When algebra breaks
Solve \(x^2 = x\). The school method: move everything to one side, \(x^2 - x = 0\), factor, \(x(x - 1) = 0\), and conclude \(x = 0\) or \(x = 1\). Two solutions.
Now try it in \(\mathbb{Z}/6\). Checking all six elements: \(0^2 = 0\), \(1^2 = 1\), \(3^2 = 9 \equiv 3\), \(4^2 = 16 \equiv 4\). Four solutions: 0, 1, 3 and 4. The factoring step is still valid, \(x(x - 1) = 0\); the step that fails is the last one. For \(x = 3\) we get \(3 \cdot 2 = 6 \equiv 0\), a product of two non-zero things that equals zero.
Cancellation fails too. In \(\mathbb{Z}/6\), \(2 \cdot 1 = 2\) and \(2 \cdot 4 = 8 \equiv 2\), so \(2 \cdot 1 = 2 \cdot 4\), yet \(1 \neq 4\). You can't "divide both sides by 2".
The culprit in both cases is a zero divisor. Remove those and algebra starts behaving again.
Integral domains
Definition
An integral domain (or just domain) is a commutative ring with \(1 \neq 0\) that has no zero divisors:
\[ ab = 0 \implies a = 0 \text{ or } b = 0. \]The name says these rings behave like the integers. The condition \(1 \neq 0\) only rules out the zero ring.
| Ring | Domain? | Why |
|---|---|---|
| \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\), \(\mathbb{C}\) | Yes | A product of non-zero numbers is non-zero. |
| \(\mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{Z}\}\) | Yes | It sits inside \(\mathbb{C}\), which has no zero divisors. |
| \(\mathbb{R}[x]\), polynomials | Yes | Leading terms multiply: degrees add, so a product of non-zero polynomials is non-zero. |
| \(\mathbb{Z}/p\), \(p\) prime | Yes | Every non-zero element is a unit (Unit 2). |
| \(\mathbb{Z}/6\) | No | \(2 \cdot 3 = 0\). |
| \(\mathbb{R} \times \mathbb{R}\) | No | \((1, 0) \cdot (0, 1) = (0, 0)\). |
| Functions \(\mathbb{R} \to \mathbb{R}\) | No | A function that is zero on the left half times one that is zero on the right half gives 0. |
| \(2 \times 2\) matrices | No | Not commutative, and \(\begin{pmatrix}0&1\\0&0\end{pmatrix}^2 = 0\). |
A useful rule: any subring of a domain is a domain, because zero divisors in the small ring would also be zero divisors in the big one. That's why \(\mathbb{Z}[i]\), \(\mathbb{Z}[\sqrt 2]\) and every other ring of numbers inside \(\mathbb{C}\) is automatically a domain.
Cancellation
Theorem: domains are exactly the rings where you can cancel
A commutative ring with \(1 \neq 0\) is a domain if and only if, whenever \(a \neq 0\),
\[ ab = ac \implies b = c. \]Show proof
Suppose the ring is a domain and \(ab = ac\) with \(a \neq 0\). Then \(a(b - c) = 0\), so \(b - c = 0\), that is \(b = c\).
Conversely, suppose cancellation holds and \(ab = 0\) with \(a \neq 0\). Then \(ab = a \cdot 0\), and cancelling \(a\) gives \(b = 0\). So there are no zero divisors. ∎
Notice what cancellation does not say: it doesn't say \(a\) has an inverse. In \(\mathbb{Z}\) you can cancel the 2 in \(2b = 2c\), but \(\tfrac12\) isn't an integer. Cancelling is weaker than dividing.
A big payoff: in a domain, a polynomial can't have more roots than its degree (proved in Unit 4). Without that guarantee, strange things happen. Here are the solutions of \(x^2 = 1\) in two rings:
\(\mathbb{Z}/8\)four square roots of 1
\(\mathbb{Z}/7\)just \(1\) and \(6 = -1\)
In a domain, \(x^2 = 1\) means \((x - 1)(x + 1) = 0\), so \(x = 1\) or \(x = -1\). Nothing else is possible. In \(\mathbb{Z}/8\), \((3 - 1)(3 + 1) = 2 \cdot 4 = 8 \equiv 0\) with neither factor zero.
Fields
Definition
A field is a commutative ring with \(1 \neq 0\) in which every non-zero element has a multiplicative inverse.
