Algebraic Number Theory – III. A Note on Lattices, Computations, Embeddings

There are some beautiful connections between lattice geometry and the study of {\mathcal{O}_K}. We will discuss this subject, the geometry of numbers, in greater detail eventually, once we have introduced the ideal class group. But presently, let us look at one nice application.

1. Counting lattice points

Lemma 3.1: Let {A \le \mathbb{Z}^n} be a rank {n} submodule, with integral basis {\mathcal{B} = \{b_1, b_2, \cdots, b_n\}}. If {M} is the matrix whose rows are the vectors of {\mathcal{B}}, then

\displaystyle |\det(M)| = [\mathbb{Z}^n : A]

In particular, {[\mathbb{Z}^n : A]} is finite.

Proof: Define the region

\displaystyle D = \left\{\sum_{i=1}^{n} \gamma_i b_i \, : \, \gamma_i \in [0,1)\right\}

This is the fundamental domain of the lattice {L[A]} associated with {A}. The picture to keep in mind is a tiling of {\mathbb{R}^n} by copies of the parallelotope {D}.

Every point in {\mathbb{Z}^n} falls in one of these copies of {D}, and can be translated onto a unique lattice point inside {D} by integer multiples of the vectors in {\mathcal{B}}. In other words, under the equivalence relation {x\sim y} iff {x-y\in A}, each equivalence class has a unique representative in {D}. Hence, the index {[\mathbb{Z}^n : A]} equals the cardinality of {D\cap \mathbb{Z}^n} (which we henceforth denote {|D\cap \mathbb{Z}^n|}). This quantity is finite since {D} is a bounded region.

Now, consider the case that all the vectors in {\mathcal{B}} are integer multiples of the standard basis vectors for {\mathbb{Z}^n}. Then {D} is a “box,” whose volume is exactly {|D\cap \mathbb{Z}^n|}. Since the volume of {D} is exactly {|\det(M)|}, in this case, we are done.

Otherwise, imagine dynamically changing the vectors {\{b_1, \cdots, b_n\}} by the application of elementary row operations to {M}. With each operation, we get a new fundamental domain {D}, whose volume is the same as the volume of the original {D}. It is a somewhat tricky exercise in visualization to see that these operations also preserve the quantity {|D\cap \mathbb{Z}^n |}. Since the vectors in {\mathcal{B}} are linearly independent, one can put {M} into diagonal form via these row reductions, thus reducing to the first case. \Box

Let {N} be a rank {n} submodule of {\mathcal{O}_K}. Suppose that {\mathcal{B}_1} is an integral basis for {N} and {\mathcal{B}_2} is an integral basis for {\mathcal{O}_K}. There is a natural isomorphism {\varphi: \mathcal{O}_K \rightarrow \mathbb{Z}^n} that sends each point to its coordinates in {\mathcal{B}_2}. Then {A:= \varphi(N)} is a rank {n} submodule of {\mathbb{Z}^n}. By the lemma,

\displaystyle [\mathcal{O}_K : N] = [\mathbb{Z}^n : A] = |\det(C)|

where {C} is the change of basis matrix for {\mathcal{B}_1} to {\mathcal{B}_2}.

Using the identity on discriminants, one gets the following fact.

Proposition 3.2: If {N \subset \mathcal{O}_K} is a rank {n} submodule, such that {N} and {\mathcal{O}_K} have integral bases {\mathcal{B}_1} and {\mathcal{B}_2} respectively, then

\displaystyle \Delta(\mathcal{B}_1) = [\mathcal{O}_K: N]^2 \Delta(\mathcal{B}_2) = [\mathcal{O}_K : N ]^2 \delta_K

One may specialize to the case of {\mathfrak{a} \subset \mathcal{O}_K} an ideal. If {\mathcal{O}_K} has integral basis {\{b_1, \cdots, b_n\}}, then for any {\alpha \in \mathfrak{a}}, there is the linearly independent set {\{b_1 \alpha, \cdots, b_n \alpha\} \subset \mathfrak{a}}. Hence {\mathfrak{a}} is a rank {n} submodule of {\mathcal{O}_K}.

Proposition 3.3: If {\mathfrak{a} \subset \mathcal{O}_K} is an ideal, the quotient ring {\mathcal{O}_K / \mathfrak{a}} is finite.

This proposition has some immediate and important algebraic corollaries.

Corollary 3.4:

  1. {\mathcal{O}_K} is a Noetherian ring, i.e. if {\mathfrak{a}_1 \subset \mathfrak{a}_2 \subset \cdots} is an ascending chain of ideals, then there is some {N} such that {\mathfrak{a}_k = \mathfrak{a}_N} for all {k\ge N}.
  2. Every prime ideal in {\mathcal{O}_K} is maximal.

