12 Ideals in number fields
12.1 Ideals
Let \(K\) be a number field. We’re going to study ideals in the ring of integers of \(K.\) (The zero ideal is an ideal, but it’s not very interesting, so henceforth, when we say “ideal” we always mean nonzero ideal.)
Definition 12.1 • Notation for ideals
For any commutative ring \(R\) and elements \(x_1, \dots, x_k\) of \(R,\) write \(\langle x_1, \dots, x_k \rangle_R\) for the set \(\{r_1 x_1 + \dots + r_k x_k : r_1, \dots, r_k \in R\},\) which is an ideal of \(R\) (the ideal generated by the \(x_i\)). We omit the subscript \(R\) if it’s obvious from context.
Notice that any \(\alpha \in \mathcal{O}_K\) gives us an ideal – the principal ideal \(\langle \alpha\rangle = \{ \alpha x : x \in \mathcal{O}_K\}.\) However, since integer rings aren’t always PIDs, there can be more ideals which aren’t of this form.
Example 12.2
Let \(R = \mathbb{Z}[\sqrt{-5}] = \mathcal{O}_{\mathbb{Q}(\sqrt{-5})},\) and let \(I\) be the ideal \(\langle 2, 1 - \sqrt{-5}\rangle\) of \(R.\) We claim this ideal is not principal.
Assume for contradiction that \(\alpha\) is a generator. Then \(\alpha\) must divide 2, so \(N(\alpha) \mid N(2) = 4\); and also \(N(\alpha) \mid N(1 - \sqrt{-5}) = 6.\) So \(N(\alpha)\) must be 1 or 2.
If \(N(\alpha)\) were equal to 1, then \(I\) would be the unit ideal. But this is not possible, since every element of \(I\) has the form \(x + y \sqrt{-5}\) with \(x = y \bmod 2\) (exercise!), so \(1 \notin I.\)
Hence \(N(\alpha)\) must be 2. But the equation \(x^2 + 5y^2 = 2\) obviously has no solutions, so we have a contradiction. 0◻
Exercise 12.3
Generalise the above! Show that if \(d \in \mathbb{N}_+\) is square-free with \(d \ne 3 \bmod 4,\) \(p \nmid d\) is a prime such that \(\left(\frac{-d}{p}\right) = 1,\) and \(t\) is a square root of \(-d \bmod p,\) then the ideal \(\langle p, \sqrt{-d} - t\rangle\) is principal in \(\mathbb{Z}[\sqrt{-d}]\) if and only if \(x^2 + d y^2 = p\) has an integer solution. Can you formulate an analogue for \(d = 1 \bmod 4\)? What about \(d < 0\)?
Definition 12.4 • Product of ideals
Let \(I\) and \(J\) be ideals in \(\mathcal{O}_K.\) Then we define \[I J = \{ i \cdot j : i \in I, j \in J\}.\]
You should check that ideal multiplication is compatible with element multiplication, i.e. \(\langle \alpha \rangle \langle \beta \rangle = \langle \alpha \beta \rangle.\) Moreover, \(\langle \alpha\rangle = \langle \beta \rangle\) iff \(\alpha\) and \(\beta\) are associates. So we get maps \[\left(\mathcal{O}_K - \{0\}\right)\ \twoheadrightarrow\ \frac{\left(\mathcal{O}_K - \{0\}\right)}{\{ \text{units}\}} \ \hookrightarrow\ \{ \text{nonzero ideals} \},\tag{\dag}\] which are compatible with multiplication (and send the identity to the identity). If \(\mathcal{O}_K\) is a PID (in particular if it’s Euclidean), then the second map is a bijection.
The moral of the next few sections will be that there is always a notion of “unique prime factorisation” for ideals. When \(\mathcal{O}_K\) is a PID, we get unique factorisation for elements from this using the bijectivity of the second map in (\(\dag\)). Conversely, when \(\mathcal{O}_K\) is not a PID, we never have unique prime factorisation in \(\mathcal{O}_K\); but the non-principal ideals are precisely the “extra stuff” we need to add to get unique factorisation back again.
12.2 Factoring ideals
Remember that an ideal \(I\) in any (commutative) ring \(A\) is said to be a prime ideal if \(I \ne A,\) and for all \(x, y \in A\) we have \(xy \in I \Rightarrow x \in I\) or \(y \in I.\) This obviously generalises the definition of prime elements: an element is prime iff the principal ideal it generates is a prime ideal.
Proposition 12.5
Let \(I\) be a non-zero ideal in \(\mathcal{O}_K,\) for \(K\) a number field. Then \(I\) is prime if and only if it is maximal, i.e. \(I \ne \mathcal{O}_K\) and there is no ideal \(J\) such that \(I \supsetneqq J \supsetneqq \mathcal{O}_K.\)
Proof. We know that \(I\) is prime iff \(R = \mathcal{O}_K / I\) is an integral domain (this is just rewriting the definition).
We claim that
this quotient \(R\) is finite,
a finite integral domain is automatically a field.
