Introduction to ring theory

November 27, 2023.

Questions:

What is a ring?
What kinds of elements are in a ring?
What is an integral domain, and what is a field?
How do rings compare to groups?
What operations can be done on rings?

In 1892, David Hilbert, while working with algebraic number theory, coined the term “Zahlring” (number ring) , which was later shortened to “Ring”, and finally translated in English to “ring” to refer to the structure we’re about to introduce.

$(R,+,\cdot)$ is a ring, defined as some underlying set $R$ whose elements have a notion of addition ( $+$ ) and multiplication ( $\cdot$ ). There are various definitions of a ring (see the appendix) but for our purposes we choose the strongest definition, where the following axioms must hold for all rings:

$(R,+)$ forms an (additive) abelian group with identity $0$ (called the zero element).
$(R,\cdot)$ forms a (multiplicative) commutative monoid with identity $1$ (called unity).
$\cdot$ distributes over $+$ .

One fact that holds for all rings is the binomial theorem, which we can derive from these axioms. We adopt a common notation for repeated addition and repeated multiplication:

$kr$ (where $k$ is an integer) is $r$ added to itself $k$ times.
$r^k$ (where $k$ is an integer) is $r$ multiplied by itself $k$ times.

Binomial Theorem: For any

r,s\in R

for a ring

R

(r+s)^n=\sum_{k=0}^n{n\choose k}r^ks^{n-k}

for all nonnegative

n

Proof by induction on $n$ .

Base case $n=0$ : $(r+s)^0=1={0\choose 0}r^0s^0$ . Note that this only exists since we have a multiplicative identity $1$ .
Inductive case $n>0$ : The inductive hypothesis gives $(r+s)^{n-1}=\sum_{k=0}^{n-1}{n-1\choose k}r^ks^{n-1-k}$ . Note that commutativity of the product lets us express every term of this sum as $c\cdot r^is^j$ for some integers $i,j$ and some integer coefficient $c={i+j-1\choose i}$ .
Now observe $\begin{aligned} (r+s)^n&=(r+s)(r+s)^{n-1}\\ &=r(r+s)^{n-1}+s(r+s)^{n-1}&\text{ by distributivity}\\ &=r\sum_{i=0}^{n-1}{n-1\choose i}r^is^{n-1-i}+s\sum_{j=0}^{n-1}{n-1\choose j}r^js^{n-1-j}&\text{ by IH}\\ &=\sum_{i=0}^{n-1}{n-1\choose i}r^{i+1}s^{n-1-i}+\sum_{j=0}^{n-1}{n-1\choose j}r^js^{n-j}\\ &=\sum_{i=1}^{n}{n-1\choose i-1}r^is^{n-i}+\sum_{j=0}^{n-1}{n-1\choose j}r^js^{n-j}\\ &=\sum_{i=0}^{n}{n-1\choose i-1}r^is^{n-i}+\sum_{j=0}^n{n-1\choose j}r^js^{n-j}&\text{ since }{\textstyle {n-1\choose -1}={n-1\choose n}=0}\\ &=\sum_{i=0}^{n}\left[{n-1\choose i-1}+{n-1\choose i}\right]r^is^{n-i}\\ &=\sum_{i=0}^{n}{n\choose i}r^is^{n-i}&\text{ by Pascal's identity} \end{aligned}$

In this section, we go over some examples of rings.

Possibly the ring that is the most ring of them all is the integers $\ZZ$ . In fact the axioms above are basically modeled after integer addition and multiplication, which is why it is very easy to check that $\ZZ$ is a ring.

More interestingly, $\ZZ_p$ , the integers mod $p$ , form a ring. Just as with integers, addition and multiplication work with the elements $\bar{a}$ of $\ZZ_p$ (called residue classes) where $\bar{a}=\bar{b}$ if $a,b$ differ by a factor of $p$ .

Other examples include the rationals $\QQ$ , the real numbers $\RR$ , and the complex numbers $\CC$ . There is also the zero ring (or trivial ring) $0$ , the ring containing just a zero element $0$ .

In this section, we study operations on rings.

Just like with groups, we can operate on rings as mathematical objects in their own right.

First, the direct product of two rings $R\times S$ is the same as with groups – the elements of $R\times S$ are all the pairs $(r_R,s_S)$ , with addition and multiplication defined pointwise. (When we have multiple rings like $R$ and $S$ , we typically distinguish their elements by adding the ring’s name as a subscript. For example, $1_R$ is the multiplicative identity in $R$ , and $0_S$ is the additive identity in $S$ .)