Proof: Easy exercise. \Box

2. Computing rings of integers

The problem of actually computing rings of integers is, in general, rather challenging. For extensions {K/\mathbb{Q}} of small degree, usually one can compute the ring of integers by hand, using case-work and divisibility considerations. As {[K:\mathbb{Q}]} gets larger, the calculations get more unwieldy, and one needs bigger machines to do the work. Right now, however, it will be instructive to play with the smaller machines in our possession.

Example 3.5: Let us compute the ring of integers of {K = \mathbb{Q}(\sqrt{-5})}. Obvious candidate is {\mathbb{Z}[\sqrt{-5}]}, so we consider the {\mathbb{Z}}-submodule {N\subset \mathcal{O}_K} generated by {1} and {\sqrt{-5}}. Then one has the discriminant

\displaystyle \Delta(1, \sqrt{-5}) = \det \begin{pmatrix} \text{Tr}_{K/\mathbb{Q}}(1^2) & \text{Tr}_{K/\mathbb{Q}}(1\cdot \sqrt{-5}) \\ \text{Tr}_{K/\mathbb{Q}}(\sqrt{-5}\cdot 1) & \text{Tr}_{K/\mathbb{Q}}( \sqrt{-5}^2) \\ \end{pmatrix} = \det \begin{pmatrix} 2 & 0 \\ 0 & -10 \end{pmatrix} = -20

By the discriminant identity,

\displaystyle -20 = \Delta(N) = [\mathcal{O}_K : N]^2 \delta_K

which implies {[\mathcal{O}_K : N] = 2} or {1}.

Suppose {[\mathcal{O}_K : N] = 2}. Then the quotient group {\mathcal{O}_K / N} has order {2}, so there is the chain of inclusions

\displaystyle 2\mathcal{O}_K \subset N \subset \mathcal{O}_K

For any {a+ b\sqrt{-5} \in \mathcal{O}_K}, where {a} and {b} are rational, this chain requires {a} and {b} are both integers or half-integers.

The inclusion {N \subset \mathcal{O}_K} is proper, so there must be some {x = a+b\sqrt{-5} \in \mathcal{O_K} \setminus N} such that at least one of {a} or {b} is a half-integer. But then {N_{K/\mathbb{Q}} (x)} is not an integer, contrary to our observation that the restriction of norm to {\mathcal{O}_K} has range {\mathbb{Z}}.

Hence {[\mathcal{O}_K : N] = 1}, and the ring of integers of {\mathbb{Q}(\sqrt{-5})} is exactly {\mathbb{Z}[\sqrt{-5}]}\Box

The general approach here is that if {K= \mathbb{Q}[\alpha]}, with {[K:\mathbb{Q}] = n}, then one hopes that the set {\mathcal{B} = \{1, \alpha, \cdots, \alpha^{n-1}\}} forms an integral basis for {\mathcal{O}_K}. If {N = \text{span}(\mathcal{B} )}, then using the discriminant identity

\displaystyle \Delta(1, \alpha, \cdots, \alpha^{n-1} ) = [\mathcal{O}_K :N]^2 \delta_K

Optimal scenario is when {\Delta(\mathcal{B})} is squarefree, so that immediately {[\mathcal{O}_K : N] = 1}. If {\Delta(\mathcal{B})} is not square-free, the attack might be salvaged as in our example above. But otherwise, one either needs more machinery, or needs to be looking at algebraic integers outside {N}.

In any case, number-theoretic information about the discriminant is evidently useful for these computations. To gain more of this information, we will need an alternative characterization of the discriminant.

3. Embeddings into {\mathbb{C}} and Stickelberger’s Theorem

Let us review some field theory.

We are interested in classifying the injective field homomorphisms {K \hookrightarrow \mathbb{C}}. Since {K/\mathbb{Q}} is a finite extension, Artin’s primitive element theorem produces an {\alpha \in \mathbb{C}} such that {K \cong \mathbb{Q}(\alpha)}. Assuming {[K:\mathbb{Q}] = n}, the minimal polynomial {\text{Irr}_{\alpha, \mathbb{Q}}} has exactly {n} roots {\alpha = \alpha_1, \alpha_2, \cdots, \alpha_n}. These are all distinct, and none are rational.

If {\sigma : K \rightarrow \mathbb{C}} is any homomorphism, {\sigma} fixes {\mathbb{Q}}, so we need only worry about the action of {\sigma} on {\alpha}. As {\text{Irr}_{\alpha, \mathbb{Q}}(\sigma(\alpha)) = \sigma(\text{Irr}_{\alpha, \mathbb{Q}}(\alpha)) = 0}, we find that {\sigma} must send {\alpha} to one of {\alpha_1, \cdots, \alpha_n}. Hence, there are exactly {n} embeddings {K \hookrightarrow \mathbb{C}}.

