10 Arithmetic in number fields
10.1 Number fields
Remember the following definition from Algebra:
Definition 10.1
For \(\alpha \in \mathbb{C},\) we say \(\alpha\) is an algebraic number if there is a non-constant polynomial \(f(X) \in \mathbb{Q}[X]\) with \(f(\alpha) = 0\); and we write \(\overline{\mathbb{Q}}\) for the set of all algebraic numbers.
Moreover, you saw that:
If \(\alpha\) is algebraic, then there is a unique “simplest” polynomial that it satisfies – the minimal polynomial of \(\alpha,\) which is the smallest-degree monic \(f\) with \(f(\alpha) = 0.\)
For any \(\alpha \in \mathbb{C},\) there is a unique smallest subfield10 \(\mathbb{Q}(\alpha) \subset \mathbb{C}\) containing \(\alpha,\) and \(\alpha\) is algebraic if and only if \(\mathbb{Q}(\alpha)\) is finite-dimensional over \(\mathbb{Q}.\)Pedantic notational remark: by definition \(\mathbb{Q}(\alpha)\) is the smallest subfield of \(\mathbb{C}\) containing \(\mathbb{Q}\) and \(\alpha,\) while \(\mathbb{Q}[\alpha]\) is the smallest subring of \(\mathbb{C}\) containing \(\mathbb{Q}\) and \(\alpha\); we have \(\mathbb{Q}[\alpha] \subseteq \mathbb{Q}(\alpha),\) because every subfield is a subring, and equality holds iff \(\alpha\) is algebraic (e.g. \(\tfrac{1}{\pi} \notin \mathbb{Q}[\pi].\)) In this module we only care about the algebraic case, so it doesn’t matter if we write \(\mathbb{Q}(\alpha)\) or \(\mathbb{Q}[\alpha]\); we’re going to standardize on \(\mathbb{Q}(\alpha).\)
If \(\alpha\) is algebraic, then \(\mathbb{Q}(\alpha)\) has basis \(\{1, \alpha, \dots, \alpha^{d-1}\}\) where \(d\) is the degree of its minimal polynomial.
\(\overline{\mathbb{Q}}\) is a field.
We’re going to study “little pieces” of \(\overline{\mathbb{Q}},\) rather than all of \(\overline{\mathbb{Q}}\) at once:
Definition 10.2
A number field is a subfield of \(\mathbb{C}\) which is finite-dimensional as a \(\mathbb{Q}\)-vector space.
Example 10.3
The field \(\mathbb{Q}(i) = \{ a + b i : a, b \in \mathbb{Q}\}\) is a number field, with \([\mathbb{Q}(i) : \mathbb{Q}] = 2.\)
Note that any number field must be contained in \(\overline{\mathbb{Q}}\): if \(K\) is a number field, and \(\alpha \in K,\) then \(\mathbb{Q}(\alpha) \subseteq K.\) Since \(K\) has finite dimension, so does \(\mathbb{Q}(\alpha),\) hence \(\alpha\) is algebraic. Conversely, for any \(\alpha \in \overline{\mathbb{Q}},\) the field \(\mathbb{Q}(\alpha)\) is a number field.
What’s less obvious, but true, is that every number field can be written in this form: for any number field \(K \subset \mathbb{C},\) we can find some \(\alpha \in K\) such that \(K = \mathbb{Q}(\alpha)\) (a “primitive element” for \(K\)).
Example 10.4
Let’s start with \(\mathbb{Q},\) and let \(K = \mathbb{Q}(i)\) be the extension of \(\mathbb{Q}\) generated by \(i\); and then let \(L\) be the extension of \(K\) generated by \(\sqrt{2}\) (which is not in \(K\)).
Then \(L\) is an extension of \(\mathbb{Q}\) of degree 4: a \(\mathbb{Q}\)-vector-space basis is given by \(\{1, i, \sqrt{2}, \sqrt{-2}\}.\)
Clearly none of these basis elements is a primitive element, but one can check that \(\alpha = i + \sqrt{2}\) is a primitive element: the powers of \(i + \sqrt{2}\) are a basis of \(L.\)
Exercise 10.5
For \(\alpha\) as in the example, write each of \(1, \alpha, \alpha^2, \alpha^3, \alpha^4\) in terms of the basis \(\{1, i, \sqrt{2}, \sqrt{-2}\}.\) Hence verify that \(\{1, \alpha, \alpha^2, \alpha^3\}\) span \(L\) as a \(\mathbb{Q}\)-vector space, and calculate the minimal polynomial of \(\alpha.\)
Remark 10.6 • Non-examinable
Here is a sketch of why every number field has a primitive element. It follows from Galois theory (cf. Algebra script) that for any number field \(K,\) there are only finitely many possible subfields \(K'\) with \(\mathbb{Q}\subseteq K' \subsetneq K.\) So the union of these subfields can’t be the whole of \(K,\) and we can choose an \(\alpha \in K\) which isn’t contained in any smaller field. This must be a primitive element for \(K.\)
(This also makes it clear that primitive elements are very non-unique; in some sense “most” elements of \(K\) are primitive elements.)
