Download Ideals

11-8-2008
Ideals
• If R is a ring, an additive subgroup of R is a subset of R which is closed under addition, contains 0,
and is closed under taking additive inverses.
• A subring is a subset S of a ring R which is an additive subgroup of R which is closed under multiplication.
• A left ideal is a subset I of a ring R which is an additive subgroup of R such that if x ∈ I and r ∈ R,
then rx ∈ I. A right ideal is a subset I of a ring R which is an additive subgroup of R such that
if x ∈ I and r ∈ R, then xr ∈ I. A two-sided ideal is a subset I of a ring R which is an additive
subgroup of R such that if x ∈ I and r ∈ R, then rx, xr ∈ I.
• If R is a commutative ring and x ∈ R, the principal ideal generated by x is the set
hxi = {rx | r ∈ R}.
• An integral domain R is called a principal ideal domain (or PID for short) if every ideal in R is
principal.
A subgroup of a group is a subset of the group which is a group in its own right, using the operation
it inehrits 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 S ⊂ R such that:
(a) S is closed under addition: If a, b ∈ S, then a + b ∈ S.
(b) The zero element of R is in S: 0 ∈ S.
(c) S is closed under additive inverses: If a ∈ S, then −a ∈ S.
(d) S is closed under multiplication: If a, b ∈ S, then ab ∈ S.
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 S ⊂ R such that:
(a) S is closed under addition: If a, b ∈ S, then a + b ∈ S.
(b) The zero element of R is in S: 0 ∈ S.
(c) S is closed under additive inverses: If a ∈ S, then −a ∈ S.
(d) If a ∈ R and b ∈ S, then ab ∈ S and ba ∈ S. 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
1
element in the ring.
ring
ring
subring
a
b
subring
then
a
b
ring
ring
ideal
a
b
ab
ideal
a
then
b
ab
ba
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 ab ∈ S for a ∈ R
and b ∈ S, I get left ideals. Likewise, if I only require that ba ∈ S for a ∈ R and b ∈ S, I get right ideals.
From now on, if I just say “ideal”, I will mean a two-sided ideal.
If R is commutative, then ab = ba, so you only need to check that one of ab, ba, 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 {0} are ideals.
Proof. R is a group under addition, and as such I’ve already proved that R (the whole group) and {0} (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 x ∈ R and r ∈ R, then xr, rx ∈ R, because R is closed under multiplication (being the whole
ring!). Therefore, R is an ideal.
For {0}, take 0 ∈ {0} — what other choice do you have? — and r ∈ R. Then
r · 0 = 0 ∈ {0} and
0 · r = 0 ∈ {0}.
Therefore, {0} 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 {0}.
Example. (The integers as a subset of the reals) Z is a subring of R: 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. √
√
/ Z.
However, Z is not an ideal in R. For example, 2 ∈ R and 3 ∈ Z, but 2 · 3 ∈
Example. (An ideal in the ring of integers) The subset
2Z = {. . . , −4, −2, 0, 2, 4, . . .}
2
is an ideal in the ring of integers.
More generally, the ideals in Z are exactly the subsets nZ = {all multiples of n}, where n ∈ Z.
Example. (An ideal in a product ring) In the ring Z4 × Z4 , consider the subset
I = {(0, 0), (1, 1), (2, 2), (3, 3)}.
It’s easy to check that I is a subring of Z4 × Z4 . First, I contains the additive identity (0, 0).
Next, a typical element of I has the form (n, n). The additive inverse is
−(n, n) = (−n, −n) = (4 − n, 4 − n),
which is an element of I.
If you add two elements of I, you get an element of I:
(a, a) + (b, b) = (a + b, a + b).
(Of course, you’ll reduce a + b mod 4, but the two components remain the same.)
Finally, if you multiply two elements of I, you get an element of I:
(a, a)(b, b) = (ab, ab).
However, I is not an ideal; for example, (2, 2) ∈ I, but
(3, 0) · (2, 2) = (2, 0) ∈
/ I.
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 a ∈ R. The principal ideal generated by a is
hai = {ra | r ∈ R}.
Lemma. Let R be a commutative ring, and let a ∈ R. Then hai is a two-sided ideal in R.
Proof. First, 0 = 0 · a ∈ h. If ra ∈ hai, then −(ra) = (−r)a ∈ hai. Finally, if ra, sa ∈ hai, then
ra + sa = (r + s)a ∈ hai.
Thus, hai is an additive subgroup of R.
If ra ∈ hai and s ∈ R, then
s(ra) = (sr)a ∈ hai and
(ra)s = (rs)a ∈ hai.
Therefore, hai is a two-sided ideal.
Example. (A principal ideal in the ring of real polynomials) In R[x], the following set is an ideal:
hx2 + 4i = {(x2 + 4) · f (x) | f (x) ∈ R[x]}.
It’s the set consisting of all multiples of x2 + 4. For example, here are some elements of hx2 + 4i:
(2x + 5) · (x2 + 4),
(−πx50 +
√
2) · (x2 + 4),
3
0 = 0 · (x2 + 4).
Definition. An integral domain R is called a principal ideal domain (or PID for short) if every ideal in
R is principal.
The integers Z and polynomial rings over fields are examples of principal ideal domains.
Let’s see how this works for a polynomial ring. Consider the set
I = {a(x) · (x2 − 4) + b(x) · (x2 − x − 2) | a(x), b(x) ∈ Q[x]}.