Second, we can adjoin a new element to a ring. Basically this means adding the element to the ring and taking the closure under addition and multiplication. For instance, the ring $R[e]$ is the result of adjoining $e\notin R$ to ring $R$ . Here are some examples:

$\QQ[\sqrt{2}]$ where $\sqrt{2}\in\RR$
$\ZZ[i]$ where $i\in\CC$
$\RR[i]=\CC$

We’ll make heavy use of this when we start talking about polynomial rings, but for now it’s just good to keep in mind that we can do this.

In this section, we define the units of a ring.

We mentioned that the integers $\ZZ$ are the prototypical example of a ring. Integer addition and multiplication are commutative, we have an additive identity $0$ and a multiplicative identity $1$ , there are additive inverses, and multiplication distributes over addition. So all the ring axioms hold on the integers.

What about multiplicative inverses? Do they exist for the integers? The multiplicative inverse of an integer $z$ is a value $z^{-1}$ such that $z\cdot z^{-1}=1$ . But the only integers that can satisfy this are $z=1$ and $z=-1$ . So it seems like only unit values of the integers can be multiplicative inverses.

Generalizing from integers to arbitrary rings $R$ , invertible elements of a ring (with respect to multiplication) are called its units. The units of any ring $R$ form a multiplicative group, denoted $R^\times$ or sometimes $R^*$ . Note that $R^\times$ is never a ring, because rings require the additive identity $0$ , and $0$ is never a unit.

Theorem:

0

is never a unit of a ring

R

To be a unit, we’d have to have $0\cdot 0^{-1}=1$ , but that’s impossible since $0$ multiplied by anything is zero.

Theorem:

u,v

are units iff their product

uv

is a unit.

If $u,v$ are units then $(uv)(v^{-1}u^{-1})=u(vv^{-1})u^{-1}=u1u^{-1}=1$ shows that $uv$ is a unit.
If $uv$ is a unit then $u(v(uv)^{-1})=(uv)(uv)^{-1}=1$ shows $u$ is a unit, and a symmetric argument shows $v$ is a unit.

A ring comprised of only zero and units is called a field. We know that all nonzero elements in a field are invertible and have the identity element $1$ , so the nonzero elements of a field $F$ actually form a multiplicative group $F^\times$ .

Theorem: The multiplicative group

F^\times

of a field

F

has order

|F|-1

The multiplicative group takes only the units of $F$ , of which there are $|F|-1$ since everything but zero is a unit in $F$ .

Fields have many special properties, which we’ll dive into in depth in a different exploration.

In this section, we introduce zero divisors.

Unlike the integers, however, in a ring it is possible for $a\cdot b=0$ for nonzero $a,b$ . We call such $a,b$ zero divisors because they effectively divide zero into two nonzero elements.

Zero divisors are generally undesirable, because their presence implies a lack of cancellability in the ring. In other words, when we have $a\cdot b=a\cdot c$ , we’d like to claim that $b=c$ as a result. But this relies on the fact that the map $x\mapsto a\cdot x$ is injective, which is not the case when $a$ is a zero divisor, because then both $a\cdot b=0$ and $a\cdot 0=0$ . So you can think of zero divisors as uncancellable elements in a ring.

A ring with no (nonzero) zero divisors is called an integral domain. Essentially, integral domains are exactly the rings that have cancellability, which is desirable. We also enforce the requirement that nonzero elements exist in integral domains, so as a special case, the zero ring $0$ is not an integral domain.

Let’s see some examples:

Theorem:

\ZZ_p

is an integral domain if

p

is prime.

Since $p$ is prime, when $p\mid ab$ for $a,b\in\ZZ_p$ then either $p\mid a$ or $p\mid b$ .
In $\ZZ_p$ , this translates to: when $ab\equiv 0\mod p$ , then either $a\equiv 0\mod p$ or $b\equiv 0\mod p$ .
So $a,b$ cannot be both zero divisors $(ab\equiv 0\mod p)$ and nonzero $(a,b\not\equiv 0\mod p)$ .
Therefore there are no nonzero zero divisors, making $\ZZ_p$ an integral domain.

Theorem: The direct product of nonzero rings cannot be an integral domain.

The result of a direct product of nonzero rings always contains the elements $(1,0)$ and $(0,1)$ .
Since $(1,0)(0,1)=(0,0)$ , both are zero divisors.
Since there are zero divisors, the direct product is not an integral domain.

Let’s see how zero divisors and units interact.

Theorem: No element can be both a zero divisor and a unit.

