Exploration 1: Rings and ideals

November 28, 2023.

Questions:

How do you take quotients of rings?
What is an ideal?
How do you work with ideals?
What do the elements in an ideal imply about the ideal?

Recall from group theory that a normal subgroup can be used to define a quotient group, by sending every element of the normal subgroup to the group identity.

If we do the same thing with rings, we reach a slight problem. Taking $R/H$ where $H$ is a normal subgroup of $R$ ’s additive group fails certain ring axioms. Let’s see why.

Given a ring $R$ , let’s say that $H$ is a subgroup of the additive subgroup within $R$ . $H$ is normal, so by definition, $R/H$ (sending elements of $H$ to $0$ ) is a quotient group. To be a quotient ring, however, we need to define multiplication in $R/H$ .

Recall that the elements of $R/H$ are equivalence classes $[a]$ under the relation $a\sim b$ iff $b-a\in H$ . Being a quotient group means we already have addition defined as $[a]+[b]=[a+b]$ . To see what multiplication $[x][a]$ must be, we’ll have to rely on the ring axioms:

Multiplicative identity: if $1a=a=a1$ in $R$ , then $[1][a]=[a]=[a][1]$ in $R/H$ .
Multiplication is associative: if $(ab)c=a(bc)$ in $R$ , then $([a][b])[c]=[a]([b][c])$ in $R/H$ .
Multiplication is distributive: if $a(b+c)=ab+ac$ in $R$ , then $[a]([b]+[c])=[a][b]+[a][c]$ in $R/H$ .

So we get three constraints. The first constraint implies that the definition $[a][b]=[ab]$ could work, and in fact this definition satisfies all the constraints:

Multiplicative identity: $[1][a]=[1a]=[a]=[a1]=[a][1]$
Multiplication is associative: $([a][b])[c]=[(ab)c]=[a(bc)]=[a]([b][c])$
Multiplication is distributive: $[a]([b]+[c])=[a(b+c)]=[ab+ac]=[a][b]+[a][c]$

So the definition $[a][b]=[ab]$ seemingly allows $R/H$ satisfies the ring axioms. The only step left is to show well-definedness of $[a][b]=[ab]$ .

Therefore: In order for

R/H

to be a ring,

H

must be a subgroup of

R

’s additive subgroup, and additionally

xH=H

for all

x\in R

Recall that $[x]+[a]=[x+a]$ only works because we showed $x+[a]=[x+a]$ , making it well defined. In the same vein, for $[x][a]=[xa]$ , we need to show that $x[a]=[xa]$ for all $x\in [x]$ . This is the same as proving that $xa\sim xb$ iff $a\sim b$ . Let’s see how this works out: $\begin{aligned} &xa\sim xb\\ \iff& xb\in [xa]\\ \iff& xb-xa\in H\\ \iff& x(b-a)\in H\\ \iff& b-a\in H\\ \iff& b\in[a]\\ \iff& a\sim b \end{aligned}$ Note that the cancellation step $x(b-a)\in H\iff b-a\in H$ requires $xH=H$ . This imposes $xH=H$ as a requirement for multiplication in $R/H$ to be well-defined.

These “additive subgroups that absorb multiplication” are known as ideals. Just as you need a normal subgroup to quotient a group, you need an ideal to quotient a ring. The elements of a quotient ring are still known as cosets, just like for quotient groups in group theory.

Theorem:

I

is an ideal of

R

iff

I

is nonempty, and that for any

a,b\in I

and

r\in R

, we have

a-b\in I

and

ra\in I

We just showed that an ideal $I$ is an additive subgroup of $R$ that absorbs multiplication by elements $r\in R$ .
$a-b\in I$ implies closure under additive inverse (let $a=0$ ) and addition (let $b=-b$ ), thus $I$ is an additive subgroup of $R$ .
$ra\in I$ implies that $I$ absorbs multiplication by elements $r\in R$ .

In this section, we classify a ring by its ideals.

In general, to find all the ideals of a ring, you need to find all the additive subgroups of the ring’s additive group. Then the ideals are those additive subgroups that also absorb product. Let’s see some examples.

$0$ and $R$ are always ideals of every ring $R$ , where $0$ refers to the zero ring that only contains the zero element $0$ .

Therefore: Proper ideals do not contain

1

What happens if an ideal contains $1$ ? Since every product with $1$ must be in the ideal, that means every element of the ring is in the ideal, thus the ideal must be equal to the ring itself.

Therefore: Proper ideals do not contain units.

What happens if an ideal contains a unit? Then $a^{-1}a=1$ is also in the ideal, which means the ideal must contain $1$ . But as we just observed, proper ideals don’t contain $1$ .

Now we are ready to explore specific kinds of ideals and how they interact with rings.

First, there is the ideal generated by an integer, $nR$ , containing every element in $R$ added to itself $n$ times. These ideals are generally only found in finite rings, or the ring of integers.

