Last time, I discussed Dedekind’s Criterion. In this post I aim to introduce Dirichlet’s Unit Theorem.

Let L be a number field (finite extension of the field of rational numbers).

We will define a **complex embedding of L** as being a field homomorphism:

Then a **real embedding of L ** is similarly defined as being a field homomorphism:

We note that the complex embeddings can be given in pairs:

where we obtain one embedding by taking the complex conjugate of the other embedding in the pair.

Hence, given a number field L, we can label its real embeddings as

and its complex embeddings as

Noting that the degree, *n*, of the field extension L/Q (where Q is the field of rational numbers) is equal to the total number of embeddings and hence we deduce that ** n = 2s + r**. The fact that there is this equivalence is non-trivial, so don’t worry if you don’t immediately see why this is! In order to prove this we must use tools from

**Galois Theory**, such as something called the ‘Tower Law’, and therefore I will emit the proof from this blog post.

*Let me know if you’re interested in seeing Galois Theory*

*related posts in the future!*

Letting be the units in the ring of algebraic integers of L we are now ready to state Dirichlet’s Unit Theorem.

## Dirichlet’s Unit Theorem

The proof of this is rather complicated and involves defining a map, whose kernel is this finite cyclic group, which takes into a lattice. Considering lattices are involved in the proof, you can start to see where the isomorphism with the integers comes from.

## Why is this useful?

Take the case where r = 0 and s = 1, i.e. L is an imaginary quadratic number field:

As *r + s – 1 = 0*, we get that is a finite group. By explicitly defining the isomorphism we can deduce that the finite cyclic group is simply {+/- 1}, and is generated by one element – the **fundamental unit, **i.e.

I’ve purposefully skipped a lot of the details as what is important is the **result**. The proof requires a bit of *magic* and is not terribly enlightening. However, the result is extremely important – using this, for the case of imaginary quadratic fields, we can completely specify , provided we find the fundamental unit.

## Finding the Fundamental Unit

It turns out that finding the fundamental unit in this case is extremely straightforwards. There are various different cases.

- If
**d = 2, 3 (mod 4)**we characterise the fundamental unit*u*in the following way: let*b*be the least positive integer such that db^{2}+ 1 or db^{2}– 1 is of the form a^{2}for some natural number*a*. Then:

- If
**d = 1 (mod 4) and d is not equal to 5:**let*b*be the least positive integer such that db^{2}+ 4 or db^{2}– 4 is of the form a^{2}for some natural number*a*. Then:

**If d = 5:**

(If you would like the proof as to **WHY** this is true, leave me a comment below and I’ll be happy to make a post!)

#### Example: d = 2

Then 𝑏 = 1 works since 2 − 1 = 1^{2 }so **1 + √2 is a fundamental unit.**

In order to get a lot of the symbols I had to LaTex some of the sections so sorry if the formatting is a little off! Next time we will start solving some equations using algebra!

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