Primes, rethought
The school definition of a prime, "a number with no factors other than 1 and itself", describes how a number breaks up. There's a second property of primes, proved by Euclid, that describes how they divide:
Euclid's lemma is the version that generalises well, because it can be stated purely in terms of the ideal \((p)\): "if a product lands in \((p)\), one of the factors was already in \((p)\)."
Definition
An ideal \(P\) of a commutative ring \(R\) is prime if \(P \neq R\) and
\[ ab \in P \implies a \in P \text{ or } b \in P. \]| Ideal | Prime? | Reason |
|---|---|---|
| \((7)\) in \(\mathbb{Z}\) | Yes | Euclid's lemma. |
| \((6)\) in \(\mathbb{Z}\) | No | \(2 \cdot 3 \in (6)\), but \(2, 3 \notin (6)\). |
| \((0)\) in \(\mathbb{Z}\) | Yes | \(ab = 0\) forces \(a = 0\) or \(b = 0\). |
| \((x)\) in \(\mathbb{Z}[x]\) | Yes | \(fg\) has constant term 0 only if \(f\) or \(g\) does. |
| \((x^2)\) in \(\mathbb{R}[x]\) | No | \(x \cdot x \in (x^2)\), but \(x \notin (x^2)\). |
Notice the third row: the zero ideal is prime exactly when the ring has no zero divisors. That's the first hint of the link with domains.
Prime ideals give domains
Theorem
\(R/P\) is an integral domain if and only if \(P\) is a prime ideal.
Show proof
In \(R/P\), an element \(a + P\) is zero exactly when \(a \in P\). So "\((a + P)(b + P) = 0\) implies \(a + P = 0\) or \(b + P = 0\)" translates word for word into "\(ab \in P\) implies \(a \in P\) or \(b \in P\)". The condition \(P \neq R\) matches \(1 \neq 0\) in the quotient. ∎
Intuition: setting \(P\) to zero creates no zero divisors precisely when \(P\) never contains a product without containing a factor. Setting \(6 = 0\) is bad, because \(2 \cdot 3\) becomes 0 while 2 and 3 survive.
Maximal ideals give fields
Definition
An ideal \(M\) is maximal if \(M \neq R\) and there's no ideal strictly between them: if \(M \subseteq J \subseteq R\) for an ideal \(J\), then \(J = M\) or \(J = R\).
In \(\mathbb{Z}\), \((n) \subseteq (m)\) means \(m \mid n\). So the ideals just below the top are generated by numbers whose only divisors are 1 and themselves: the primes. The school definition of a prime ("can't be broken up") is exactly maximality.
Theorem
\(R/M\) is a field if and only if \(M\) is a maximal ideal.
Show proof
From Unit 6: a commutative ring is a field exactly when its only ideals are \(\{0\}\) and itself. By the correspondence theorem, the ideals of \(R/M\) match the ideals of \(R\) containing \(M\). So \(R/M\) has only two ideals exactly when only two ideals contain \(M\), namely \(M\) and \(R\). That's maximality.
A direct version, which shows how inverses arise: take \(a \notin M\). The ideal \(M + (a) = \{m + ra\}\) is strictly bigger than \(M\), so by maximality it's all of \(R\), and contains 1: \(1 = m + ra\) for some \(m \in M\), \(r \in R\). In the quotient, \(m\) becomes 0, so \(ra = 1\). The class of \(r\) is the inverse of the class of \(a\). ∎
The direct proof is Bézout's identity in disguise. For \(R = \mathbb{Z}\), \(M = (p)\) and \(a\) not divisible by \(p\): \(1 = mp + ra\), and \(r\) is the inverse of \(a\) mod \(p\), exactly the calculation from Unit 2.
Prime versus maximal
Every field is a domain, so every maximal ideal is prime. The converse fails, and a table makes the difference vivid:
| Ideal | Quotient | Prime? | Maximal? |
|---|---|---|---|
| \((5) \subset \mathbb{Z}\) | \(\mathbb{Z}/5\), a field | Yes | Yes |
| \((6) \subset \mathbb{Z}\) | \(\mathbb{Z}/6\), zero divisors | No | No |
| \((0) \subset \mathbb{Z}\) | \(\mathbb{Z}\), a domain | Yes | No |
| \((x) \subset \mathbb{Z}[x]\) | \(\mathbb{Z}\), a domain | Yes | No: \((x) \subsetneq (2, x)\) |
| \((2, x) \subset \mathbb{Z}[x]\) | \(\mathbb{Z}/2\), a field | Yes | Yes |
| \((x^2 + 1) \subset \mathbb{R}[x]\) | \(\mathbb{C}\), a field | Yes | Yes |
| \((x^2 - 1) \subset \mathbb{Q}[x]\) | \(\mathbb{Q} \times \mathbb{Q}\) | No | No |
Think of a chain of ideals as a ladder going up from \((0)\) to \(R\). In \(\mathbb{Z}\), prime ideals sit on just two rungs: \((0)\) at the bottom and the \((p)\) at the top. In \(\mathbb{Z}[x]\) there's an extra rung in between: \((0) \subsetneq (x) \subsetneq (2, x)\). The length of the longest chain of primes is the dimension of a ring. \(\mathbb{Z}\) has dimension 1, like a line; \(\mathbb{Z}[x]\) has dimension 2, like a plane. The picture at the end of this unit makes that literal.
