Rings and Fields

Number Systems and Axioms#

So far we have worked with the familiar number systems $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ and $\mathbb{R}$. These all behave nicely, but they share one problem for computational purposes; they are all infinite, whilst digital storage is finite. The plan is to eventually build finite number systems that still behave like the ones we know. To do that, we first need to describe our familiar systems as generally as possible, so that we can recognise the same properties in unfamiliar ones.

The properties we care about are called axioms; base assumptions used to build up an algebraic system. Suppose $R$ is a set of elements equipped with the operations of addition and multiplication. Then $R$ may or may not satisfy the following axioms.

$$a(b+c) = ab+ac \quad \text{and} \quad (a+b)c = ac+bc.$$

Basically, each axiom captures one piece of "normal" arithmetic behaviour. Closure says the operation never kicks you out of the set; associativity says bracketing doesn't matter; commutativity says order doesn't matter; identity says there is an element that does nothing ($0$ for addition, $1$ for multiplication); inverses say every element can be undone (which is what subtraction and division really are); and distributivity is the one axiom that ties the two operations together.

It is important to note that the "$0$" and "$1$" in the identity axioms are just names for whichever elements do nothing; they do not have to literally be the numbers 0 and 1. Also notice the asymmetry in the inverses row; every element needs an additive inverse, but only nonzero elements are ever expected to have a multiplicative inverse (you can never divide by zero).

Rings#

Depending on which of the axioms a system satisfies, it earns different names.

Basically, a non-unital ring is a system where addition works perfectly (you can add and subtract freely), and multiplication merely exists and is well behaved with brackets; there is no requirement for a $1$, for $ab = ba$, or for division.

Unital Rings#

Be aware that some other texts use this as the definition of a ring; always check which convention is in play before quoting a result.

Commutative Unital Rings#

All of the rings we study in this course are unital and commutative. Unless otherwise stated, whenever the word "ring" is used in this course, it is shorthand for "commutative unital ring".

So the running total for a (commutative unital) ring is: everything in the axioms table except multiplicative inverses. In other words, a ring is a place where you can add, subtract and multiply, but not necessarily divide.

Examples and Non-Examples of Rings#

Verifying all of the axioms for these is routine and self explanatory so I'm not gonna write it out.

Before doing any non-examples, here is the shortcut that saves you from checking all the axioms every single time. If $S$ is a subset of a known ring $R$ and uses the same addition and multiplication, then associativity, commutativity and distributivity hold in $S$ automatically (they are inherited from $R$); you only ever need to check closure, the identities, and inverses. This is why most "is this a ring?" questions come down to hunting for the one or two axioms that fail; and you should always check the cheap axioms (closure and identity) first.

Example. Explain why $\mathbb{Z}^+$ is not a ring.
$\mathbb{Z}^+ = \{1,2,3,\dots\}$ is a subset of the ring $\mathbb{Z}$, so we only need to test closure, identities and inverses. The additive identity axiom already fails; we need an element $z \in \mathbb{Z}^+$ with $a + z = a$ for all $a$, which forces $z = 0$, but

$$0 \notin \mathbb{Z}^+.$$

Additive inverses fail too; for $1 \in \mathbb{Z}^+$ we would need $-1 \in \mathbb{Z}^+$, which is false. Therefore, $\mathbb{Z}^+$ fails the additive identity and additive inverse axioms, so it is not a ring.

Example. Explain why $\mathbb{N}$ is not a ring.
This time the additive identity is fine, since $0 \in \mathbb{N}$ (it is important to remember that the naturals contains 0). However, additive inverses fail; consider $1 \in \mathbb{N}$. We need some $x \in \mathbb{N}$ with

$$\begin{align*} 1 + x &= 0, \\ x &= -1, \end{align*}$$

but $-1 \notin \mathbb{N}$. Therefore, $\mathbb{N}$ is not a ring, because it lacks additive inverses; basically you cannot subtract inside $\mathbb{N}$.

Example. Explain why $2\mathbb{Z} = \{\dots,-4,-2,0,2,4,\dots\}$ is a non-unital commutative ring.
Every element of $2\mathbb{Z}$ has the form $2m$ for some $m \in \mathbb{Z}$. Since $2\mathbb{Z} \subseteq \mathbb{Z}$ with the usual operations, associativity, commutativity and distributivity are inherited; we check the rest. For any $2m, 2n \in 2\mathbb{Z}$,

$$\begin{align*} 2m + 2n &= 2(m+n) \in 2\mathbb{Z} \text{, (closed under addition)}, \\ (2m)(2n) &= 2(2mn) \in 2\mathbb{Z} \text{, (closed under multiplication)}, \\ 0 &= 2 \cdot 0 \in 2\mathbb{Z} \text{, (additive identity)}, \\ -(2m) &= 2(-m) \in 2\mathbb{Z} \text{, (additive inverses)}. \end{align*}$$

