§15 Ideals and Quotient Rings

§15.1 Ideals

You might think as follows: if $S\subset R$ is a “normal” subring, then $R/S$ should be some kind of ring. This is a blind analogy with groups. However, this analogy is wrong.

Definition 15.1 Let $R$ be a commutative ring. A subset $I\subset R$ is called an ideal if

(1) $I$ is a subgroup under addition, and

(2) $x\in I$ implies $rx\in I$ for every $r\in R$ .

Remark Notice that (2) implies that if $x,y\in I$ , then

xy\in I.

So this looks like a closure condition for being a subobject.

However, $I$ need not contain the multiplicative identity of $R$ , so $I$ is certainly not necessarily a subring.

Heuristically, condition (2) really says that $I$ absorbs every element of $R$ under multiplication.

Proposition 15.1 For every nonzero integer $n$ , let

n\mathbb{Z}\subset\mathbb{Z}

be the set of integers divisible by $n$ .

Then

n\mathbb{Z}

is an ideal in the ring $\mathbb{Z}$ .

Proof:

(1) $n\mathbb{Z}$ contains $0$ , and if two numbers are divisible by $n$ , then their sum is also divisible by $n$ .

Similarly, if $a$ is divisible by $n$ , then $-a$ is also divisible by $n$ .

Therefore, $n\mathbb{Z}$ is a subgroup under addition.

(2) Finally, if $r$ is any integer and $x$ is divisible by $n$ , then

rx

is also divisible by $n$ .

~\tag*{$\square$}

Remark Since $R$ is an Abelian group, every subgroup $I$ is normal.

Therefore there exists an Abelian group

R/I.

Proposition 15.2 Let $R$ be a commutative ring and let

I\subset R

be an ideal.

Then the operation

\times: R/I\times R/I \to R/I, \qquad \overline{r}\cdot\overline{s} = \overline{rs}

together with the usual addition on $R/I$ , makes $R/I$ into a commutative ring.

Proof: We need to prove that this operation does not depend on the choice of representatives

r\in\overline{r}

and

s\in\overline{s}.

Let

r^{\prime}=r+x

and

s^{\prime}=s+y,

where

x,y\in I.

This means

\overline{r^{\prime}} = \overline{r} \in R/I,

and

\overline{s^{\prime}} = \overline{s} \in R/I.

Then

r^{\prime}s^{\prime} = (r+x)(s+y) = rs+xs+ry+xy.

Notice that the last three terms lie in $I$ because $I$ is an ideal.

Therefore their sum also lies in $I$ , since $I$ is a subgroup.

Hence

\overline{r^{\prime}s^{\prime}} = \overline{rs}.

Thus the operation is well defined.

We already know that

(R/I,+)

is an Abelian group.

Therefore, we need to prove that

(R/I,\times)

is an Abelian monoid and that multiplication distributes over addition.

First, multiplication is associative because

(\overline{a}\overline{b})\overline{c} = \overline{ab}\,\overline{c} = \overline{(ab)c} = \overline{a(bc)} = \overline{a}(\overline{b}\overline{c}).

The key step here uses the fact that

(R,\times)

is associative.

Multiplication is commutative because

\overline{a}\overline{b} = \overline{ab} = \overline{ba} = \overline{b}\overline{a},

again using the commutativity of $(R,\times)$ .

The multiplicative identity is

\overline{1}:

\overline{1}\,\overline{a} = \overline{1a} = \overline{a}, \qquad \overline{a}\,\overline{1} = \overline{a1} = \overline{a}.

Finally, multiplication distributes over addition because

\overline{a} (\overline{b}+\overline{c}) = \overline{a(b+c)} = \overline{ab+ac} = \overline{a}\,\overline{b} + \overline{a}\,\overline{c}.

~\tag*{$\square$}

Thus, in order to obtain new and interesting rings, we can look for ideals and then take quotient rings.

Example 15.1 The ring

\mathbb{Z}/n\mathbb{Z}

is the quotient of $\mathbb{Z}$ by the ideal

I=n\mathbb{Z}.

Example 15.2

\mathbb{Z}\subset\mathbb{Q}

is a subgroup and is in fact also a subring, but it is certainly not an ideal.

Indeed, if $x$ is an integer and $r$ is a rational number, then

rx

need not be an integer.

In fact, subrings are usually not ideals.

§15.2 Examples of Ideals and Quotient Rings

Definition 15.2 Let

x\in R

be an element of a commutative ring.

The ideal generated by $x$ is the set of all elements of the form

rx,

where $r\in R$ .

We denote this ideal by

(x).

Proposition 15.3 This is an ideal.

Proof: Let

I=(x).

$I$ is closed under addition because

rx+sx = (r+s)x \in I.

It contains the additive identity because

0x=0.

It contains additive inverses because

-(rx) = (-r)x.

Thus $I$ is a subgroup under addition.

Finally, if

s\in R

and

rx\in I,

then

s(rx) = (sr)x \in I.

~\tag*{$\square$}

Example 15.3 Let

R=\mathbb{R}[t]

be the polynomial ring in one variable.

Consider the ideal $I$ generated by the polynomial

t^2+1.

Thus

I = \{ f(t) ~\text{such that}~ f(t)=g(t)(t^2+1) ~\text{for some polynomial}~ g(t)\in\mathbb{R}[t] \}.