A zero divisor $a$ satisfies $ab=0$ for some nonzero $b$ . But if $a$ is a unit, then no such $b$ exists since left-multiplying $ab=0$ by $a^{-1}$ gives $b=0$ .

Corollary: Every field is an integral domain.

An integral domain must have no nonzero zero divisors. But every nonzero in a field is a unit by definition, and therefore not a zero divisor.

Theorem: Every finite integral domain is a field.

Take any nonzero $a\in R$ , so that the set $\{a,a^2,a^3,\ldots\}$ is not all zeros.
Since we’re in a finite ring, $\{a,a^2,a^3,\ldots\}$ eventually repeats, so that there is some $a^n$ equal to $a^m$ where $n>m$ .
Since we’re in an integral domain, we can cancel $a^m$ from both sides of $a^n=a^m$ , producing $a^{n-m}=1$ since $n>m$ .
Note that $n-m>0$ , so this can be rewritten as $a\cdot a^{n-m-1}=1$ , proving that $a$ is a unit.
Since every nonzero $a\in R$ is a unit, $R$ must be a field.

In this section, we introduce idempotents.

An idempotent in a ring $R$ is an element $e\in R$ such that $e^2=e$ , and therefore $e^k=e$ for all $k\ge 1$ . For every ring, $0$ and $1$ are trivially idempotent since $0^2=0$ and $1^2=1$ .

Theorem: Every nontrivial idempotent is a zero divisor.

For idempotents $e$ that aren’t $0$ or $1$ , we can show that since $e^2=e$ , we have $e^2-e=0$ and therefore $e(e-1)=0$ by distributivity. Since $e\ne 1$ implies $e-1$ is nonzero, and $e\ne 0$ , $e$ and $e-1$ must both be zero divisors.

Therefore the nontrivial idempotents are a special subset of zero divisors. This means that if a nontrivial idempotent exists in a ring, the ring is not an integral domain.

Studying the idempotents themselves gives rise to some interesting structures:

Theorem: The idempotents of a ring form a partially ordered set.

A partially ordered set (poset) is a set where a partial order $a\le b$ is defined for all elements $a,b$ , so that $\le$ satifies

reflexivity $\forall a\ldotp a\le a$ ,
antisymmetry $\forall a,b\ldotp a\le b\land b\le a\implies a=b$ , and
transitivity $\forall a,b,c\ldotp a\le b\land b\le c\implies a\le c$ .

On the idempotents, define $e\le f$ iff $ef=e$ or $ef=f$ . The idea is that all factors are “less than” their products. This satisfies the requirements:

Reflexivity: $ee=e^2=e$ , therefore $e\le e$ .
Antisymmetry: Assuming $ef=e$ and $fe=f$ we get $e=ef=fe=f$ .
Transitivity: Assuming $ef=f$ end $fg=g$ we get $eg=efg=fg=g$ .

Thus the idempotents form a poset.

Theorem: The idempotents of a ring form a boolean algebra.

A boolean algebra is a poset together with the operators $\lnot,\lor,\land$ corresponding to negation, disjunction, and conjunction respectively, as well as two distinguished elements $0$ and $1$ , all satisfying the following:

$\lnot x=0$ iff $x=1$ , $\lnot x=1$ iff $x=0$
$x\lor y=0$ iff $x=y=0$ , otherwise $x\lor y=1$
$x\land y=1$ iff $x=y=1$ , otherwise $x\land y=0$
$\lor$ and $\land$ are commutative
$\lor$ and $\land$ are associative

Define:

negation $\lnot e$ as $1-e$ .
disjunction $e\lor f$ as $e+f-ef$ .
conjunction $e\land f$ as $ef$ .
$0$ as the additive identity $0$ .
$1$ as the multiplicative identity $1$ .

Then:

$\lnot e=1-e$ , which is $0$ if $e=1$ and $1$ if $e=0$ .
$e\lor f=e+f-ef$ , which is $0$ if $e=f=0$ and $1$ otherwise.
$e\land f=ef$ , which is $1$ if $e=f=1$ and $0$ otherwise.
Commutativity can be shown by observing that both $e+f-ef$ and $ef$ are unchanged when you swap $e$ and $f$ .
Associativity of $\land$ comes from associativity of the product in a ring. Associativity of $\lor$ is harder but routine: $\begin{aligned} e\lor (f\lor g) &=e\lor (f+g-fg)\\ &=e+(f+g-fg)-e(f+g-fg)\\ &=e+f+g-fg-(ef+eg-efg)\\ &=e+f-ef+g-(eg+fg-efg)\\ &=(e+f-ef)+g-g(e+f-ef)\\ &=(e+f-ef)\lor g\\ &=(e\lor f)\lor g \end{aligned}$

Corollary: Every finite ring contains

2^k

idempotents for some

k

The proof of this isn’t really a ring theory proof, so I left it out. But it is a property of boolean algebras that every boolean algebra is isomorphic to the power set of a $k$ -element set, therefore every finite ring contains $2^k$ idempotents.