The only axiom left is the multiplicative identity, and this is the one that fails. Suppose some $e \in 2\mathbb{Z}$ acted as an identity; then in particular $e \cdot 2 = 2$, which forces $e = 1$, but $1 \notin 2\mathbb{Z}$ as $1$ is odd. Notice how we did not just say "$1$ is not in the set"; we showed that no element of $2\mathbb{Z}$ can do the job of $1$. Therefore, $2\mathbb{Z}$ satisfies every ring axiom except the multiplicative identity, making it a non-unital commutative ring; a structure that fails exactly one axiom.

Example. Is the set of odd integers $\{\dots,-3,-1,1,3,\dots\}$ a ring?
No, and it dies at the very first axiom. Checking closure under addition,

$$\begin{align*} 1, 3 &\in \{\text{odd integers}\}, \\ 1 + 3 &= 4, \end{align*}$$

and $4$ is even, so the set is not closed under addition. (It also has no additive identity, since $0$ is even.) Therefore, the odd integers are not a ring; there is no point checking any further axioms once closure fails, which is why closure should always be your first check.

Example. Is $M_{2\times 2}(\mathbb{R})$, the set of $2\times 2$ real matrices, a commutative ring?
Matrix addition and multiplication are closed, associative and distributive, the zero matrix is the additive identity, $-A$ is the additive inverse of $A$, and the identity matrix $I$ is a multiplicative identity; so $M_{2\times 2}(\mathbb{R})$ is a unital ring. The axiom that fails is multiplicative commutativity. Take

$$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}.$$

Then

$$\begin{align*} AB &= \begin{pmatrix} 1\cdot 1 + 1 \cdot 1 & 1 \cdot 0 + 1 \cdot 1 \\ 0 \cdot 1 + 1 \cdot 1 & 0 \cdot 0 + 1 \cdot 1 \end{pmatrix} \\ &= \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}, \\ BA &= \begin{pmatrix} 1\cdot 1 + 0 \cdot 0 & 1 \cdot 1 + 0 \cdot 1 \\ 1 \cdot 1 + 1 \cdot 0 & 1 \cdot 1 + 1 \cdot 1 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}, \end{align*}$$

so $AB \neq BA$. Therefore, $M_{2 \times 2}(\mathbb{R})$ is a unital ring but not a commutative one; under this course's convention (where "ring" means commutative unital ring), it does not count as a ring at all. A single explicit counterexample is all you ever need to disprove an axiom, but proving an axiom holds requires an argument for all elements.

Fields#

Basically, a field is a system where all four operations work; you can add, subtract, multiply, and divide by anything nonzero, since dividing by $a$ just means multiplying by $a^{-1}$. Remember that $0$ is exempt from the multiplicative inverse axiom; no field lets you divide by zero.

The full hierarchy is worth memorising, since exam questions usually ask you to place a given structure on it:

$$\boxed{\text{fields} \subseteq \text{commutative unital rings} \subseteq \text{unital rings} \subseteq \text{non-unital rings}.}$$

Each step to the left adds exactly one requirement: a $1$, then commutativity of multiplication, then multiplicative inverses.

Notice that $\mathbb{Z}$ has been dropped from the list compared to the rings fact; that is the whole point of the next example.

Example. Explain why $\mathbb{Z}$ is not a field.
$\mathbb{Z}$ is a commutative unital ring, so the only axiom in question is multiplicative inverses. Consider $2 \in \mathbb{Z}$; we need some $x \in \mathbb{Z}$ with

$$\begin{align*} 2x &= 1, \\ x &= \frac{1}{2}, \end{align*}$$

but $\frac{1}{2} \notin \mathbb{Z}$. Therefore, $\mathbb{Z}$ fails the multiplicative inverse axiom and is not a field; it is "only" a ring.

Example. Show that the only elements of $\mathbb{Z}$ that have multiplicative inverses in $\mathbb{Z}$ are $\pm 1$.
Suppose $a,b \in \mathbb{Z}$ with $ab = 1$. Taking absolute values,

$$\begin{align*} |a||b| &= 1, \end{align*}$$

where $|a|$ and $|b|$ are positive integers (neither can be $0$, since $0 \cdot b = 0 \neq 1$). If $|a| \geq 2$, then $|a||b| \geq 2 > 1$; so we are forced into $|a| = |b| = 1$, i.e. $a = \pm 1$. Both actually work, since $1 \cdot 1 = 1$ and $(-1)(-1) = 1$. Therefore, $\mathbb{Z}$ doesn't just barely fail the field axioms; every integer other than $\pm 1$ is missing an inverse. (Elements of a ring that do have multiplicative inverses are called units; so the units of $\mathbb{Z}$ are exactly $\pm 1$, while in a field every nonzero element is a unit.)

