Ideals

$$\langle x\rangle = \{rx \mid r \in R\}.$$


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 $S
   \subset R$ such that:

(a) S is closed under addition: If $a, b
   \in S$ , then $a + b \in S$ .

(b) The zero element of R is in S: $0 \in
   S$ .

(c) S is closed under additive inverses: If $a \in S$ , then $-a \in S$ .

(d) S is closed under multiplication: If $a, b \in S$ , then $ab \in S$ .

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 $S \subset R$ such that:

(a) S is closed under addition: If $a, b
   \in S$ , then $a + b \in S$ .

(b) The zero element of R is in S: $0 \in
   S$ .

(c) S is closed under additive inverses: If $a \in S$ , then $-a \in S$ .

(d) If $a \in R$ and $b \in S$ , then $ab \in
   S$ and $ba \in S$ . 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 {\it any} element in the ring.

$$\hbox{\epsfysize=2in \epsffile{ideal1.eps}}$$

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 $ab \in S$ for $a \in R$ and $b \in S$ , I get left ideals. Likewise, if I only require that $ba \in S$ for $a
   \in R$ and $b \in S$ , 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 $x \in R$ and $r \in R$ , then $xr, rx
   \in R$ , because R is closed under multiplication (being the whole ring!). Therefore, R is an ideal.

For $\{0\}$ , take $0 \in \{0\}$ --- what other choice do you have? --- and $r \in R$ . Then

$$r\cdot 0 = 0 \in \{0\} \quad\hbox{and}\quad 0\cdot r = 0 \in \{0\}.$$

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) $\integer$ is a subring of $\real$ : 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, $\integer$ is not an ideal in $\real$ . For example, $\sqrt{2} \in \real$ and $3 \in \integer$ , but $\sqrt{2}\cdot 3 \notin \integer$ .


Example. ( An ideal in the ring of integers) The subset

$$2\integer = \{\ldots, -4, -2, 0, 2, 4, \ldots\}$$

is an ideal in the ring of integers.

More generally, the ideals in $\integer$ are exactly the subsets $n\integer = \{\hbox{all multiples of n}\}$ , where $n \in \integer$ .


Example. ( An ideal in a product ring) In the ring $\integer_4 \times \integer_4$ , consider the subset

$$I = \{(0,0), (1,1), (2,2), (3,3)\}.$$

It's easy to check that I is a subring of $\integer_4 \times \integer_4$ . 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, $(2,2) \in I$ , but

$$(3,0)\cdot (2,2) = (2,0) \notin I.$$

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


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

$$\langle a\rangle = \{ra \mid r \in R\}.$$

Lemma. Let R be a commutative ring, and let $a \in R$ . Then $\langle
   a\rangle$ is a two-sided ideal in R.

Proof. First, $0
   = 0\cdot a \in \langle$ . If $ra \in \langle a\rangle$ , then $-(ra) = (-r)a \in
   \langle a\rangle$ . Finally, if $ra, sa \in \langle a\rangle$ , then $ra + sa
   = (r + s)a \in \langle a\rangle$ .

Thus, $\langle a\rangle$ is an additive subgroup of R.

If $ra \in \langle a\rangle$ and $s \in R$ , then

$$s(ra) = (sr)a \in \langle a\rangle \quad\hbox{and}\quad (ra)s = (rs)a \in \langle a\rangle.$$

Therefore, $\langle a\rangle$ is a two-sided ideal.


Example. ( A principal ideal in the ring of real polynomials) In $\real[x]$ , the following set is an ideal:

$$\langle x^2 + 4\rangle = \{(x^2 + 4)\cdot f(x) \mid f(x) \in \real[x]\}.$$

It's the set consisting of all multiples of $x^2 + 4$ . For example, here are some elements of $\langle x^2 +
   4\rangle$ :

$$(2x + 5)\cdot (x^2 + 4), \quad (-\pi x^{50} + \sqrt{2})\cdot (x^2 + 4), \quad 0 = 0\cdot (x^2 + 4).\quad\halmos$$


Definition. An integral domain R is called a principal ideal domain (or PID for short) if every ideal in R is principal.

The integers $\integer$ and polynomial rings over fields are examples of principal ideal domains.