One can show using Remark 11.20 that for any \(0 \ne \alpha \in \mathcal{O}_K\) we have \(\# (\mathcal{O}_K / \alpha) = |\operatorname{Nm}_{K/\mathbb{Q}}(\alpha)|.\)
To prove (b), suppose \(R\) is an integral domain and \(0 \ne x \in R.\) Then multiplication by \(x\) is a map \(R \to R\) which is injective, by the integral-domain assumption. But an injection from a finite set to itself must be a bijection; so 1 is in the image and hence \(x\) is invertible.
To finish the proof, we note that for any commutative ring \(A\) and ideal \(I\) of \(A,\) the ideal \(I\) is maximal iff \(A / I\) is a field (exercise). So \[( \text{$I$ prime}) \iff (\text{$R$ int. domain}) \iff (\text{$R$ field}) \iff (\text{$I$ maximal}).\]
Corollary 12.6
Let \(0 \ne \alpha \in \mathcal{O}_K.\) Then:
\(\alpha\) is a prime element iff there is no ideal strictly containing \(\langle \alpha \rangle\) except the unit ideal.
\(\alpha\) is indecomposable iff there is no principal ideal strictly containing \(\langle \alpha \rangle\) except the unit ideal.
Proof. The first assertion is just the previous proposition applied to \(\langle \alpha \rangle.\) The second is obvious, since \(\langle \beta \rangle \supset \langle \alpha \rangle\) iff \(\beta \mid \alpha.\)
In particular, if \(\mathcal{O}_K\) is a PID, then prime elements and indecomposable elements coincide (something you saw without proof in the Algebra module).
Theorem 12.7 • Dedekind
Let \(I, J\) be ideals in \(\mathcal{O}_K\) with \(I \subseteq J.\) Then there exists an ideal \(H\) such that \(I = H J.\)
This is surprisingly hard, and we’re not going to prove it in this course. For a proof see Stewart & Tall.
Remark 12.8
This theorem would be false if we replaced \(\mathcal{O}_K\) with a ring like \(\mathbb{Z}[\sqrt{-3}],\) which isn’t equal to the full ring of integers of its parent number field.
Corollary 12.9
Multiplying by a non-zero ideal is injective: that is, if \(H, I, J\) are (nonzero!) ideals of \(\mathcal{O}_K,\) and \(HI = HJ,\) then \(I = J.\)
Proof. Firstly, we suppose \(H\) is principal, say \(H = \langle x \rangle.\) Then \(HI\) is exactly the set of elements \(x i : i \in I,\) and similarly \(H J.\) Since multiplication by \(x\) is injective, it follows that \(I = \{ y : xy \in HI\} = \{ y : xy \in HJ\} = J.\)
For a general ideal \(H,\) we choose a non-zero element \(x \in H.\) Then \(H \supseteq \langle x \rangle,\) so \(\langle x \rangle = H' H\) for some \(H'.\) So if \(H I = H J\) then \(H' H I = H' H J,\) i.e. \(\langle x \rangle I = \langle x \rangle J,\) and the previous paragraph shows that \(I = J.\)
Theorem 12.10 • Unique factorisation of ideals
Any nonzero ideal of \(\mathcal{O}_K\) is equal to a product of finitely many prime ideals, and its expression in this form is unique up to ordering.
Proof. For any \(I,\) there are finitely many ideals containing \(I,\) since they biject with the ideals of the finite quotient ring \(\mathcal{O}_K / I.\) Hence we can find one which is maximal (not contained in any other ideal). Let \(P\) be such an ideal. Then \(P\) divides \(I,\) so \(I = P J\) for some \(J.\)
Clearly \(J\) can’t be equal to \(I,\) since if \(I = PI\) then \(\mathcal{O}_K = P,\) a contradiction. So \(J\) is strictly larger than \(I.\) By induction on the size of \(\mathcal{O}_K / I,\) we may assume that \(J\) is a product of maximal ideals, hence so is \(I.\)
The proof of uniqueness proceeds exactly as before.
Example 12.12 • important
Let’s use what we know about unique factorization of ideals to understand better how unique factorisation of elements fails in \(\mathbb{Z}[\sqrt{-5}].\)
Remember that we had two different factorisations of 6 into indecomposable elements: \[6 = 2 \cdot 3 = (1 + \sqrt{-5}) \cdot (1 - \sqrt{-5}),\] One checks that the ideals \[\begin{aligned} \mathfrak{p} &= \langle 2, 1 + \sqrt{-5}\rangle,\\ \mathfrak{q}_1 &= \langle 3, 1 + \sqrt{-5}\rangle,\\ \mathfrak{q}_2 &= \langle 3, 1 - \sqrt{-5}\rangle \end{aligned}\] are all prime; but none of them can be principal, since that would contradict the indecomposability of \(2\) and \(3\) in \(\mathbb{Z}[\sqrt{-5}].\)
Now, one can show (exercise!) \[\begin{aligned} \mathfrak{p}^2 &= \langle 2\rangle, & \mathfrak{q}_1 \mathfrak{q}_2 &= \langle 3 \rangle, \\ \mathfrak{p} \mathfrak{q}_1 &= \langle 1 + \sqrt{-5}\rangle, & \mathfrak{p} \mathfrak{q}_2 &= \langle 1 - \sqrt{-5}\rangle. \end{aligned}\]
So the (unique) factorisation of the ideal \(\langle 6 \rangle\) is \[\langle 6 \rangle = \mathfrak{p}^2 \mathfrak{q}_1 \mathfrak{q}_2,\] and the rival factorisations of the element 6 into indecomposables correspond to the ways of grouping the factors into subsets whose product is principal: \[\langle 6 \rangle = (\mathfrak{p}^2) (\mathfrak{q}_1 \mathfrak{q}_2) = (\mathfrak{p} \mathfrak{q}_1) (\mathfrak{p} \mathfrak{q}_2).\]
Exercise 12.11
Compute the factorisation of \(\langle 21 \rangle\) into prime ideals in \(\mathbb{Z}[\sqrt{-5}].\) Hence show that there are exactly 3 distinct factorisations of \(21\) into indecomposable elements, up to units and re-ordering.