In a field you can add, subtract, multiply, and divide by anything except zero, and all the familiar rules hold. Everything you learned about solving linear equations, Gaussian elimination, and vectors and matrices works over any field. Examples: \(\mathbb{Q}\), \(\mathbb{R}\), \(\mathbb{C}\), and \(\mathbb{Z}/p\) for every prime \(p\). The integers are not a field, because 2 has no inverse in \(\mathbb{Z}\).
Theorem
Every field is an integral domain.
Show proof
Suppose \(ab = 0\) and \(a \neq 0\). Multiply both sides by \(a^{-1}\): \(b = a^{-1}ab = a^{-1} \cdot 0 = 0\). ∎
The converse is false: \(\mathbb{Z}\) is a domain but not a field. Yet for finite rings, the converse is true. That's the most surprising result of this unit.
Finite domains are fields
Draw "multiply by \(a\)" as arrows from each element \(x\) to \(ax\). Here it is for \(a = 3\) in \(\mathbb{Z}/7\), and for \(a = 2\) in \(\mathbb{Z}/6\):
\(\times 3\) in \(\mathbb{Z}/7\)a perfect shuffle
\(\times 2\) in \(\mathbb{Z}/6\)collisions, and 1, 3, 5 are missed
The collisions on the right are exactly cancellation failing: \(2 \cdot 0 = 2 \cdot 3\). And there's a counting fact at work. If a map from a finite set to itself has no collisions, there's no room to miss anything: \(n\) arrows landing on \(n\) different places must cover all \(n\) places. This is the pigeonhole principle.
Theorem
Every finite integral domain is a field.
Show proof
Let \(D\) be a finite domain and \(a \neq 0\). Consider the map \(x \mapsto ax\) from \(D\) to \(D\). By cancellation, \(ax = ay\) forces \(x = y\), so no two elements land in the same place. A collision-free map from a finite set to itself hits every element, so some \(x\) has \(ax = 1\). That \(x\) is \(a^{-1}\). ∎
Combining this with Unit 2 gives a complete answer for \(\mathbb{Z}/n\):
Prime numbers, which you met as "numbers with no factors", turn out to be exactly the moduli that give a well-behaved number system. This is the first glimpse of a theme that runs through the whole course: primes and fields are two sides of one coin. Unit 7 makes that precise.
Going deeper: what if multiplication isn't commutative?
A ring where every non-zero element has an inverse, but multiplication need not commute, is a division ring. The quaternions (Unit 11) are an infinite example: \(ij = k\) but \(ji = -k\).
The same pigeonhole argument shows that a finite ring without zero divisors is a division ring. And then something remarkable happens. Wedderburn's little theorem (1905) says every finite division ring is automatically commutative, so it's a field. Finiteness forces commutativity. The proof uses a clever counting argument with the class equation from group theory and roots of unity.
Building fractions
The integers aren't a field, but they sit inside one: the rationals. Fractions were invented precisely to make division possible. Remarkably, the same trick works for any domain.
Think about what a fraction is. \(\tfrac12\), \(\tfrac24\) and \(\tfrac{-3}{-6}\) are different-looking pairs of integers that we agree to treat as the same number. The rule is: \(\tfrac ab = \tfrac cd\) exactly when \(ad = bc\). Then we add and multiply with the familiar formulas:
\[ \frac ab + \frac cd = \frac{ad + bc}{bd}, \qquad \frac ab \cdot \frac cd = \frac{ac}{bd}. \]Nothing here uses anything special about \(\mathbb{Z}\) except that it's a commutative ring and that \(bd \neq 0\) when \(b, d \neq 0\). That last point is exactly the domain condition: without it, adding two fractions could produce a zero denominator.
Theorem: the field of fractions
Every integral domain \(D\) sits inside a field \(\operatorname{Frac}(D)\) whose elements are fractions \(a/b\) with \(a, b \in D\), \(b \neq 0\). It's the smallest field containing \(D\).
| Domain \(D\) | Field of fractions |
|---|---|
| \(\mathbb{Z}\) | \(\mathbb{Q}\), the rational numbers |
| \(\mathbb{Z}[i]\) | \(\mathbb{Q}(i) = \{p + qi : p, q \in \mathbb{Q}\}\) |
| \(\mathbb{R}[x]\) | \(\mathbb{R}(x)\), rational functions such as \(\dfrac{x^2 + 1}{x - 3}\) |
| Any field \(F\) | \(F\) itself: it already has all its fractions |
Going deeper: the careful construction
To build \(\operatorname{Frac}(D)\) from nothing, take all pairs \((a, b)\) with \(b \neq 0\), and declare \((a, b) \sim (c, d)\) when \(ad = bc\). One checks this is an equivalence relation. Transitivity is where the domain condition is needed: from \(ad = bc\) and \(cf = de\), multiply to get \(adf = bcf = bde\), so \(d(af - be) = 0\), and since \(d \neq 0\), \(af = be\).