It’s straightforward to show that I is an ideal. I’ll show that in fact I is principal — that is, it actually
consists of all multiples of a mystery polynomial f (x).
What could f (x) be? Well, if I take a(x) = 1 and b(x) = 0, I see that x2 − 4 is in I. Likewise, a(x) = 0
and b(x) = 1 shows that x2 − x − 2 is in I. So if everything in I is a multiple of f , then in particular these
two polynomials must be multiples of f — or what is the same, f divides x2 − 4 and x2 − x − 2.
Note that
x2 − 4 = (x − 2)(x + 2) and x2 − x − 2 = (x − 2)(x + 1).
Now I can see something which divides x2 − 4 and x2 − x − 2, namely x − 2. I’m going to guess that
f (x) = x − 2 is my mystery polynomial.
In the first place,
a(x) · (x2 − 4) + b(x) · (x2 − x − 2) = a(x) · (x − 2)(x + 2) + b(x) · (x − 2)(x + 1).
So x − 2 divides everything in I.
Now I want to show that anything divisible by x−2 is in I. So suppose x−2 | g(x), or g(x) = (x−2)h(x)
for some h(x). Why is g(x) ∈ I?
The key is to observe that x − 2 is the greatest common divisor of x2 − 4 and x2 − x − 2. Thus, I can
write x − 2 as a linear combination of x2 − 4 and x2 − x − 2. Here’s one:
x − 2 = (x2 − 4) − (x2 − x − 2).
Hence,
g(x) = (x2 − 4) − (x2 − x − 2) h(x) = h(x) · (x2 − 4) − h(x) · (x2 − x − 2).
The last expression is in I, since it’s a linear combination of x2 − 4 and x2 − x − 2. So g(x) ∈ I, as I
wanted to show.
Therefore, I is principal:
I = hx − 2i.
Now you can see how to do this in a more general case. If you have the ideal
{a1 (x)f1 (x) + · · · + an (x)fn (x) | a1 (x), . . . , an (x) ∈ F [x]}
it will be generated by the single element (f1 (x), . . . , fn (x)), the greatest common divisor of the f ’s.
Example. (Finding a generator for a principal ideal) Consider the ring Z[x] of polynomials with integer
coefficients. Let
I = hx, x + 2i = {a(x)(x + 2) + b(x)x | a(x), b(x) ∈ Z[x]}.
I is an ideal in Z[x]. It consists of all linear combinations (with polynomial coefficients) of x + 2 and x.
For example, the following polynomials are elements of I:
(x2 + 5x + 1)(x + 2) + (x117 − 89)(x),
(−2x + 3)(x + 2) + 47x,
4
(1)(x + 2) + (0)(x),
(0)(x + 2) + (1)(x).
I’ll let you verify that I satisfies the axioms for an ideal. Taking this for granted, I’ll show that I is not
principal — that is, I does not consist of multiples of a single polynomial p(x).
Suppose on the contrary that every element of I is a multiple of a polynomial p(x) ∈ Z[x]. Look at the
last two sample elements above;
x + 2 = (1)(x + 2) + (0)(x) and
(0)(x + 2) + (1)(x) = x
are both elements of I. Since I is an ideal, their difference (x + 2) − x = 2 is also an element of I.
By assumption, every element of I is a multiple of p(x), so 2 is a multiple of p(x). Thus, 2 = a(x)p(x)
for some polynomial a(x).
However, the only integer polynomials which divide the polynomial 2 are ±1 and ±2. So p(x) is −1, 1,
−2, or 2.
x is also an element of I, so x is a multiple of p(x). Of the possibilities −1, 1, −2, or 2, only −1 and 1
divide x. So p(x) = 1 or p(x) = −1.
However, remember that elements of I have the form a(x)(x + 2) + b(x)(x). The constant term of this
polynomial is the constant term of a(x) times 2 — that is, the constant term must be divisible by 2. Since
neither 1 nor −1 are divisible by 2, it follows that p(x) can’t be 1 or −1.
This contradiction shows that there is no such p(x): The ideal I is not principal.
Consequently, Z[x] is not a principal ideal domain.
Definition. Let I1 , . . . , In be ideals in a ring R. The ideal sum is
n
X
k=1
Ik = {x1 + · · · + xn | xk ∈ Ik }.
Definition. Let I and J be ideals in a ring R. The ideal product is
IJ = {x1 y1 + · · · + xn yn | xi ∈ I, yi ∈ J}.
Thus, IJ consists of all finite sums of products xy, x ∈ I, y ∈ J.
Proposition. Let R be a ring.
(a) Suppose R has an identity and I is an ideal. If 1 ∈ I, then I = R.
(b) The intersection I ∩ J of (left, right, two-sided) ideals I and J is a (left, right, two-sided) ideal.
(c) If I1 , . . . , In 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.
(e) If R is a field, the only ideals are {0} and R.
Proof. I’ll prove the first and last statements by way of example. Let I be an ideal in a ring with 1. I ⊂ R,
so I need to prove R ⊂ I. Let r ∈ R. Now 1 ∈ I, so by the definition of an ideal, r = r · 1 ∈ I. Therefore,
R ⊂ I, so R = I.
For the last statement, let R be a field, and let I ⊂ R be an ideal. Assume I 6= {0}, and find x 6= 0 in I.
Since R is a field, x is invertible; since I is an ideal, 1 = x−1 · x ∈ I. Therefore, I = R by the first property.
c 2008 by Bruce Ikenaga
5