Second, every element $a\in R$ can generate a principal ideal $(a)$ containing every possible product with $a$ . $0$ and $R$ are both principal ideals, generated by $0$ and $1$ respectively, and so they are sometimes written $(0)$ and $(1)$ .

Sometimes $aR$ , the product of every element in $R$ with $a$ , is used to denote the principal ideal $(a)$ because it’s the same thing. For example:

Theorem:

\ZZ_p

, the integers mod

p

, is isomorphic to

\ZZ/p\ZZ

, the integers quotiented by the principal ideal

(p)

The integers mod $p$ essentially set $p=0$ , which is the same as what happens when you send the principal ideal $(p)$ to $0$ .

Principal ideals $(a)$ have an important property that if the generator $a$ factors into non-units $bc$ , each factor $b$ or $c$ generates a strictly larger principal ideal.

Theorem: Given a principal ideal

(a)

and a non-unit

b

, the ideal

(ab)

is strictly contained in

(a)

Since $a$ generates $ab$ , it is clear that $(ab)\subseteq(a)$ . The case that $(ab)=(a)$ is not possible — this implies that $ab$ generates $a$ , i.e. there is some $b^{-1}$ such that $abb^{-1}=a$ , implying that $b$ is a unit. Therefore, if $b$ is a non-unit, $(ab)\subsetneq(a)$ .

Third, let’s look at prime ideals. They are similar to prime numbers: if $p$ divides a product $ab$ , then either $p$ divides $a$ or $p$ divides $b$ . By replacing “ $p$ divides” with “ $P$ contains”, you get a prime ideal $P$ , an ideal with the property that if $ab$ is in a prime ideal, either $a$ or $b$ is also in the prime ideal.

Theorem: Quotienting by a prime ideal results in an integral domain.

Let $[a],[b]$ be two nonzero cosets in the quotient. Let $P$ be a prime ideal, so that $ab\in P$ implies $a\in P$ or $b\in P$ .
Translating this into coset terms, we have $[ab]=[0]$ implies $[a]=[0]$ or $[b]=[0]$ , both of which are false since $[a]$ and $[b]$ are defined to be nonzero.
But that means $[a][b]=[0]$ has no nonzero solutions, i.e. there are no zero divisors.
No zero divisors is the definition of an integral domain.

Finally, we have one last type of ideal: a maximal ideal, one that isn’t contained in a larger (proper) ideal. It is possible that there are no maximal ideals (in certain infinite rings), or more than one maximal ideal.

Theorem: Quotienting by a maximal ideal results in a field.

Let $M$ be a maximal ideal of a ring $R$ .
We must show that every nonzero coset $[a]$ in the quotient $R/M$ is a unit, i.e. there is always some coset $[b]$ such that $[b][a]=[1]$ .
Note that since $[a]\ne [0]$ , $a\notin M$ . Therefore the ideal $\<a,M\>$ , which is generated by $M$ plus an element $a\notin M$ , is a larger ideal.
But since $M$ is maximal, no larger proper ideal contains $M$ . So $\<a,M\>$ must not be proper – it must be the whole ring $R$ , and therefore $1\in\<a,M\>$ .
This means that we somehow obtained $1$ by adding some multiple of $a$ to an element $m\in M$ . That is, $ka+m=1$ for some $k\in R$ .
In coset terms, we have $[k][a]+[m]=[1]$ . Since $m\in M$ , $[m]=[0]$ , so this simplifies to $[k][a]=[1]$ .
This implies that every element $[a]$ of the quotient is a unit, and therefore the quotient is a field.

A final note: a simple ring is one where the only ideals are $0$ and $R$ . Since we’re only talking about commutative rings, we actually find that simple rings and fields coincide.

Theorem: The simple rings are exactly the fields.

Since an ideal containing a unit must be $R$ itself, and fields only contain zero and units, fields only have the ideals $\{0\}$ and $R$ .
Conversely, a ring with only ideals $\{0\}$ and $R$ implies that any nontrivial principal ideal $(a)$ is equal to $R$ , which contains $1$ . That means some multiple of $a$ is equal to $1$ , and $ka=1$ implies that every (nonzero) element $a$ is a unit, therefore the ring only contains zero and units and is therefore a field.

Corollary: All maximal ideals are prime.

In this section, we learn how to manipulate ideals.

First, let’s talk about the sum of two ideals. It is an ideal.

Theorem: The sum of two ideals

A+B

, defined as the set

\{a+b\mid a\in A, b\in B\}

, is an ideal.

To prove it is an ideal, we need to prove it is a normal subgroup of the additive group of the ring, and that it absorbs product.
The group product $A+B$ of two normal subgroups $A,B$ is a normal subgroup. (proof)
Since $r(A+B)=rA+rB=A+B$ , we know $A+B$ also absorbs product.
Therefore $A+B$ is an ideal.

What about the union and intersection of two ideals?