If every element in a ring is idempotent, then the whole ring is a boolean algebra, so we call it a boolean ring.

In this section, we introduce nilpotents.

Another special case of zero divisors are those elements that, when raised to a suitable power, become zero. These are elements $a$ that satisfy $\exists k\ldotp a^k=0$ , and they are called nilpotents. The zero element $0$ is always a nilpotent in every ring.

Theorem: A nonzero element

e

cannot be both idempotent and nilpotent.

An idempotent element $e$ is still itself when raised to an arbitrary power $e^k=e$ . That means that, unless $e=0$ , powers of $e$ never become zero, therefore $e$ cannot be nilpotent.

Theorem: If

r,s

are nilpotent elements of a ring

R

, then

r+s

and

r-s

are also nilpotent.

Say $r^n=0$ and $s^m=0$ . Then using the binomial theorem, $(r+s)^{n+m}=\sum_{k=0}^{n+m}{n+m\choose k}r^ks^{n+m-k}$ . Ignoring the coefficient, notice that each term $r^ks^{n+m-k}$ vanishes, because if $k\ge n$ then $r^k=0$ and if $k\le n$ then $n+m-k\ge m$ thus $s^{n+m-k}=0$ . Therefore $r+s$ is nilpotent. The same argument works for $r-s$ .

Like all zero divisors, nilpotents cannot be units. However, the existence of nilpotents is special since it always implies the existence of units:

Theorem: For every nilpotent

a

in a ring

R

, the element

1-a

is a unit of

R

If $a^k=0$ for some $k$ , then let $S=\sum_{i=0}^{k-1}a^i$ be the sum of all the powers of $a$ up to $a^{k-1}$ , which is an element of $R$ . Observe that because $a^k=0$ , $aS=\left(\sum_{i=1}^{k}a^i\right)$ is exactly every element of $S$ except for $a^0=1$ . Then their difference $S-aS$ must be equal to 1, and by factoring out $S$ , we have $S(1-a)=1$ , meaning $1-a$ is a unit of $R$ .

Corollary: If

a

is nilpotent in a ring

R

and

u

is a unit,

u+ka

for all

k\in\ZZ

are also units of

R

For $u-a$ , use the same argument as above with the series $S=\sum_{i=0}^{k-1}(au^{-1})^i$ . $(u-a)(u^{-1}S)=S-(au^{-1})S=1$ This implies $(u-a)-a$ is a unit as well, and so on, thus $u+ka$ is a unit for all $k\le 0$ .

Similarly, for $u+a$ , use the series $S=\sum_{i=0}^{k-1}(-au^{-1})^i$ . $(u+a)(u^{-1}S)=S-(-au^{-1})S=1$ This implies $(u+a)+a$ is a unit as well, and so on, thus $u+ka$ is a unit for all $k\ge 0$ .

Corollary:

u+b

is a unit for every unit

u

and nilpotent

b

Assume $b^k=0$ . Then note that $u+b=u+uu^{-1}b=u(1+u^{-1}b)=u(1-r)$ where $r=-u^{-1}b$ is also nilpotent: $r^k=-u^{-k}b^k=0$ .
Since $r$ is nilpotent, $1-r$ is a unit, so $u(1-r)=u+b$ is a unit.

In this section, we introduce the characteristic of a ring.

The above proof that $\forall k\in\ZZ\ldotp u+ka$ is a unit seems to imply that every nilpotent can produce an infinite number of units. However, this isn’t always true — some rings are finite. Can you think of one?

You might have come up with the ring of integers mod $n$ , denoted $\ZZ_n$ . Specifically let’s try $\ZZ_8$ , so that $2$ is a nilpotent element because $2^3=0$ in $\ZZ_8$ . Using our formula $1-k2$ , this implies that $1,3,5,7$ are units of the ring. But because $1+2+2+2+2=1$ , the ring “loops back” on itself at some point, so we don’t get any additional units.