Example. Explain why $\mathbb{Q}[i] = \{a+bi : a,b \in \mathbb{Q},\ i^2 = -1\}$ is a field.
$\mathbb{Q}[i]$ is a subset of the field $\mathbb{C}$ with the same operations, so associativity, commutativity and distributivity are inherited; we check the rest. For closure, given $a+bi, c+di \in \mathbb{Q}[i]$,

$$\begin{align*} (a+bi) + (c+di) &= (a+c) + (b+d)i, \\ (a+bi)(c+di) &= ac + adi + bci + bdi^2 \\ &= (ac - bd) + (ad+bc)i, \end{align*}$$

and since $\mathbb{Q}$ is closed under addition and multiplication, all of the components $a+c$, $b+d$, $ac-bd$ and $ad+bc$ are rational; so both results stay in $\mathbb{Q}[i]$. The identities are $0 = 0 + 0i$ and $1 = 1 + 0i$, and the additive inverse of $a+bi$ is $(-a)+(-b)i$, all clearly in $\mathbb{Q}[i]$.

The interesting axiom is multiplicative inverses. Given $a + bi \neq 0$ (so $a$ and $b$ are not both zero), we rationalise using the conjugate:

$$\begin{align*} \frac{1}{a+bi} &= \frac{1}{a+bi} \cdot \frac{a-bi}{a-bi} \\ &= \frac{a-bi}{a^2 - (bi)^2} \\ &= \frac{a-bi}{a^2+b^2} \\ &= \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i. \end{align*}$$

Both components are rational, and the denominator $a^2 + b^2 > 0$ because $a,b$ are not both zero; so the inverse exists inside $\mathbb{Q}[i]$. Therefore, $\mathbb{Q}[i]$ satisfies every axiom and is a field, with

$$\boxed{(a+bi)^{-1} = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i.}$$

Example. Find the multiplicative inverse of $2+3i$ in $\mathbb{Q}[i]$, and verify it.
Using the boxed formula with $a=2$, $b=3$ (so $a^2+b^2 = 13$),

$$\begin{align*} (2+3i)^{-1} &= \frac{2}{13} - \frac{3}{13}i. \end{align*}$$

Checking:

$$\begin{align*} (2+3i)\left( \frac{2}{13} - \frac{3}{13}i \right) &= \frac{4}{13} - \frac{6}{13}i + \frac{6}{13}i - \frac{9}{13}i^2 \\ &= \frac{4}{13} + \frac{9}{13} \\ &= 1. \end{align*}$$

Therefore, the inverse of $2+3i$ is $\frac{2}{13} - \frac{3}{13}i$, as required.

Example. Explain why $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} : a,b \in \mathbb{Q}\}$ is a field.
This runs exactly like $\mathbb{Q}[i]$; it is a subset of the field $\mathbb{R}$, closure and identities are straightforward (e.g. $(a+b\sqrt{2})(c+d\sqrt{2}) = (ac+2bd) + (ad+bc)\sqrt{2}$, which has rational components), so the real content is multiplicative inverses. Given $a + b\sqrt{2} \neq 0$, we rationalise with the conjugate $a - b\sqrt{2}$:

$$\begin{align*} \frac{1}{a+b\sqrt{2}} &= \frac{1}{a+b\sqrt{2}} \cdot \frac{a-b\sqrt{2}}{a-b\sqrt{2}} \\ &= \frac{a-b\sqrt{2}}{a^2-2b^2} \\ &= \frac{a}{a^2-2b^2} - \frac{b}{a^2-2b^2}\sqrt{2}. \end{align*}$$

The subtle step is showing the denominator is nonzero; this is where irrationality earns its keep. Suppose $a^2 - 2b^2 = 0$ with $a,b$ not both zero. If $b = 0$ then $a^2 = 0$ forces $a = 0$ too, a contradiction; so $b \neq 0$, and then

$$\begin{align*} a^2 &= 2b^2, \\ \left( \frac{a}{b} \right)^2 &= 2, \\ \sqrt{2} &= \left| \frac{a}{b} \right|, \end{align*}$$

which says $\sqrt{2}$ is rational; a contradiction. Hence the denominator is never zero and every nonzero element has an inverse in $\mathbb{Q}[\sqrt{2}]$. Therefore, $\mathbb{Q}[\sqrt{2}]$ is a field. As a quick sanity check with clean numbers, the inverse of $1 + \sqrt{2}$ is

$$\begin{align*} \frac{1 - \sqrt{2}}{1^2 - 2 \cdot 1^2} &= \frac{1-\sqrt{2}}{-1} \\ &= -1 + \sqrt{2}, \end{align*}$$

and indeed $(1+\sqrt{2})(-1+\sqrt{2}) = -1 + \sqrt{2} - \sqrt{2} + 2 = 1$.