Building fields
Now apply this to polynomials. Over a field \(F\), every ideal of \(F[x]\) is principal (Unit 8 proves this using long division), and \((f) \subseteq (g)\) means \(g \mid f\). So an ideal \((f)\) is maximal exactly when \(f\) has no factors except constants and constant multiples of itself. That's irreducibility.
Theorem
For a field \(F\) and a non-constant \(f \in F[x]\):
\[ F[x]/(f) \text{ is a field} \iff f \text{ is irreducible over } F. \]Its elements are represented by the polynomials of degree less than \(\deg f\).
The parallel with \(\mathbb{Z}\) is exact: \(\mathbb{Z}/(n)\) is a field when \(n\) is prime, and \(F[x]/(f)\) is a field when \(f\) is irreducible. Primes and irreducibles are the same idea in two rings.
Example: inverting in \(\mathbb{Q}[x]/(x^3 - 2)\)
\(x^3 - 2\) is irreducible over \(\mathbb{Q}\) (it's a cubic with no rational root), so the quotient is a field: a copy of \(\mathbb{Q}(\sqrt[3]{2})\) with \(x\) playing \(\sqrt[3]{2}\). What is the inverse of \(1 + x\)? Use the identity \((x + 1)(x^2 - x + 1) = x^3 + 1\). In the quotient \(x^3 = 2\), so the right side is \(3\). Therefore
\[ \frac{1}{1 + \sqrt[3]{2}} = \frac{\sqrt[3]{4} - \sqrt[3]{2} + 1}{3}. \]This is "rationalising the denominator", which you may have learned as a trick. It's really the field structure of a quotient ring. In general, Euclid's algorithm in \(F[x]\) finds the inverse.
The most important application is to finite fields. Over \(\mathbb{Z}/2\), the polynomial \(x^2 + x + 1\) is irreducible (Unit 4's sieve). So \((\mathbb{Z}/2)[x]/(x^2 + x + 1)\) is a field. Its elements are \(0, 1, \alpha, \alpha + 1\), where \(\alpha\) is the class of \(x\), and the rule \(\alpha^2 = \alpha + 1\) (since \(-1 = 1\)) handles all multiplication:
Addition
| + | 0 | 1 | α | α+1 |
|---|---|---|---|---|
| 0 | 0 | 1 | α | α+1 |
| 1 | 1 | 0 | α+1 | α |
| α | α | α+1 | 0 | 1 |
| α+1 | α+1 | α | 1 | 0 |
Multiplication
| × | 0 | 1 | α | α+1 |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | α | α+1 |
| α | 0 | α | α+1 | 1 |
| α+1 | 0 | α+1 | 1 | α |
Every non-zero element has an inverse (each non-zero row has exactly one purple 1), so this is a genuine field with four elements, written \(\mathbb{F}_4\). It is not \(\mathbb{Z}/4\), which has the zero divisor 2. Unit 10 is devoted to fields like this one.
Roots on demand
\(x^2 - 2\) has no root in \(\mathbb{Q}\). Historically, \(\sqrt2\) was found by geometry. But we can also manufacture a root algebraically: in \(K = \mathbb{Q}[x]/(x^2 - 2)\), the class of \(x\) squares to 2. This works for any polynomial over any field.
Theorem: Kronecker (1887)
Every non-constant polynomial over a field \(F\) has a root in some larger field \(K \supseteq F\).
Show proof
Let \(f\) have an irreducible factor \(g\). Then \(K = F[x]/(g)\) is a field. The map \(F \to K\) sending each constant to its class is a homomorphism out of a field, hence one-to-one (Unit 5), so \(K\) contains a copy of \(F\). Let \(\alpha\) be the class of \(x\). Then \(g(\alpha)\) is the class of \(g\), which is 0. So \(\alpha\) is a root of \(g\), hence of \(f\). ∎
Repeat the construction and you get a field where \(f\) splits into linear factors completely. No geometry, no limits, no approximation: roots are created by pure algebra. This is how the complex numbers looked to 19th-century algebraists, and it's how every finite field is built.
Going deeper: do maximal ideals always exist?
In \(\mathbb{Z}\), every proper ideal sits inside a maximal one, because every \(n > 1\) has a prime factor. Is that true in every ring? Yes, but for enormous rings the proof needs Zorn's lemma, an axiom equivalent to the axiom of choice: start with a proper ideal and keep enlarging it; Zorn's lemma guarantees that this process has a top. This result is Krull's theorem (1929). It implies, for instance, that every non-zero ring has at least one field as a quotient.
A map of the primes
Here's a way of thinking about rings that drove much of 20th-century mathematics. Treat each prime ideal of a ring as a point. The set of points is called the spectrum, \(\operatorname{Spec} R\).