10.2 Algebraic integers
We’d like to find more examples of rings like \(\mathbb{Z}[i]\) and \(\mathbb{Z}[\omega]\) above, which have interesting factorisation theories attached to them. Number fields themselves are not interesting in this way (in a field, every non-zero element is a unit). We want to pick out those algebraic numbers which “don’t have any denominators” in some sense, just like \(\mathbb{Z}[i]\) inside \(\mathbb{Q}(i).\)
It turns out the good definition is the following:
Definition 10.7
We say \(\alpha \in \mathbb{C}\) is an algebraic integer if there exists a monic polynomial \(f(X) \in \mathbb{Z}[X]\) with \(f(\alpha) = 0.\) We write \(\bar{\mathbb{Z}}\) for the set of algebraic integers.
Note the similarity to the definition of “algebraic number”; but here it really matters that \(f\) be monic. (Exercise: show that for any algebraic number \(\alpha,\) we can find an \(f \in \mathbb{Z}[X],\) usually not monic, with \(f(\alpha) = 0.\))
Example 10.8
Clearly we have \(\mathbb{Z}\subseteq \bar{\mathbb{Z}},\) since for any \(n \in \mathbb{Z},\) \(f(X) = X - n\) is a monic polynomial that it satisfies. Moreover, if \(n \in \mathbb{Z}\) then \(\sqrt{n} \in \bar{\mathbb{Z}}.\)
Less obviously, \(\omega = \tfrac{-1 + \sqrt{-3}}{2} \in \bar{\mathbb{Z}},\) since it satisfies \(X^2 + X + 1 = 0.\)
Exercise 10.9
Show that if \(\alpha \in \overline{\mathbb{Z}},\) then \(\sqrt{\alpha} \in \overline{\mathbb{Z}}.\)
Proposition 10.10
For any algebraic number \(\alpha,\) there exists some \(N \in \mathbb{N}_+\) such that \(N\alpha \in \bar{\mathbb{Z}}.\)
Proof. Exercise. (Hint: if \(f(X) = X^n + a_{n-1} X^{n-1} + \dots + a_0 \in \mathbb{Q}[X]\) is the minimal polynomial of \(\alpha,\) and \(\beta = N\alpha\) for some \(N,\) then what is the minimal polynomial of \(\beta\)?)
What’s less obvious is how one would show that anything is not an algebraic integer! Fortunately, we have the following criterion:
Proposition 10.11
An algebraic number \(\alpha \in \mathbb{C}\) is an algebraic integer if and only if its minimal polynomial has integer coefficients.
Proof. Let \(f \in \mathbb{Q}[X]\) be the minimal polynomial of \(\alpha.\) If \(f \in \mathbb{Z}[X],\) then clearly \(f\) is an algebraic integer.
Conversely, suppose \(f\) does not have integer coefficients, but there is some (larger-degree) monic integral polynomial \(h\) with \(h(\alpha) = 0.\) Then we must have \(h(X) = f(X) g(X)\) for some \(g \in \mathbb{Q}[X].\)
Let \(C\) be the least common multiple of the denominators of the coefficients of \(f,\) so that \(C f \in \mathbb{Z}[X],\) and similarly \(D\) for \(g.\) Then we clearly have \((C f) (D g) = (CD)h.\) Now let \(p\) be a prime dividing \(CD.\) Clearly at least one coefficient of \(C f\) is not divisible by \(p\) (since otherwise \(C / p\) would be the LCM of the denominators). Similarly at least one of the coefficients of \(D g\) is not divisible by \(p.\) So \(Cf \bmod p\) and \(D g \bmod p\) are non-zero in \(\mathbb{F}_p[X].\) But their product \(C D h\) is zero, since \(p \mid CD\) and \(h\) has integral coefficients. This contradicts the fact that \(\mathbb{F}_p[X]\) is an integral domain. So \(CD\) must in fact be 1, i.e. both \(f\) and \(g\) are integral.
Example 10.12
If \(x \in \mathbb{Q}- \mathbb{Z},\) then \(x\) is not an algebraic integer. (That is, we have \(\bar{\mathbb{Z}} \cap \mathbb{Q}= \mathbb{Z}\)).
The number \(\tfrac{1 + \sqrt{3}}{2}\) is not an algebraic integer: it is a root of the polynomial \(x^2 - x - \tfrac{1}{2},\) and since it clearly isn’t in \(\mathbb{Q},\) this must be the minimal polynomial.
It follows that a rational number is an algebraic integer iff it’s an integer in the usual sense.
Exercise 10.13 • Warning!
