There are some beautiful connections between lattice geometry and the study of . 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 be a rank submodule, with integral basis . If is the matrix whose rows are the vectors of , then
In particular, is finite.
Proof: Define the region
This is the fundamental domain of the lattice associated with . The picture to keep in mind is a tiling of by copies of the parallelotope .
Every point in falls in one of these copies of , and can be translated onto a unique lattice point inside by integer multiples of the vectors in . In other words, under the equivalence relation iff , each equivalence class has a unique representative in . Hence, the index equals the cardinality of (which we henceforth denote ). This quantity is finite since is a bounded region.
Now, consider the case that all the vectors in are integer multiples of the standard basis vectors for . Then is a “box,” whose volume is exactly . Since the volume of is exactly , in this case, we are done.
Otherwise, imagine dynamically changing the vectors by the application of elementary row operations to . With each operation, we get a new fundamental domain , whose volume is the same as the volume of the original . It is a somewhat tricky exercise in visualization to see that these operations also preserve the quantity . Since the vectors in are linearly independent, one can put into diagonal form via these row reductions, thus reducing to the first case.
Let be a rank submodule of . Suppose that is an integral basis for and is an integral basis for . There is a natural isomorphism that sends each point to its coordinates in . Then is a rank submodule of . By the lemma,
where is the change of basis matrix for to .
Using the identity on discriminants, one gets the following fact.
Proposition 3.2: If is a rank submodule, such that and have integral bases and respectively, then
One may specialize to the case of an ideal. If has integral basis , then for any , there is the linearly independent set . Hence is a rank submodule of .
Proposition 3.3: If is an ideal, the quotient ring is finite.
This proposition has some immediate and important algebraic corollaries.
- is a Noetherian ring, i.e. if is an ascending chain of ideals, then there is some such that for all .
- Every prime ideal in is maximal.
Proof: Easy exercise.
2. Computing rings of integers
The problem of actually computing rings of integers is, in general, rather challenging. For extensions of small degree, usually one can compute the ring of integers by hand, using case-work and divisibility considerations. As 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 . Obvious candidate is , so we consider the -submodule generated by and . Then one has the discriminant
By the discriminant identity,
which implies or .
Suppose . Then the quotient group has order , so there is the chain of inclusions
For any , where and are rational, this chain requires and are both integers or half-integers.
The inclusion is proper, so there must be some such that at least one of or is a half-integer. But then is not an integer, contrary to our observation that the restriction of norm to has range .
Hence , and the ring of integers of is exactly .
The general approach here is that if , with , then one hopes that the set forms an integral basis for . If , then using the discriminant identity
Optimal scenario is when is squarefree, so that immediately . If 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 .
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 and Stickelberger’s Theorem
Let us review some field theory.
We are interested in classifying the injective field homomorphisms . Since is a finite extension, Artin’s primitive element theorem produces an such that . Assuming , the minimal polynomial has exactly roots . These are all distinct, and none are rational.
If is any homomorphism, fixes , so we need only worry about the action of on . As , we find that must send to one of . Hence, there are exactly embeddings .
One splits this collection of embeddings into the real embeddings and complex embeddings. The real embeddings are those for which is a real root of and the complex embeddings are those for which is not real. The complex embeddings come in pairs, as conjugate embeddings. We denote the number of real embeddings , and the number of pairs of complex embeddings , so that .
These embeddings provide another way to reason about the algebraic conjugates of an element . For the same reason as with , the images , , are algebraic conjugates of . However, need not have degree over , so some of these images may be equal. In particular, suppose . There are exactly embeddings , and each restriction is one of the . Conversely, consider the proposition.
Proposition 3.6: Suppose is a field of characteristic zero. Let be a finite degree extension and an injective homomorphism. Then there are exactly extensions of .
Proof: Essentially the same as the above proof that there are embeddings .
Hence, the collection consists of precisely copies of the set of algebraic conjugates of . This observation allows one to reformulate the notions of field norm and trace as follows:
Given an integral basis for , using these equations, one may rewrite the trace pairing matrix as , where . Taking determinants, one gets
We now have the requisite tools to prove Stickelberger’s theorem.
Theorem 3.7 (Stickelberger): or .
Proof: We may write
Observe that every fixes , meaning the only algebraic conjugate of is itself. Hence, . Similarly, .
However, since the basis , any algebraic conjugate of a basis element is also in . Specifically, it follows that and are algebraic integers. Hence, and . In conclusion,
Using Stickelberger’s theorem, one can immediately determine the ring of integers for every quadratic extension , where is square-free.
If or , then one can compute . Setting as the submodule generated by and , there is the equality
By Stickelberger, is divisible by 4, so .
Otherwise, when , note that is an algebraic integer, the root of . Then one computes , which is squarefree. Hence, in this case, .
To be fair, this is not so much easier than just working “hands-on” as we did above with the case . Nevertheless, it illustrates the principle that there are often multiple ways to perform these computations.