The spectrum of \(\mathbb{Z}\) is a line of points: one for each prime \((2), (3), (5), \dots\), plus a special point \((0)\) that lies "everywhere at once". The spectrum of \(\mathbb{Z}[x]\) is two-dimensional. Above each prime \(p\) stands a vertical line: the primes of \((\mathbb{Z}/p)[x]\). And each irreducible polynomial traces a curve across the picture:
The pattern of splitting (at \(p = 5, 13, 17, \dots\), the primes \(\equiv 1 \pmod 4\)) is the subject of Unit 9, where it decides which primes are sums of two squares. This geometric way of seeing rings, algebraic geometry, is previewed in Unit 11.
Practice set
Mixed questions covering the whole unit. Try each one before opening the answer.
1. Which ideals of \(\mathbb{Z}/12\) are prime? Which are maximal? What are the quotients?
Show answer
\((2)\) and \((3)\), and they're both maximal: \((\mathbb{Z}/12)/(2) \cong \mathbb{Z}/2\) and \((\mathbb{Z}/12)/(3) \cong \mathbb{Z}/3\). The others fail: \((0)\) isn't prime since \(3 \cdot 4 = 0\); \((4)\) isn't since \(2 \cdot 2 \in (4)\); \((6)\) isn't since \(2 \cdot 3 \in (6)\).
2. Is \((x^2 + 1)\) maximal in \(\mathbb{Q}[x]\)? In \(\mathbb{C}[x]\)? In \((\mathbb{Z}/5)[x]\)? In \((\mathbb{Z}/7)[x]\)?
Show answer
\(\mathbb{Q}[x]\): yes, no rational roots. \(\mathbb{C}[x]\): no, \((x - i)(x + i)\). \((\mathbb{Z}/5)[x]\): no, \((x - 2)(x - 3)\). \((\mathbb{Z}/7)[x]\): yes, no square root of \(-1\) mod 7, giving a field with 49 elements.
3. In \(\mathbb{R}[x, y]\) (polynomials in two variables), show that \((x)\) is prime but not maximal, and \((x, y)\) is maximal.
Show answer
Setting \(x = 0\) gives \(\mathbb{R}[x, y]/(x) \cong \mathbb{R}[y]\), a domain but not a field, so \((x)\) is prime but not maximal. Setting \(x = y = 0\) gives \(\mathbb{R}[x, y]/(x, y) \cong \mathbb{R}\), a field, so \((x, y)\) is maximal. Geometrically, \((x)\) is the \(y\)-axis (a line) and \((x, y)\) is the origin (a point).
4. In \(\mathbb{Q}[x]/(x^3 - 2)\), find the inverse of \(x\).
Show answer
\(x \cdot x^2 = x^3 = 2\), so \(x \cdot \tfrac{x^2}{2} = 1\). The inverse is \(\tfrac12 x^2\): indeed \(1/\sqrt[3]{2} = \sqrt[3]{4}/2\).
5. In \(\mathbb{F}_4\), compute \(\alpha^3\). Then list the powers \(\alpha, \alpha^2, \alpha^3\).
Show answer
\(\alpha^2 = \alpha + 1\), so \(\alpha^3 = \alpha^2 + \alpha = 2\alpha + 1 = 1\). The powers are \(\alpha, \alpha + 1, 1\): all three non-zero elements. So \(\alpha\) generates the units, just like the primitive roots of Unit 2.
6. Prove directly that a maximal ideal is prime, without using quotient rings.
Show answer
Suppose \(ab \in M\) and \(a \notin M\). Then \(M + (a) = R\), so \(1 = m + ra\). Multiply by \(b\): \(b = mb + rab\). Both \(mb\) and \(r(ab)\) are in \(M\), so \(b \in M\).
Summary and glossary
- A prime ideal contains a factor whenever it contains a product. \(R/P\) is a domain exactly when \(P\) is prime.
- A maximal ideal has nothing between it and \(R\). \(R/M\) is a field exactly when \(M\) is maximal.
- Maximal implies prime, but not conversely: \((0)\) in \(\mathbb{Z}\), \((x)\) in \(\mathbb{Z}[x]\).
- \(F[x]/(f)\) is a field exactly when \(f\) is irreducible. This builds \(\mathbb{C}\), \(\mathbb{Q}(\sqrt[3]{2})\) and \(\mathbb{F}_4\).
- Kronecker: every polynomial has a root in some bigger field.
- Prime ideals can be pictured as points of a space, the spectrum.
- Prime ideal
- \(P \neq R\) with \(ab \in P \Rightarrow a \in P\) or \(b \in P\).
- Maximal ideal
- A proper ideal with no proper ideal above it.
- Euclid's lemma
- A prime dividing a product divides a factor.
- \(\mathbb{F}_4\)
- The field with four elements.
- Kronecker's theorem
- Roots can always be adjoined by a quotient.
- Krull dimension
- The length of the longest chain of prime ideals.
- Spectrum
- \(\operatorname{Spec} R\): the set of prime ideals, viewed as a space.