Theorem: The intersection of two ideals

A\cap B

is an ideal.

To prove it is an ideal, we need to prove it is a normal subgroup of the additive group of the ring, and that it absorbs product.
The intersection $A\cap B$ of two normal subgroups $A,B$ is a normal subgroup. (proof)
Since $rA=A$ and $rB=B$ , then by definition of $A\cap B$ , we know $A\cap B$ also absorbs product as well ( $r(A\cap B)=A\cap B$ ).
Therefore $A\cap B$ is an ideal.

Theorem: The union of two ideals

A\cap B

is not always an ideal.

Consider the counterexample $(2)\cup (3)$ in $\ZZ$ , which contains all multiples of $2$ and $3$ . But this subset of $\ZZ$ is not closed under addition (e.g. $2+3=5$ ) and is therefore not an ideal.

In this section, we explore what kinds of elements an ideal can contain.

We’ve already established that a proper ideal cannot contain units. What are some other limitations on what an ideal can contain?

First, consider: when does an ideal contain a zero divisor?

Therefore: Nontrivial ideals always contain a zero divisor, if one exists.

Let’s say that the ring contains a zero divisor $r$ and some ideal $A$ .
Since product with a zero divisor results in either zero or a zero divisor, $rA$ contains nothing but zero and zero divisors. Then since ideals absorb product, $rA\subseteq A$ .
If $rA$ has a zero divisor, then that zero divisor is in $A$ and we are done. Otherwise $rA=\{0\}$ , but that means $ra=0$ for all $a\in A$ , meaning for all nonzero $a\in A$ , $a$ is a zero divisor.
Either way, as long as $A$ is nontrivial (not the zero ring), then it contains a zero divisor.

Next, when does an ideal contain an idempotent?

Therefore: Prime ideals contain an idempotent element, if one exists.

Let’s say that the ring contains an idempotent $a$ and some ideal $A$ .
$a^2=a$ implies $0=a(1-a)$ . For a prime ideal, this is interesting: since all ideals contain $0$ , a prime ideal must contain one of the factors of $0$ , i.e. either $a$ or $(1-a)$ .
Both are idempotent: $a$ is by definition, and $(1-a)^2=a^2-2a+1=a-2a+1=1-a$ .

Finally, when does an ideal contain a nilpotent?

Theorem: Every prime ideal

P

contains every nilpotent element.

Like every ideal, $P$ contains $0$ . Since $0$ is nilpotent, every ideal contains at least one nilpotent element. So let $r$ be an arbitrary nilpotent element so that $r^n=0$ for some $n$ .
This means $0$ can be factored into $r\cdot r^{n-1}$ . By the property of prime ideals, either $r\in P$ (so that we are done), or $r^{n-1}\in P$ , in which case we recursively apply this logic for $r^{n-1}$ . Eventually you get to $r\cdot r$ which proves that $r\in P$ .
Since $r$ is an arbitrary nilpotent element, this shows that every nilpotent element is in $P$ .

There is a deeper result here, though it requires Zorn’s lemma to prove.

Theorem: The intersection of all prime ideals in

R

is exactly the set

N

of all nilpotent elements of

R

Since every prime ideal contains $N$ , so does the intersection of all prime ideals. To prove that $N$ contains the intersection of all prime ideals, we must prove every element of the intersection $r$ is nilpotent. Assume towards contradiction that $r$ is not nilpotent.
Then the set $S=\{s\in R\mid r^ns=0\text{ for some }n\ge 1\}$ is a nonempty set (since $s=0$ is in $S$ ) of proper ideals (since $s=1$ can’t be in $S$ , as it would imply $r^n=0$ , and including $1$ implies not being a proper ideal).
Note that $S$ cannot contain $r$ or any of its powers $r^i$ , because otherwise $r^nr^i=0$ would imply $r$ is nilpotent, violating our assumption.
Zorn’s lemma (for ideals): Every nonempty set of proper ideals $S$ ordered by inclusion includes some maximal ideal $M$ .
- Zorn’s lemma states that every poset that defines an upper bound for every nonempty chain in the poset contains a maximal element.
- Ideals ordered by inclusion form a typical poset. Then a chain of ideals looks like $I_1\subseteq I_2\subseteq I_3\subseteq\ldots$ .
- The union of a chain of ideals is an ideal as well. It’s an upper bound since it contains every ideal in the chain. Therefore every nonempty chain has an upper bound.
- Then by Zorn’s lemma, every nonempty set of proper ideals contains a maximal element, which is a maximal ideal.
So by Zorn’s lemma, $S$ contains a maximal ideal $M$ .
Since all maximal ideals are prime, this means $M$ is a prime ideal that doesn’t contain $r$ .
But this contradicts the fact that $r$ is in the intersection of prime ideals. Therefore $r$ must be nilpotent in order to be in the intersection of prime ideals.
This means the intersection of prime ideals is exactly all the nilpotent elements of $R$ .

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