Let's see how this works for a polynomial ring. Consider the set

$$I = \{a(x)\cdot (x^2 - 4) + b(x)\cdot (x^2 - x - 2) \mid a(x), b(x) \in \rational[x]\}.$$

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 $x^2 - 4$ is in I. Likewise, $a(x) = 0$ and $b(x) =
   1$ shows that $x^2 - x - 2$ 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 $x^2 -
   4$ and $x^2 - x - 2$ .

Note that

$$x^2 - 4 = (x - 2)(x + 2) \quad\hbox{and}\quad x^2 - x - 2 = (x - 2)(x + 1).$$

Now I can see something which divides $x^2 - 4$ and $x^2 - x - 2$ , namely $x - 2$ . I'm going to guess that $f(x) = x - 2$ is my mystery polynomial.

In the first place,

$$a(x)\cdot (x^2 - 4) + b(x)\cdot (x^2 - x - 2) = a(x)\cdot (x - 2)(x + 2) + b(x)\cdot (x - 2)(x + 1).$$

So $x - 2$ divides everything in I.

Now I want to show that anything divisible by $x - 2$ is in I. So suppose $x - 2 \mid g(x)$ , or $g(x) = (x -
   2)h(x)$ for some $h(x)$ . Why is $g(x) \in I$ ?

The key is to observe that $x - 2$ is the greatest common divisor of $x^2 - 4$ and $x^2 - x - 2$ . Thus, I can write $x - 2$ as a linear combination of $x^2 - 4$ and $x^2 - x - 2$ . Here's one:

$$x - 2 = (x^2 - 4) - (x^2 - x - 2).$$

Hence,

$$g(x) = \left[(x^2 - 4) - (x^2 - x - 2)\right]h(x) = h(x)\cdot (x^2 - 4) - h(x)\cdot (x^2 - x - 2).$$

The last expression is in I, since it's a linear combination of $x^2 - 4$ and $x^2 - x - 2$ . So $g(x) \in I$ , as I wanted to show.

Therefore, I is principal:

$$I = \langle x - 2\rangle.$$

Now you can see how to do this in a more general case. If you have the ideal

$$\{a_1(x)f_1(x) + \cdots + a_n(x)f_n(x) \mid a_1(x), \ldots, a_n(x) \in F[x]\}$$

it will be generated by the single element $(f_1(x), \ldots, f_n(x))$ , the greatest common divisor of the f's.


Example. ( Finding a generator for a principal ideal) Consider the ring $\integer[x]$ of polynomials with integer coefficients. Let

$$I = \langle x, x + 2\rangle = \{a(x)(x + 2) + b(x)x \mid a(x), b(x) \in \integer[x]\}.$$

I is an ideal in $\integer[x]$ . It consists of all linear combinations (with polynomial coefficients) of $x +
   2$ and x. For example, the following polynomials are elements of I:

$$(x^2 + 5x + 1)(x + 2) + (x^{117} - 89)(x), \quad (-2x + 3)(x + 2) + 47x, \quad (1)(x + 2) + (0)(x), \quad (0)(x + 2) + (1)(x).$$

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 $p(x) \in \integer[x]$ . Look at the last two sample elements above;

$$x + 2 = (1)(x + 2) + (0)(x) \quad\hbox{and}\quad (0)(x + 2) + (1)(x) = x$$

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 $\pm 1$ and $\pm 2$ . 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, $\integer[x]$ is not a principal ideal domain.


Definition. Let $I_1$ , ..., $I_n$ be ideals in a ring R. The ideal sum is

$$\sum_{k=1}^n I_k = \{x_1 + \cdots + x_n \mid x_k \in I_k\}.$$

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

$$IJ = \{x_1y_1 + \cdots + x_ny_n \mid x_i \in I, y_i \in J\}.$$

Thus, $IJ$ consists of all finite sums of products $xy$ , $x \in I$ , $y \in J$ .

Proposition. Let R be a ring.

(a) Suppose R has an identity and I is an ideal. If $1 \in I$ , then $I = R$ .

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

(c) If $I_1$ , ..., $I_n$ 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. $I \subset R$ , so I need to prove $R
   \subset I$ . Let $r \in R$ . Now $1 \in I$ , so by the definition of an ideal, $r = r\cdot 1 \in I$ . Therefore, $R \subset
   I$ , so $R = I$ .

For the last statement, let R be a field, and let $I \subset R$ be an ideal. Assume $I
   \ne \{0\}$ , and find $x \ne 0$ in I. Since R is a field, x is invertible; since I is an ideal, $1 = x^{-1}\cdot x \in I$ . Therefore, $I = R$ by the first property.


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