The equivalence classes are the fractions. One then checks that the addition and multiplication formulas don't depend on which representatives you pick (just like the well-definedness check for \(\mathbb{Z}/n\) in Unit 2), that the ring axioms hold, and that \(b/a\) is the inverse of a non-zero \(a/b\). Finally \(a \mapsto a/1\) places \(D\) inside the new field.
Characteristic
Start at 0 and keep adding 1: \(1, 1 + 1, 1 + 1 + 1, \dots\). In \(\mathbb{Q}\) you climb forever. In \(\mathbb{Z}/6\), after six steps you're back at 0. This "how many 1s make 0" number is a basic fingerprint of a ring.
Definition
The characteristic of a ring \(R\) is the smallest positive \(n\) with \(\underbrace{1 + 1 + \dots + 1}_{n} = 0\). If no such \(n\) exists, the characteristic is \(0\).
| Ring | Characteristic |
|---|---|
| \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\), \(\mathbb{C}\), \(\mathbb{R}[x]\) | 0 |
| \(\mathbb{Z}/n\) | \(n\) |
| \((\mathbb{Z}/5)[x]\), polynomials with coefficients mod 5 | 5, though it's infinite |
| \(\mathbb{Z}/2 \times \mathbb{Z}/3\) | 6: you need \(6 = \operatorname{lcm}(2, 3)\) steps to return to \((0, 0)\) |
Theorem
The characteristic of an integral domain (in particular, of a field) is either \(0\) or a prime number.
Show proof
Write \(\bar m\) for \(1 + 1 + \dots + 1\) (\(m\) times). By the distributive law, \(\bar a \cdot \bar b = \overline{ab}\). Suppose the characteristic is \(n > 0\). Since \(1 \neq 0\), \(n \ge 2\). If \(n = ab\) with \(1 < a, b < n\), then \(\bar a \cdot \bar b = \bar n = 0\). But \(\bar a\) and \(\bar b\) are non-zero, since \(n\) is the smallest count that gives 0. That's a pair of zero divisors, impossible in a domain. So \(n\) is prime. ∎
So every field falls into one of two worlds. In characteristic 0, the elements \(\bar 1, \bar 2, \bar 3, \dots\) and their quotients form a copy of \(\mathbb{Q}\) inside the field. In characteristic \(p\), the elements \(\bar 0, \bar 1, \dots, \overline{p-1}\) form a copy of \(\mathbb{Z}/p\). Every field is built on top of \(\mathbb{Q}\) or some \(\mathbb{Z}/p\), its prime subfield.
The freshman's dream
A classic beginner's mistake is writing \((a + b)^2 = a^2 + b^2\). In characteristic 2, it's not a mistake: \((a + b)^2 = a^2 + 2ab + b^2\), and \(2ab = 0\). Does this generalise? Look at the binomial coefficients, Pascal's triangle, and ask which inner entries of row \(n\) are divisible by \(n\):
Theorem: the freshman's dream
In any commutative ring of prime characteristic \(p\),
\[ (a + b)^p = a^p + b^p. \]Show proof
By the binomial theorem, \((a + b)^p = \sum_k \binom pk a^k b^{p-k}\). For \(0 < k < p\),
\[ \binom pk = \frac{p!}{k!\,(p-k)!}. \]The prime \(p\) divides the numerator, but it doesn't divide any of the factors \(1, \dots, k\) or \(1, \dots, p - k\) of the denominator, all smaller than \(p\). So \(p\) divides \(\binom pk\), and in characteristic \(p\) that coefficient is \(0\). Only the \(k = 0\) and \(k = p\) terms survive. ∎
Here is a lovely payoff: a second proof of Fermat's little theorem. In \(\mathbb{Z}/p\), the dream gives \((a + 1)^p = a^p + 1\). Start from \(0^p = 0\) and step up one at a time: \(1^p = 1\), \(2^p = (1 + 1)^p = 1 + 1 = 2\), \(3^p = 2^p + 1 = 3\), and so on. So \(a^p = a\) for every \(a \in \mathbb{Z}/p\). Cancelling one \(a \neq 0\) gives \(a^{p-1} = 1\).