12.3 The class group
We’re now going to cook up an algebraic object which measures how badly ideals can fail to be principal (and thus how badly unique factorisation fails for elements).
Definition 12.13
A fractional ideal of \(\mathcal{O}_K\) is a subset of \(K\) of the form \[\mathcal{O}_K x_1 + \dots + \mathcal{O}_K x_n\] for some \(x_1, \dots, x_n \in K\) (not all of which are zero).
Thus, a fractional ideal contained in \(\mathcal{O}_K\) is just an ideal, but things like \(\tfrac{1}{2} \mathcal{O}_K\) are also fractional ideals.
Note that one can multiply fractional ideals to get new fractional ideals; and it follows from Dedekind’s theorem that every fractional ideal has an inverse. Along with some easy checks for associativity etc, this shows that fractional ideals form an abelian group.
Definition 12.14
The class group of \(K\) is the quotient \[\operatorname{Cl}_K = \frac{\{\text{fractional ideals}\}}{\{\text{principal fractional ideals}\}}.\]
We’ll now state one of the most important theorems in algebraic number theory:
Theorem 12.15
For any number field \(K,\) the class group \(\operatorname{Cl}_K\) is finite.
There is a simpler proof in the special case of imaginary quadratic fields, i.e. \(\mathbb{Q}(\sqrt{-D})\) with \(D > 0,\) which would be a nice project for a bachelor thesis.
Example 12.16
Going back to Example 12.12, the ideal \(\mathfrak{p}\) is not principal (since \(x^2 + 5y^2 = 2\) has no solutions) but \(\mathfrak{p}^2\) is principal, so \([\mathfrak{p}]\) is a nontrivial element of \(\operatorname{Cl}_K\) of order 2. Since \(\mathfrak{p}\mathfrak{q}_1\) and \(\mathfrak{p}\mathfrak{q}_2\) are principal, all three of the ideals \(\{ \mathfrak{p}, \mathfrak{q}_1, \mathfrak{q}_2\}\) all lie in this nontrivial ideal class.
It turns out that this is the only non-trivial element of the class group, so \(\operatorname{Cl}_K \cong C_2.\)
12.4 Cyclotomic fields, and Fermat’s Last Theorem
Definition 12.17
The \(n\)-th cyclotomic field is the number field \(\mathbb{Q}(\zeta_n),\) where \(\zeta_n = \exp(2 \pi i / n).\)
This is indeed a number field, because \((\zeta_n)^n = 1,\) so \(\zeta_n\) is algebraic. One can check that the ring of integers is equal to \(\mathbb{Z}[\zeta_n].\)
Theorem 12.18 • Kummer
Let \(p\) be an odd prime, and suppose that \(p\) does not divide the order of the class group of the \(p\)-th cyclotomic field \(\mathbb{Q}(\zeta_p).\) Then there are no solutions to Fermat’s equation \(x^n + y^n = z^n\) with \(n\) divisible by \(p.\)
The idea of Kummer’s proof was to write \(y^n = x^n - z^n\) and factor this in \(\mathbb{Z}[\zeta_p]\) as \((x - z)(x - \zeta_p z) \dots (x - \zeta_p^{p-1} z).\) For simplicity, suppose \(xyz \ne 0 \bmod p\); then one can show that the factors on the right are pairwise coprime.
If \(\mathbb{Z}[\zeta_p]\) were a PID, then – by considering prime factorisations – each of the terms must itself be a \(p\)-th power (up to units); and this eventually gives enough information to deduce that no such \(x, y, z\) exist.
Kummer realised that one can push through the same argument as long as the class group has prime-to-\(p\) order (it doesn’t have to be trivial), and this seems to hold for “most” primes \(p\) – there are very few primes such that the class group is trivial, but lots for which it has prime-to-\(p\) order. So this proves Fermat’s Last Theorem for a large set of exponents \(n,\) although not all of them.
Remark 12.19
Several earlier mathematicians had tried to make such an argument assuming that unique factorisation worked in \(\mathbb{Z}[\zeta_p],\) which is of course false in general. Kummer invented the whole machine of ideal theory and class groups in order to sort out the mess!