What is the ring

R/I?

Proposition 15.4 The ring

\mathbb{R}[t]/(t^2+1)

is isomorphic to

\mathbb{C}.

Isn't that cool?

In general, when you have a ring $R$ and quotient its polynomial ring by some polynomial equation, you are effectively “adjoining” to $R$ an element satisfying that polynomial equation.

This is the beginning of Galois theory.

§15.3 Geometric Interpretation of Ideals

Question: How should you think about ideals?

Algebraically: an ideal is a subgroup closed under multiplication by arbitrary elements of the ring.

That is,

I\subset R,

with

$rx\in I$ for every

x\in I, \qquad r\in R;

$(I,+)\subset(R,+)$ is a subgroup.

You may find this rather uninspiring.

So let us look at the geometric interpretation.

Let

R = \{ \text{continuous functions from some space }X\text{ to }\mathbb{R} \}.

This $R$ is a ring because

the sum of continuous functions is continuous;
the product of continuous functions is continuous;
the zero function is the additive identity:

(0+f)(x) = 0(x)+f(x) = f(x),

0+f=f.

Similarly,

f=f+0.

The constant function

1:x\mapsto1_{\mathbb{R}}

is the multiplicative identity:

(1\cdot f)(x) = 1(x)\cdot f(x) = 1_{\mathbb{R}}\cdot f(x) = f(x),

1\cdot f=f.

Similarly,

f\cdot1=f.

The function $-f$ sends

x\mapsto-f(x)

and is the additive inverse of $f$ .

We have

\begin{aligned} f\cdot(g+h): x\mapsto f(x)((g+h)(x)) &= f(x)(g(x)+h(x))\\ &= f(x)g(x)+f(x)h(x), \end{aligned}

f(g+h) = fg+fh.

Associativity is also easy to verify.

Thus

R = \{\text{continuous functions}\}.

Let

Y\subset X

be a subset and define

I_Y = \{ \text{functions }f\in R ~\text{such that}~ f(y)=0, \forall y\in Y \}.

That is,

I_Y = \{ \text{functions vanishing on }Y \}.

Proposition 15.5

I_Y\subset R

is an ideal.

Proof: Let

f_1,f_2\in I_Y.

Then for every

y\in Y,

(f_1+f_2)(y) = f_1(y)+f_2(y) = 0+0 = 0.

Therefore,

f_1+f_2 \in I_Y.

Similarly,

(-f_1)(y) = -f_1(y) = -0 = 0.

Hence

-f_1\in I_Y.

Notice that this also implies

0\in I_Y.

More directly,

0(y)=0, \qquad \forall y\in Y,

0\in I_Y.

Therefore,

I_Y\subset R

is a subgroup.

We only need to verify that it is closed under multiplication by elements of $R$ .

Given

g\in R, \qquad f\in I_Y,

we have

(\underbrace{g\cdot f}_{\in R})(y) = \underbrace{g(y)\cdot f(y)}_{\in R} = g(y)\cdot0 = 0.

Therefore,

gf\in I_Y.

~\tag*{$\square$}

Conclusion: Every subset

Y\subset X

gives rise to an ideal

I_Y\subset R.

Remark It is far from obvious that this example should be viewed as a paradigm.

For example, how should one interpret

\mathbb{Z}

as “functions on some space $X$ ”?

How is the ideal

p\mathbb{Z}

given by some “subset” of $X$ ?

In any case, this way of thinking has had an enormous influence on areas such as differential geometry and number theory.

Imagine being able to discuss the geometry of prime numbers!

Moreover, if

Y\subset X,

then we should be able to talk about functions on $Y$ —another ring.

Philosophy: Let $Y$ give rise to the ideal

I_Y.

Then

\{ \text{functions on }Y \} \cong R/I_Y.

Example 15.4 (We now move away from all continuous functions and consider only polynomial functions.)

Let

R = \mathbb{R}[x,y] = \{ \text{polynomial functions on }\mathbb{R}^2 \}.

Let

Y = \{ (x,y) ~\text{such that}~ y^2-x=0 \}.

Then

I_Y = \{ \text{polynomials }f ~\text{such that}~ f(y^2,y)=0 \} \xlongequal{\text{not obvious}} (y^2-x),

where

(y^2-x)

is the ideal generated by the element

y^2-x\in R.

Roughly speaking, if

f(x,y)

vanishes on $Y$ , then it must factor through

y^2-x.

Then

\{ \text{algebraic/polynomial functions on }Y \} \cong R/(y^2-x).

Why is this?

If $f_1$ and $f_2$ are functions on $X$ , then they restrict to functions on $Y$ .

But

f_1(y) = f_2(y), \qquad \forall y\in Y

$\Leftrightarrow$

f_1(y)-f_2(y) = 0, \qquad \forall y\in Y

$\Leftrightarrow$

f_1-f_2 \in I_Y.

That is, $f_1$ and $f_2$ define the same function on $Y$ if and only if

[f_1] = [f_2] \in R/I_Y.

Contents

§15 Ideals and Quotient Rings

§15.1 Ideals

§15.2 Examples of Ideals and Quotient Rings

§15.3 Geometric Interpretation of Ideals