Give a counterexample to show that is not true that if \(\alpha\) is an algebraic integer, then every monic polynomial that \(f\) satisfies has to have integral coefficients.
10.3 Arithmetic with algebraic integers
For doing arithmetic with algebraic integers, the following characterisation is useful:
Proposition 10.14
\(\alpha \in \mathbb{C}\) is an algebraic integer if and only if \(\mathbb{Z}[\alpha]\) is finitely generated as an abelian group.
Proof. If \(\alpha\) satisfies a polynomial \(f(X) = X^n + a_{n-1} X^{n - 1} + \dots,\) then \(\alpha^{n}\) is in the \(\mathbb{Z}\)-span of \(1, \dots, \alpha^{n-1},\) and by induction one can show that \(\alpha^{n+1},\) \(\alpha^{n + 2}\) etc are also in this span.
Conversely, if this group is finitely generated, then each generator can only mention finitely many powers of \(\alpha,\) so there is some \(N\) such that \(\{1, \dots, \alpha^N\}\) is a generating set. Hence \(\alpha^{N + 1}\) is in the \(\mathbb{Z}\)-span of \(\{1, \dots, \alpha^N\},\) giving a monic integral polynomial that \(\alpha\) satisfies.
Corollary 10.15
If \(\alpha, \beta\) are algebraic integers then so are \(\alpha \pm \beta\) and \(\alpha \beta.\)
Proof. Suppose \(\alpha,\) \(\beta\) satisfy polynomials of degree \(M, N\) respectively. Consider the subgroup of \(\mathbb{C}\) generated by \(\{ \alpha^i \beta^j : 0 \leqslant i < N, 0 \leqslant j < M\}.\) This is finitely generated and contains \(\alpha^r \beta^s\) for all \(r, s \in \mathbb{N},\) so in particular it contains \((\alpha \beta)^j\) and \((\alpha \pm \beta)^k\) for all \(j, k.\) Since a subgroup of a finitely generated abelian group is finitely generated, the result follows.
Thus the set \(\overline{\mathbb{Z}}\) of all algebraic integers is a subring of of \(\mathbb{C}.\)
Remark 10.16
Note that the above proofs are not constructive: we’ve proved that \(\alpha \pm \beta\) and \(\alpha \beta\) satisfy monic polynomials in \(\mathbb{Z}[X],\) but we haven’t shown how to explicitly write down those polynomials.
Exercise 10.17
Find a monic polynomial \(f(X) \in \mathbb{Z}[X]\) with \(f(\sqrt{2} + \sqrt{3}) = 0.\)
10.4 Rings of integers
Definition 10.18
If \(K\) is a number field, then we define \(\mathcal{O}_K,\) the ring of integers of \(K\), as \(K \cap \bar{\mathbb{Z}}.\)
Note that if \(\alpha\) is an algebraic integer, \(\mathbb{Z}[\alpha]\) is contained in the ring of integers of \(\mathbb{Q}(\alpha),\) but it might be smaller. For instance, \(\mathbb{Z}[\sqrt{-3}]\) is not the ring of integers of \(\mathbb{Q}(\sqrt{-3}),\) because it doesn’t contain \(\omega.\)
Proposition 10.19 • Rings of integers of quadratic fields
Let \(d \in \mathbb{Z}\) with \(d \ne 1,\) and suppose \(d\) is not divisible by \(n^2\) for any \(n > 1\) (\(d\) is “square-free”). Then the ring of integers of \(\mathbb{Q}(\sqrt{d})\) is given by \[\mathcal{O}_{\mathbb{Q}(\sqrt{d})} = \begin{cases} \mathbb{Z}\left[\tfrac{1 + \sqrt{d}}{2}\right] & \text{if $d = 1 \bmod 4$}, \\ \mathbb{Z}[\sqrt{d}] & \text{otherwise}. \end{cases}\]
Proof. First, note that \(\tfrac{1 + \sqrt{d}}{2}\) is a root of \(X^2 - X + \tfrac{1 - d}{4},\) so it is an algebraic integer iff \(d = 1 \bmod 4.\)
Conversely, let \(\alpha = u + v \sqrt{d}\) with \(u, v \in \mathbb{Q},\) and suppose \(\alpha \in \bar{\mathbb{Z}}.\) Then \(\alpha' = u - v \sqrt{d}\) is also in \(\bar{\mathbb{Z}},\) since it satisfies the same polynomial that \(\alpha\) does; and hence \(\alpha + \alpha' = 2u \in \bar{\mathbb{Z}} \cap \mathbb{Q}= \mathbb{Z}.\) Similarly, \(\alpha - \alpha' = 2v \sqrt{d} \in \bar{\mathbb{Z}}\); thus \((2v)^2 d \in \mathbb{Z},\) but since \(d\) is squarefree, this implies that \(2v \in \mathbb{Z}.\)
So, if \(\alpha\) is an algebraic integer but doesn’t lie in \(\mathbb{Z}[\sqrt{d}],\) then we can subtract a \(\mathbb{Z}\)-linear combination of \(1\) and \(\sqrt{d}\) to deduce that one of \(\{ \tfrac{1}{2}, \tfrac{\sqrt{d}}{2}, \tfrac{1 + \sqrt{d}}{2}\}\) is an algebraic integer. Clearly \(\tfrac{1}{2}\) and \(\tfrac{\sqrt{d}}{2}\) are never algebraic integers (since \(4 \nmid d\)); and \(\tfrac{1 + \sqrt{d}}{2}\) is an algebraic integer iff \(d = 1 \bmod 4.\)
Remark 10.20
Note that \(\mathcal{O}_{\mathbb{Q}(\sqrt{d})}\) is isomorphic to \(\mathbb{Z}^2\) as an abelian group: every element can be written uniquely in the form \(a + b \lambda\) for \(a, b \in \mathbb{Z},\) where \(\lambda = \frac{1 + \sqrt{d}}{2}\) or \(\sqrt{d}\) respectively.