One splits this collection {\{\sigma_1, \cdots, \sigma_n\}} of embeddings into the real embeddings and complex embeddings. The real embeddings are those for which {\sigma(\alpha)} is a real root of {\text{Irr}_{\alpha, \mathbb{Q}}} and the complex embeddings are those for which {\sigma(\alpha)} is not real. The complex embeddings come in pairs, as conjugate embeddings. We denote the number of real embeddings {r}, and the number of pairs of complex embeddings {s}, so that {r+2s = n}.

These embeddings provide another way to reason about the algebraic conjugates of an element {x\in K}. For the same reason as with {\alpha}, the images {\sigma_1 (x)}, {\cdots}, {\sigma_n (x)} are algebraic conjugates of {x}. However, {x} need not have degree {n} over {\mathbb{Q}}, so some of these images may be equal. In particular, suppose {[\mathbb{Q}(x) : \mathbb{Q}] = d}. There are exactly {d} embeddings {\tau_1, \cdots, \tau_d : \mathbb{Q}(x) \hookrightarrow \mathbb{C}}, and each restriction {\sigma_i \vert_{\mathbb{Q}(x)}} is one of the {\tau_j}. Conversely, consider the proposition.

Proposition 3.6: Suppose {F} is a field of characteristic zero. Let {L/F} be a finite degree extension and {\tau: F \rightarrow \mathbb{C}} an injective homomorphism. Then there are exactly {[L:F]} extensions {\sigma : L \rightarrow \mathbb{C}} of {\tau}.

Proof: Essentially the same as the above proof that there are {n} embeddings {K\hookrightarrow \mathbb{C}}. \Box

Hence, the collection {\{\sigma_1 (x), \cdots, \sigma_n (x)\}} consists of precisely {n/d} copies of the set of algebraic conjugates of {x}. This observation allows one to reformulate the notions of field norm and trace as follows:

\displaystyle \text{Tr}_{K/\mathbb{Q}}(x) = \sum_{i=1}^{n} \sigma_i (x)

\displaystyle N_{K/\mathbb{Q}}(x) = \prod_{i=1}^{n} \sigma_i (x)

Given an integral basis {\mathcal{B} = \{b_1, \cdots, b_n\}} for {\mathcal{O}_K}, using these equations, one may rewrite the trace pairing matrix as {M_{\text{TrP}, \mathcal{B}, \mathcal{B}} = A^T A}, where {A = \{\sigma_i (b_j)\}}. Taking determinants, one gets

\displaystyle \Delta(\mathcal{B}) = \det(\{\text{TrP}_{K/\mathbb{Q}}(b_i, b_j)\}) = \det(\{\sigma_i (b_j)\})^2

We now have the requisite tools to prove Stickelberger’s theorem.

Theorem 3.7 (Stickelberger): {\delta_K \equiv 0} or {1 \pmod{4}}.

Proof: We may write

\displaystyle \det(\{\sigma_i (b_j)\}) = \sum_{\pi \in S_n \setminus A_n} \prod_{i=1}^{n} \sigma_i (b_{\pi(i)}) - \sum_{\pi \in A_n} \prod_{i=1}^{n} \sigma_i (b_{\pi(i)}) := P-N

Observe that every {\sigma_i} fixes {P+N}, meaning the only algebraic conjugate of {P+N} is {P+N} itself. Hence, {P+N \in \mathbb{Q}}. Similarly, {PN \in \mathbb{Q}}.

However, since the basis {\{b_1, \cdots, b_n \} \subset \mathcal{O}_K}, any algebraic conjugate of a basis element is also in {\mathcal{O}_K}. Specifically, it follows that {P+N} and {PN} are algebraic integers. Hence, {P+N} and {PN \in \mathbb{Z}}. In conclusion,

\displaystyle \delta_K = (P-N)^2 = (P+N)^2 - 4PN \equiv (P+N)^2 \equiv 0 \text{ or } 1 \pmod{4}


Using Stickelberger’s theorem, one can immediately determine the ring of integers for every quadratic extension {\mathbb{Q}(\sqrt{d})}, where {d} is square-free.

If {d\equiv 2} or {3\pmod{4}}, then one can compute {\Delta(1, \sqrt{d}) = 4d}. Setting {N} as the submodule generated by {1} and {\sqrt{d}}, there is the equality

\displaystyle 4d = [\mathcal{O}_K : N]^2 \delta_K

By Stickelberger, {\delta_K} is divisible by 4, so {[\mathcal{O}_K : N] = 1}.

Otherwise, when {d \equiv 1 \pmod{4}}, note that {(1+\sqrt{d})/2} is an algebraic integer, the root of {x^2 - x + (1-m)/4 = 0}. Then one computes {\Delta(1, (1+\sqrt{d})/2) = d}, which is squarefree. Hence, in this case, {\mathcal{O}_K = \mathbb{Z}[(1+\sqrt{d})/2]}.

To be fair, this is not so much easier than just working “hands-on” as we did above with the case {d = -5}. Nevertheless, it illustrates the principle that there are often multiple ways to perform these computations.


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