This special ring property where addition “loops back” is called the characteristic. The characteristic of a ring $R$ is the number of times you have to add $1$ to itself before you get $0$ . This makes sense in the ring of integers mod $n$ , because in that ring, adding $1$ to itself $n$ times gives $0$ . So $\ZZ_n$ has characteristic $n$ , and we write $\char\ZZ_n=n$ . For rings like the integers $\ZZ$ , however, no amount of adding $1$ to itself will give $0$ , so for those rings the characteristic is defined to be $0$ . Thus $\char\ZZ=0$ . The idea is that the only way to get $0$ is to add $1$ zero times to itself.

Since most rings we work with have characteristic $0$ , we will only mention characteristic when it matters. For instance, it turns out that characteristic $2$ rings are particularly strange. Here’s an example of why.

Theorem: In a characteristic

2

ring

R

, addition is the same as subtraction.

Since $a+a=2a=0$ for every $a\in R$ , every element in $R$ is its own additive inverse: $a=-a$ Thus $a+b=a-b=-a+b=-a-b$

While rings can be of any nonnegative characteristic in general, this is not true of integral domains.

Theorem: The characteristic of an integral domain is either

0

or a prime number

p

$(ab)1=ab=0$ is precisely the requirement for the characteristic to be a composite number $ab$ . But since integral domains have no nonzero zero divisors, it’s not possible that $ab=0$ for nonzero $a,b$ . Therefore, the characteristic is either prime or zero.

Corollary: The characteristic of a field is either $0$ or a prime number $p$ .

Important note: Recall that when we adjoin an element from one ring to another, we take the additive and multiplicative closure of the result. For closure to make sense, the two rings must be of the same characteristic. So you can only adjoin elements of one ring to another if they have the same characteristic.

In this section, we talk about subrings.

A subring is a subset of elements of a ring $R$ that satisfies the ring axioms. Additionally, it must includes $1$ as the multiplicative identity. This is enough to show it includes $0$ as the additive identity as well, since $1-1=0$ .

This last requirement is curious, but it’s certainly possible for $R$ to have subsets that are rings with a different multiplicative identity. They’re just not considered subrings.

Theorem: Every ring

R

of characteristic

0

includes a subring isomorphic to the integers

\ZZ

We can define a correspondence between the integers and a subring of $R$ . Assign every integer $n\in\ZZ$ to $n1$ , which is $1$ added to itself $n$ times. Since $\char R=0$ , each $n1$ is a unique element in $R$ . Then we know that the subset of these $n1$ elements is a ring, because we can use the ring $\ZZ$ to define addition and multiplication on the $n$ part of these elements. Thus $R$ includes $\ZZ$ as a subring.

Theorem: The intersection of two subrings is a subring of both.

The intersection is an additive group, since both subrings are additive groups, and the intersection of additive groups is also an additive group.
The intersection has $1$ from the original ring, since that’s present in both subrings.
The intersection is closed under product, since both subrings are closed under product.
The intersection inherits the multiplicative identity and distributive laws from the original ring.
Since the intersection is a subset of both given subrings, is an additive group, contains $1$ , is closed under product, and satisfies identity and distributive laws, it is a subring of both.

Theorem: Every subring of an integral domain is an integral domain.

Since integral domains don’t have zero divisors, none of its subrings can have zero divisors, so every subring is also an integral domain.

Appendix A

Our earlier definition actually breaks down into four ring axioms:

$(R,+)$ forms an (additive) group with zero element $0$ .
$(R,\cdot)$ forms a (multiplicative) monoid with unity (identity) $1$ .
$\cdot$ distributes over $+$ , and this actually implies $+$ is commutative, so $(R,+)$ must be an abelian group.
$\cdot$ is also commutative.

While we will assume all of these axioms hold for a ring, one might define more general rings by relaxing these axioms. For completeness, here are the names for some ring variants:

A rig or a semiring is a ring where $(R,+)$ is also a monoid, dropping the requirement for additive inverses (n egatives).
A rng is a ring where $(R,\cdot)$ is a semigroup rather than a monoid, dropping the requirement for $1$ (the multiplicative i dentity). A ring with a multiplicative identity is sometimes explicitly called a unital ring.
A near-ring doesn’t require $+$ or $\cdot$ to be commutative. When $\cdot$ is not commutative, then left near-rings only require left distributivity (so only products on the left distribute), and similarly for right near-rings.
Noncommutative rings are those where $\cdot$ is not commutative. Otherwise it’s a commutative ring. For our purposes, we assume all rings are commutative unless otherwise specified.

So what we call rings are technically what some people call unital commutative rings. We won’t mention these other names very much since having to deal with non-unital or non-commutative rings introduces a lot of complexity, and I’d rather get into that complexity much later.

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Exploration 1: Rings and ideals