We will prove in the next chapter that for any number field \(K,\) \(\mathcal{O}_K\) is isomorphic to \(\mathbb{Z}^d\) as an abelian group, where \(d = [K : \mathbb{Q}]\); but this requires a little more work.
Exercise 10.21
Use Propositions 10.14 and 10.19 to justify the claim we made in Chapter 8 that there are no rings “finitely larger than” \(\mathbb{Z}[\sqrt{-5}].\)
We finish this section with a useful little result which will be helpful later on:
Proposition 10.22
For any number field \(K\) and any non-zero \(\alpha \in \mathcal{O}_K,\) there exists a non-zero \(\beta \in \mathcal{O}_K\) such that \(\alpha \beta \in \mathbb{Z}.\) That is, \(\alpha\) divides some non-zero integer.
Proof. This is a disguised version of Proposition 10.10. Let \(\gamma = 1 / \alpha.\) Then \(\gamma \in \overline{\mathbb{Q}},\) so there is some \(N \in \mathbb{N}_+\) such that \(N\gamma\) is an algebraic integer. Let \(\beta = N\gamma\) for any such \(N.\) Then \(\beta = N / \alpha\) is in \(K,\) and it’s an algebraic integer, so it’s in \(\mathcal{O}_K\); and we have \(\alpha \beta = N.\)
11 Determining the integer ring
We’ll now study the ring \(\mathcal{O}_K,\) for \(K\) a number field, a bit more closely.
11.1 Norm and trace
If \(K\) is a number field, and \(x \in K,\) then we can consider the “multiplication by \(x\)” map \(M_x : K \to K,\) defined by \(M_x(y) = xy.\) This is clearly \(\mathbb{Q}\)-linear.
Definition 11.1
The norm \(\operatorname{Nm}_{K / \mathbb{Q}}(x)\) and trace \(\operatorname{Tr}_{K/\mathbb{Q}}(x)\) are the determinant and trace (in the sense of linear algebra) of \(M_x,\) viewed as a \(\mathbb{Q}\)-linear map \(K\to K.\)
One checks easily that norm is compatible with multiplication, and trace compatible with addition: \[\begin{aligned} \operatorname{Nm}_{K / \mathbb{Q}}(xy) &= \operatorname{Nm}_{K / \mathbb{Q}}(x)\operatorname{Nm}_{K / \mathbb{Q}}(y), \\ \operatorname{Tr}_{K / \mathbb{Q}}(x \pm y) &= \operatorname{Tr}_{K / \mathbb{Q}}(x) \pm \operatorname{Tr}_{K / \mathbb{Q}}(y). \end{aligned}\] Moreover, if \(x \ne 0,\) then taking \(y = x^{-1}\) in the first equation we deduce that \(\operatorname{Nm}_{K/\mathbb{Q}}(x) \ne 0,\) so \(\operatorname{Nm}_{K/\mathbb{Q}}\) is a group homomorphism \(K^\times \to \mathbb{Q}^\times.\)
Example 11.2
Let \(K = \mathbb{Q}(\sqrt{d})\) for a square-free integer \(d,\) and \(x = a + b \sqrt{d}.\) We claim that \[\operatorname{Tr}_{K / \mathbb{Q}}(x) = 2a, \qquad \operatorname{Nm}_{K / \mathbb{Q}}(x) = a^2 - d b^2.\]
To prove this, consider the basis \(\{1, \sqrt{d}\}\) of \(K.\) In this basis, the matrix of \(M_x\) is \[M_x = \begin{pmatrix} a & db \\ b & a \end{pmatrix},\] and the result is now clear.
