.
The * ideal product* is

Thus, consists of all finite sums of products , , .

* Proposition.* Let R be a ring.

(a) Suppose R has an identity and I is an ideal. If , then .

(b) The intersection of (left, right, two-sided) ideals I and J is a (left, right, two-sided) ideal.

(c) If , ..., are (left, right, two-sided) ideals, the ideal sum is a (left, right, two-sided) ideal.

(d) If I and J are (left, right, two-sided) ideals, the ideal product is a (left, right, two-sided) ideal.

* Proof.* I'll prove the first statement by way
of example. Let I be an ideal in a ring with 1. , so I need to prove . Let . Now , so by the definition of an ideal, . Therefore, , so .

Copyright 2018 by Bruce Ikenaga