Example. (Edge case.) Is the one-element set $\{0\}$, with $0 + 0 = 0$ and $0 \cdot 0 = 0$, a field under the axioms as stated?
Run through the table: closure, associativity, commutativity and distributivity are trivial since every computation outputs $0$. The additive identity is $0$; the multiplicative identity is also $0$, since $0 \cdot a = a$ holds for the only element $a = 0$ (recall the "$1$" in the axiom is just a name, and nothing says $1 \neq 0$). The additive inverse of $0$ is $0$, and the multiplicative inverse axiom only concerns nonzero elements, of which there are none; so it holds vacuously. Therefore, under the axioms exactly as stated, $\{0\}$ (called the zero ring) technically qualifies as a field; many texts add the extra axiom $1 \neq 0$ purely to rule this degenerate case out. This one is worth knowing because it tests whether you read the inverse axiom carefully, not because the zero ring is ever useful.

Finite Rings and Fields#

Recall that the entire motivation for this abstraction was to find finite systems that behave like $\mathbb{Z}$ or $\mathbb{Q}$. The classic construction is $\mathbb{Z}_n$: take the set $\{0, 1, 2, \dots, n-1\}$, and add and multiply as usual but always replace the result with its remainder on division by $n$ (which exists and is unique by the division theorem). With these operations, $\mathbb{Z}_n$ is a commutative unital ring for every $n \geq 2$; the interesting question is when it is a field.

Example. Is $\mathbb{Z}_6 = \{0,1,2,3,4,5\}$ a field?
Everything down to the multiplicative inverse axiom checks out, so we hunt for inverses. Consider $2 \in \mathbb{Z}_6$; we need $2k \equiv 1$ for some $k$, so just try all six candidates:

$$\begin{align*} 2 \cdot 0 &= 0, \\ 2 \cdot 1 &= 2, \\ 2 \cdot 2 &= 4, \\ 2 \cdot 3 &= 6 \equiv 0, \\ 2 \cdot 4 &= 8 \equiv 2, \\ 2 \cdot 5 &= 10 \equiv 4. \end{align*}$$

The outputs only ever cycle through $\{0,2,4\}$ and never hit $1$, so $2$ has no multiplicative inverse. Therefore, $\mathbb{Z}_6$ is a commutative unital ring but not a field. Intuitively, $2$ fails because it shares a common factor with $6$; every multiple of $2$ stays even, and so does its remainder after dividing by the even number $6$.

Notice something from that list that should feel deeply wrong: $2 \cdot 3 \equiv 0$ in $\mathbb{Z}_6$, even though $2 \neq 0$ and $3 \neq 0$ (such elements are called zero divisors). In a field, two nonzero elements can never multiply to give zero; if $ab = 0$ and $a^{-1}$ exists, then

$$\begin{align*} b &= 1b \\ &= (a^{-1}a)b \\ &= a^{-1}(ab) \\ &= a^{-1} \cdot 0 \\ &= 0, \end{align*}$$

so one of them was zero all along. Spotting a zero divisor is therefore an instant way to conclude a ring is not a field, without hunting for a specific element with no inverse.

Example. Is $\mathbb{Z}_5 = \{0,1,2,3,4\}$ a field?
Again only inverses are in question, and this time every nonzero element has one:

$$\begin{align*} 1 \cdot 1 &= 1, \\ 2 \cdot 3 &= 6 \equiv 1, \\ 3 \cdot 2 &= 6 \equiv 1, \\ 4 \cdot 4 &= 16 \equiv 1. \end{align*}$$

So $1^{-1} = 1$, $2^{-1} = 3$, $3^{-1} = 2$ and $4^{-1} = 4$; every nonzero element is a unit. Therefore, $\mathbb{Z}_5$ is a field, and a finite one; exactly the kind of object we set out to find. The pattern behind these two examples is that $\mathbb{Z}_n$ turns out to be a field precisely when $n$ is prime (for composite $n = ab$, the factors $a$ and $b$ are zero divisors just like $2$ and $3$ were in $\mathbb{Z}_6$); we will make heavy use of this later.

To summarise the working method: to classify a structure, march down the axioms from cheapest to most expensive; closure first, then identities and inverses, remembering that associativity, commutativity and distributivity come for free inside a known ring. To disprove an axiom, exhibit one concrete counterexample; to prove one, argue for arbitrary elements. And once all the boxes are ticked, the name of the structure just reads off how far down the field axioms you got.