Remark 11.3
Notice that this depends on \(K\): if we have two number fields \(K, L,\) and \(x \in K \cap L,\) then \(\operatorname{Tr}_{K / \mathbb{Q}}(x)\) and \(\operatorname{Tr}_{L / \mathbb{Q}}(x)\) are both well-defined, but they aren’t the same in general. So it is a little dangerous to write “\(\operatorname{Tr}(x)\)” without specifying \(K,\) although we’ll allow ourselves to do this sometimes when \(K\) is clear from context.
(Thus, for quadratic number fields, \(\operatorname{Nm}_{K / \mathbb{Q}}(x)\) is what we were calling \(N(x)\) before.)
Proposition 11.4
If \(\alpha \in \mathcal{O}_K,\) then \(\operatorname{Nm}_{K / \mathbb{Q}}(\alpha)\) and \(\operatorname{Tr}_{K / \mathbb{Q}}(\alpha)\) are in \(\mathbb{Z}.\)
Proof. Let’s suppose first that \(K = \mathbb{Q}(\alpha).\) Then the numbers \(1, \alpha, \alpha^2, \dots, \alpha^{d-1}\) are a \(\mathbb{Q}\)-basis of \(K,\) where \(d = [K: \mathbb{Q}].\) In this basis, the matrix of \(M_\alpha\) looks like \[\begin{pmatrix} 0 & 0 & 0 & \dots & \star\\ 1 & 0 & 0 & \dots & \star\\ 0 & 1 & 0 & \dots & \star \\ \vdots & \vdots & \vdots \end{pmatrix}\] where the \(\star\)’s in the right-most column are (up to sign) the coefficients of the minimal polynomial of \(\alpha.\) This is a matrix of integers; so its determinant and trace are integers.
If \(K\) is larger than \(\mathbb{Q}(\alpha),\) then one can check that \[\operatorname{Nm}_{K / \mathbb{Q}}(\alpha) = \left(\operatorname{Nm}_{L / \mathbb{Q}}(\alpha)\right)^{[K : L]}, \qquad \operatorname{Tr}_{K / \mathbb{Q}}(\alpha) =[K : L] \cdot \operatorname{Tr}_{L / \mathbb{Q}}(x)\] where \(L = \mathbb{Q}(\alpha)\); and the result follows from the previous case.
Remark 11.5
It’s not true in general that if \(x \in K,\) and \(\operatorname{Nm}_{K / \mathbb{Q}}(x)\) and \(\operatorname{Tr}_{K / \mathbb{Q}}(y)\) are in \(\mathbb{Z},\) then \(x \in \mathcal{O}_K\) (although this is true if \(K\) is quadratic).
Exercise 11.6
Prove the following refinement of Proposition 10.22: for any \(\alpha \in \mathcal{O}_K,\) the divisibility \(\alpha \mid \operatorname{Nm}_{K / \mathbb{Q}}(\alpha)\) holds in \(\mathcal{O}_K.\) [Hint: First reduce to the case \(K = \mathbb{Q}(\alpha),\) then apply the Cayley–Hamilton theorem.]
11.2 Lattices and orders
We want to understand “how big” \(\mathcal{O}_K\) is, and how it sits inside \(K,\) for an arbitrary number field \(K.\)
Definition 11.7
Let \(V\) be a finite-dimensional \(\mathbb{Q}\)-vector space. A lattice \(\mathcal{L}\) in \(V\) is a subgroup of \((V, +)\) which is finitely-generated as a group. If \(\mathcal{L}\) spans \(V\) as a \(\mathbb{Q}\)-vector space, we say \(\mathcal{L}\) is full.
One can check (see Addendum below) that any lattice in \(V\) has to be isomorphic as a group to \(\mathbb{Z}^m\) for some \(m \leqslant\dim V,\) with equality iff \(\mathcal{L}\) is full. Moreover, a subgroup of a lattice is a lattice.
