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