Colour the entries of Pascal's triangle by whether they're odd or even, that is, look at it in characteristic 2, and a fractal appears:
Going deeper: the Frobenius map
In a field \(F\) of characteristic \(p\), define \(\phi(a) = a^p\). The dream says \(\phi(a + b) = \phi(a) + \phi(b)\), and obviously \(\phi(ab) = \phi(a)\phi(b)\) and \(\phi(1) = 1\). So \(\phi\) respects all the ring operations: it's a homomorphism (Unit 5), called the Frobenius map. It's also one-to-one, since \(a^p = 0\) forces \(a = 0\) in a field. For a finite field it therefore shuffles the elements, just like multiplication by \(a\) did above. Frobenius is the key that unlocks the structure of finite fields in Unit 10.
Practice set
Mixed questions covering the whole unit. Try each one before opening the answer.
1. Which of these are integral domains: \(\mathbb{Z}/9\), \(\mathbb{Z}/11\), \(\mathbb{Z}[x]\), \(\mathbb{Z} \times \mathbb{Z}\)?
Show answer
\(\mathbb{Z}/9\): no, \(3 \cdot 3 = 0\). \(\mathbb{Z}/11\): yes, it's even a field. \(\mathbb{Z}[x]\): yes, leading coefficients multiply to something non-zero. \(\mathbb{Z} \times \mathbb{Z}\): no, \((1, 0)(0, 1) = (0, 0)\).
2. Find all solutions of \(x^2 = x\) in \(\mathbb{Z}/10\). Why doesn't this contradict anything?
Show answer
\(0, 1, 5, 6\) (\(5^2 = 25 \equiv 5\), \(6^2 = 36 \equiv 6\)). \(\mathbb{Z}/10\) isn't a domain, so a degree-2 equation may have more than 2 roots. Here \(5 \cdot 4 = 20 \equiv 0\) is the zero-divisor pair behind \(x = 5\).
3. Show that in a domain, if \(a^n = 0\) for some \(n\), then \(a = 0\). Find a non-zero \(a\) in \(\mathbb{Z}/8\) with \(a^3 = 0\).
Show answer
If \(a \neq 0\) then \(a \cdot a \neq 0\), then \(a^2 \cdot a \neq 0\), and so on: a product of non-zero elements stays non-zero. In \(\mathbb{Z}/8\), \(2^3 = 8 \equiv 0\). Elements like this are called nilpotent.
4. Using the pigeonhole idea, find the inverse of 4 in \(\mathbb{Z}/11\) by listing \(4x\) for \(x = 1, 2, 3, \dots\)
Show answer
\(4, 8, 12 \equiv 1\). So \(4^{-1} = 3\).
5. What is the characteristic of \(\mathbb{Z}/4 \times \mathbb{Z}/6\)? Is it a domain?
Show answer
Adding \((1, 1)\) repeatedly returns to \((0, 0)\) after \(\operatorname{lcm}(4, 6) = 12\) steps, so the characteristic is 12. Not prime, so it's not a domain (as you'd expect from \((1, 0)(0, 1) = 0\)).
6. In a ring of characteristic 3, simplify \((a + b)^9\) and \((a - b)^3\).
Show answer
\((a + b)^9 = ((a + b)^3)^3 = (a^3 + b^3)^3 = a^9 + b^9\). And \((a - b)^3 = a^3 + (-b)^3 = a^3 - b^3\).
Summary and glossary
- An integral domain is a commutative ring, \(1 \neq 0\), with no zero divisors. Domains are exactly the rings where non-zero factors can be cancelled.
- A field is a commutative ring where every non-zero element has an inverse. Every field is a domain.
- By the pigeonhole principle, every finite domain is a field. So \(\mathbb{Z}/n\) is a field exactly when \(n\) is prime.
- Every domain embeds in its field of fractions, just as \(\mathbb{Z}\) sits in \(\mathbb{Q}\).
- The characteristic of a domain is 0 or a prime \(p\); every field contains \(\mathbb{Q}\) or \(\mathbb{Z}/p\).
- In characteristic \(p\), \((a + b)^p = a^p + b^p\).
- Integral domain
- Commutative, \(1 \neq 0\), and \(ab = 0\) forces \(a = 0\) or \(b = 0\).
- Cancellation
- \(ab = ac\) and \(a \neq 0\) imply \(b = c\).
- Field
- Commutative ring in which every non-zero element is a unit.
- Division ring
- Like a field, but multiplication need not commute.
- Field of fractions
- The smallest field containing a domain.
- Characteristic
- How many 1s add up to 0 (or 0 if never).
- Prime subfield
- The copy of \(\mathbb{Q}\) or \(\mathbb{Z}/p\) inside every field.
- Nilpotent
- An element with some power equal to 0.
- Frobenius map
- \(a \mapsto a^p\) in characteristic \(p\).