Example 11.8
For example, \(\mathbb{Z}^2\) is obviously a full lattice in \(\mathbb{Q}^2.\) More subtly, so is \(\{ \begin{pmatrix} a \\ b \end{pmatrix} \in \mathbb{Z}^2 : a + b \text{ is even}\}\): it is generated by \(\begin{pmatrix} 1 \\ 1\end{pmatrix}\) and \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}.\)
Since number fields are finite-dimensional \(\mathbb{Q}\)-vector spaces, we can ask about lattices inside them. But a number field, unlike a general vector space, we know how to multiply things; so we can make the next definition:
Definition 11.9
An order in a number field \(K\) is a full lattice which is also a subring of \(K.\)
For instance, both \(\mathbb{Z}\) and \(\tfrac{1}{17} \mathbb{Z}\) are full lattices in \(\mathbb{Q},\) and \(\mathbb{Z}\) is an order, but \(\tfrac{1}{17} \mathbb{Z}\) is not. Moreover, \(\mathbb{Z}[i]\) is an order in \(\mathbb{Q}(i),\) and both \(\mathbb{Z}[\sqrt{-3}]\) and \(\mathbb{Z}[\omega]\) are orders in \(\mathbb{Q}(\sqrt{-3}).\)
Exercise 11.10
Show that if \(A\) is an order in \(K,\) then \(A \subseteq \mathcal{O}_K.\)
11.3 The trace dual of a lattice
The crucial construction we’ll use to understand lattices and orders in number fields is the following:
Definition 11.11
If \(K\) is a number field, and \(\mathcal{L}\) is a subgroup of \((K, +),\) then the trace dual of \(\mathcal{L}\) is defined by \[\mathcal{L}^\vee = \{ x \in K : \operatorname{Tr}_{K / \mathbb{Q}}(x y) \in \mathbb{Z}\quad \forall y \in \mathcal{L}\}.\]
Note that \(\mathcal{L}^\vee\) is also a subgroup of \((K, +)\ \) (exercise). Moreover, taking the trace dual is inclusion-reversing: if \(\mathcal{L} \subseteq \mathcal{M},\) then \(\mathcal{L}^\vee \supseteq \mathcal{M}^\vee.\)
Proposition 11.12
If \(\mathcal{L}\) is a full lattice in \(K,\) then the trace dual \(\mathcal{L}^\vee\) is also a full lattice.
Proof. This is an instance of a general result (see Addendum below) applying to any finite-dimensional \(\mathbb{Q}\)-vector space equipped with a non-degenerate quadratic form.
To apply this in our situation, we need to check that the “trace form” \((x, y) \mapsto \operatorname{Tr}_{K/\mathbb{Q}}(xy)\) is a quadratic form (which is obvious), and that it is non-degenerate. So, suppose \(x \in K\) satisfies \(\operatorname{Tr}_{K/\mathbb{Q}}(xy) = 0\) for all \(y \in K.\) If \(x \ne 0,\) then we can take \(y = x^{-1}\) and we have \(\operatorname{Tr}(xy) = \operatorname{Tr}(1) = [K : \mathbb{Q}] \ne 0,\) a contradiction. Hence we must have \(x = 0,\) showing that the trace form is non-degenerate.
Exercise 11.13
Take \(\mathcal{L} = \mathbb{Z}[i],\) considered as a lattice in \(K = \mathbb{Q}(i),\) and calculate \(\mathcal{L}^\vee.\)
Proposition 11.14
We have \(\mathcal{O}_K^\vee \supseteq \mathcal{O}_K.\)
Proof. Let \(x \in \mathcal{O}_K.\) Then for any \(y \in \mathcal{O}_K,\) we have \(xy \in \mathcal{O}_K\) (because \(\mathcal{O}_K\) is a ring), and hence \(\operatorname{Tr}(xy) \in \mathbb{Z}\) by Proposition 11.4. Thus \(x \in \mathcal{O}_K^\vee.\)
Corollary 11.15
\(\mathcal{O}_K\) is an order in \(K.\)
Proof. We know \(\mathcal{O}_K\) is a subring, so we need to show \(\mathcal{O}_K\) is a full lattice.
First we claim \(\mathcal{O}_K\) contains a full lattice. Let \(x_1, \dots, x_d\) be a \(\mathbb{Q}\)-basis of \(\mathcal{O}_K.\) If we multiply each \(x_i\) by a non-zero integer, then the new set is still a basis, so by Proposition 10.10 we can arrange that the \(x_i\) are all in \(\mathcal{O}_K.\) Thus \(\mathcal{L} = \mathbb{Z}x_1 + \dots + \mathbb{Z}x_d\) is a full lattice contained in \(\mathcal{O}_K.\) (Note there’s no reason for it to be an order.)
Now, if \(\mathcal{L}\) is a full lattice such that \(\mathcal{L} \subseteq \mathcal{O}_K,\) then \(\mathcal{L}^\vee \supseteq \mathcal{O}_K^\vee\); and \(\mathcal{L}^\vee\) is also a full lattice. Since \[\mathcal{L}^\vee \supseteq \mathcal{O}_K^\vee \supseteq \mathcal{O}_K \supseteq \mathcal{L}.\] So we have sandwiched \(\mathcal{O}_K\) between two full lattices, \(\mathcal{L}\) and \(\mathcal{L}^\vee.\) Since \(\mathcal{O}_K \supseteq \mathcal{L},\) we know that \(\mathcal{O}_K\) spans \(K\) as a \(\mathbb{Q}\)-vector space. Since \(\mathcal{O}_K \subseteq \mathcal{L}^\vee,\) we know that \(\mathcal{O}_K\) is finitely-generated. Thus \(\mathcal{O}_K\) is a full lattice (and hence an order).
Remark 11.16
Note that this corollary also gives us a way of computing \(\mathcal{O}_K.\) Assume we know a primitive element \(\alpha\) of \(K.\) By scaling if necessary, we can suppose \(\alpha \in \mathcal{O}_K.\) Then \(A = \mathbb{Z}[\alpha]\) is a full lattice (and indeed an order) contained in \(\mathcal{O}_K.\) The quotient \(A^\vee / A\) is finite (and explicitly computable); and for each element \(x + A\) of this quotient, we can determine whether \(x + A \in \mathcal{O}_K / A,\) by calculating the minimal polynomial of \(x.\)
(This is essentially what we did in the previous chapter for quadratic fields \(\mathbb{Q}(\sqrt{d}),\) taking \(A\) to be the order \(\mathbb{Z}[\sqrt{d}].\))
11.4 Addendum: Some \(\mathbb{Z}\)-linear algebra
Just for completeness, we’ll outline the proofs of a few results about subgroups of \(\mathbb{Z}^n\) which we used in this chapter. The proofs in this section are non-examinable
11.4.1 Subgroups of \(\mathbb{Z}^n\)
We begin with Theorem 4.4 of the Algebra module, which says the following:
Proposition 11.17
Let \(G\) be a subgroup of the additive group \((\mathbb{Z}, +).\) Then we have \[G = m \mathbb{Z}= \{ m z : z \in \mathbb{Z}\}\] for a uniquely determined \(m \geqslant 0.\) In particular, either \(G = \{0\}\) or \(G\) is isomorphic to \(\mathbb{Z}\) itself.
Motivated by this, what can we say about subgroups of \(\mathbb{Z}^n,\) for an arbitrary \(n \geqslant 1\)?
Theorem 11.18
Let \(H\) be a subgroup of \(\mathbb{Z}^n.\) Then there is a unique \(m \in \mathbb{N}\) such that \(H \cong \mathbb{Z}^m,\) and we have \(0 \leqslant m \leqslant n.\)
Proof of uniqueness. Note that \(H \cong \mathbb{Z}^m\) iff there exists a set of \(m\) elements \(h_1, \dots, h_m \in H\) which are independent generators, i.e. every \(x \in H\) can be written as \(x = \sum a_i h_i\) for a unique \((a_1, \dots, a_m) \in \mathbb{Z}^m.\)
Let \(W\) be the \(\mathbb{Q}\)-vector space spanned by \(H.\) Then \(h_1, \dots, h_m\) clearly span \(W\) as a \(\mathbb{Q}\)-vector space. They are also \(\mathbb{Q}\)-linearly independent, because if we had a nontrivial \(\mathbb{Q}\)-linear relation between them, we could clear denominators to get a nontrivial \(\mathbb{Z}\)-linear relation. Hence we must have \(m = \dim W,\) which clearly satisfies \(0 \leqslant m \leqslant n.\)
Proof of existence. To deduce existence, we’ll use induction on \(n.\) The result is trivial for \(n = 0,\) so assume it holds for \(n -1.\)
Given \(H \subseteq \mathbb{Z}^n,\) consider the “forget the last entry” map \(\mathbb{Z}^n \to \mathbb{Z}^{n-1}.\) The image \(\bar{H}\) of \(H\) is a subgroup of \(\mathbb{Z}^{n-1},\) so (by the induction hypothesis) we can find an independent generating set \(\bar{h}_1, \dots, \bar{h}_r,\) for some \(r \leqslant n-1.\) Choose arbitrary elements \(h_1, \dots, h_r\) of \(H\) mapping to \(\bar{h}_1, \dots, \bar{h}_r.\) Then any \(h \in H\) can be uniquely written as \(\sum_{i = 1}^r a_i h_i + (0, \dots, 0, x),\) for some \((a_1, \dots, a_r) \in \mathbb{Z}^r\) and \(x \in \mathbb{Z}.\)
Now consider the subgroup \(X = \{x \in \mathbb{Z}: (0, \dots, 0, x) \in H\}.\) This is a subgroup of \(\mathbb{Z},\) so it must be either \(\{0\},\) or \(d \mathbb{Z}\) for some \(d \geqslant 1.\) If \(X = \{0\},\) then \(h_1, \dots, h_r\) are an independent generating set of \(H.\) If \(X = d\mathbb{Z}\) for \(d \geqslant 1,\) then we set \(h_{r + 1} = (0, \dots, 0, d)\); then \((h_1, \dots, h_{r+1})\) are an independent generating set.
What can we say about subgroups \(H \cong \mathbb{Z}^n\) which are isomorphic to \(\mathbb{Z}^n\)? Of course, this doesn’t imply that \(H\) is the whole of \(\mathbb{Z}^n\) (as we’ve already seen for \(n = 1\)). What we can say is the following:
Theorem 11.19
For a subgroup \(H \subseteq \mathbb{Z}^n,\) the following are equivalent:
\(H\) is isomorphic to \(\mathbb{Z}^n\);
the index \([\mathbb{Z}^n : H]\) is finite.
Proof. If \([\mathbb{Z}^n : H]\) is finite, of size \(d\) say, then every element of the quotient \(\mathbb{Z}^n / H\) has order dividing \(d\) (“element order divides group order”); so \(d v \in H\) for every \(v \in \mathbb{Z}^n.\) In particular, \(H\) contains \(d e_j\) for each \(j,\) and thus spans \(\mathbb{Q}^n.\) So it must be isomorphic to \(\mathbb{Z}^n.\)
Careful: the abelian property is needed here – there exists a famous example of an infinite non-abelian group, the modular group, generated by two elements of order 2 and 3 respectively.
Remark 11.20
One can show that if \(H\) is a finite-index subgroup of \(\mathbb{Z}^n,\) and \(h_1, \dots, h_n\) is an independent generating set of \(H,\) then we have \[[\mathbb{Z}^n : H] = |\det A|,\] where \(A\) is the matrix with the \(h_i\) as rows.
11.4.2 Lattices in \(\mathbb{Q}\)-vector spaces
Now suppose \(V\) is a finite-dimensional \(\mathbb{Q}\)-vector space; without loss of generality \(V = \mathbb{Q}^n\) for some \(n.\)
If \(\mathcal{L}\) is a finitely-generated subgroup of \(\mathbb{Q}^n,\) then we have \(\mathcal{L} \subseteq N^{-1} \mathbb{Z}^n\) for some \(N \geqslant 1\) (it suffices to take the LCM of the denominators of any generating set of \(\mathcal{L}\)). Since multiplying by \(N\) is an isomorphism \(N^{-1} \mathbb{Z}^n \cong \mathbb{Z}^n,\) we conclude that \(\mathcal{L}\) is isomorphic to \(\mathbb{Z}^m\) for some \(0 \leqslant m \leqslant n,\) as before.
Remark 11.21
Not all subgroups of \((V, +)\) are lattices: for instance, \(V\) itself is a subgroup of \(V,\) but it is not a lattice (except in the trivial case \(V = \{0\}\)).
(Exercise: can you find a proper subgroup of \((\mathbb{Q}, +)\) which is not a lattice?)
11.4.3 Duals of lattices
This is not strictly needed, it’s just for notational simplicity.
Let’s now assume the pairing on \(V\) is non-degenerate, i.e. if \(x \in V\) satisfies \(\langle x, y \rangle = 0\) for all \(y \in V,\) then \(x = 0.\)
Proposition 11.22
If \(\mathcal{L}\) is a full lattice, then so is \(\mathcal{L}^\vee.\)
Proof. Let \(\mathbf{v} = (v_1, \dots, v_d)\) be an (ordered) independent generating set of \(\mathcal{L}\); then it is also a \(\mathbb{Q}\)-basis of \(V,\) since \(\mathcal{L}\) is full. Let \(M\) be the matrix with \((i, j)\) entry \(\langle v_i, v_j\rangle\) (the matrix of the bilinear form).
Since the pairing \(\langle-,- \rangle\) is non-degenerate, \(M\) is non-singular, so it has an inverse \(M^{-1}.\)
Let \(b_i\) be the \(i\)-th row of \(M^{-1}\); and let \(w_i = b_1 v_1 + \dots + b_d v_d\) be the vector whose coordinates in the basis \(\mathbf{v}\) are \(b_i.\) Then \(\mathbf{w} = (w_1, \dots, w_d)\) is also a basis of \(V,\) and one computes that \[\langle w_i, v_j\rangle = \begin{cases} 1 & \text{if $i = j$}\\ 0 & \text{otherwise}.\end{cases}\]
Hence, if we write an arbitrary \(x\in V\) as \(x = \sum c_i w_i\) for some \(\vec{c} \in \mathbb{Q}^d,\) we have \(x \in \mathcal{L}^\vee\) iff \(c_i \in \mathbb{Z}\) for all \(i.\) Thus \(\mathcal{L}^\vee\) is precisely the \(\mathbb{Z}\)-linear combinations of the basis \(\mathbf{w},\) showing that it is a full lattice.
Remark 11.23
This is related to the notion of dual bases from Linear Algebra II. More precisely, you saw in that module that a nondegenerate bilinear form defines an isomorphism from \(V\) to its dual space \(V^*.\) You also saw that for any basis \(\mathbf{v} = (v_1, \dots, v_n)\) of \(V\) there is a dual basis \((\nu_1, \dots, \nu_n)\) of \(V^*\) with \(\nu_i(v_j) = \delta_{ij}.\) The basis \(\mathbf{w}\) in the above proof, satisfying \(\langle w_i, v_j \rangle = \delta_{ij},\) is given by transporting the dual basis \(\pmb{\nu}\) along the isomorphism \(V^* \cong V.\)
Exercise 11.24
Show that if \(\mathcal{L}\) is a full lattice, then \(\mathcal{L}^\vee{}^\vee = \mathcal{L}.\)