## 3: Dirichlet’s Unit Theorem

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 db2 + 1 or db2 – 1 is of the form a2 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 db2 + 4 or db2 – 4 is of the form a2 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 = 1so 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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## MATHS BITE: Brouwer Fixed Point Theorem

The Brouwer Fixed Point Theorem is a result in topology that has proven to be extremely useful in mathematics.

### 2D

Suppose you take two sheets of paper with one lying directly above the other. If you crumple the top sheet and place it on top of the other sheet, then Brouwer’s theorem states that there is at least one point that has not moved, i.e. there is at least one point on the top sheet that is directly above the corresponding point on the bottom sheet.

### 3D

Take a cup of coffee, and mix it around. The theorem states that after mixing there must be some point in the coffee, which is in the exact spot that it was before the mixing. Furthermore, if you try to move that point out of its original position then you will inevitably move another point into its original position.

### Formally

A continuous function from an n-ball into an n-ball must have a fixed point. Note that continuity of the function is essential.

## F.T.A. via Complex Analysis

Although this requires a bit of knowledge on Complex Anlaysis, I recently discovered this new way to prove the Fundamental Theorem of Algebra and I couldn’t help but share it.

First of all, what is the Fundamental Theorem of Algebra (FTA)? This very important (hence the name!) result states that:

Every non-constant polynomial with complex coefficients has a complex root.

In order to prove this, we must first be aware of Liouville’s Theorem:

Every bounded, entire function is constant.

Definitions

Bounded: a function on a set X is said to be bounded if there exists a real number M such that

for all x in X.

Entire: An entire function is a holomorphic function on the entire complex plane.

Liouville’s theorem is proved using the Cauchy integral formula for a disc, one of the most important results in Complex Analysis. Although I will not describe how to prove it or what it states in this blog post, I encourage you to read about here it as it is truly a remarkable result.

Now armed with Liouville’s Theorem we can prove the FTA.

### Proof

Let P(z) = zn + cn-1zn-1 + … + c1z + c0 be a polynomial of degree n > 0. Then |P(z)| –> ∞ as |z| –> ∞, so there exists R such that |P(z)| > 1 for all z with |z| > R.

Consider f(z) = 1/P(z). If P has no complex zeros then f is entire. So, as f is continuous, f is bounded on {|z| ≤ R}.

As |f(z)| < 1 when |z| > R, f is a bounded entire function, so by Liouville’s Theorem f is constant, which is a contradiction.

The only thing we assumed was that P had no complex zeros, and so we contradicted this fact. Hence, P must have at least one complex zero. Amazing right!

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## Ramsey’s Theorem

Ramsey’s theorem is a fundamental theorem in combinatorial mathematics, and initiated the combinatorial theory called Ramsey Theory. This theory aims to seek regularity in disorder: “general conditions for the existence of substructures with regular properties“.

Ramsey’s theorem states that:

“for each pair of positive integers k and l there exists an integer R(k,l) such that any graph with R(k,l) nodes contains a clique with at least k nodes or an independent set with at least l nodes.

Clique:

A clique of agraph G is a complete subgraph of G.

Independent Set:

An independent vertex set of a graph is a subset of the vertices such that no two vertices in the subset represent an edge of the graph. Below are some examples:

Another formulation of the theorem states that “there exists a least positive integer R(rs) for which every blue-red edge colouring of the complete graph on R(rs) vertices contains a blue clique on vertices or a red clique on s vertices”. Extending this theorem, we can apply this to n colours rather than just 2.

### Ramsey Numbers

The numbers R(r,s) in Ramsey’s theorem are called Ramsey numbers. Ramsey numbers give the solution to the party problem that asks:

“the minimum number of guests, R(mn), that must be invited so that at least m will know each other or at least n will not know each other”

Connecting Ramsey numbers to graph theory, the Ramsey number is the minimum number of vertices, v = R(m,n), needed in order for all undirected graphs of order v to contain a clique of order m or an independent set of order n. Ramsey’s theorem tells us that this exists for all m and n.

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## Last Digit

How would we go about finding the last digit of very large numbers, such as ?

There are a variety of different tools that we can use, including modular arithmetic and the Chinese remainder theorem, which I am going to talk about in today’s post.

### Pattern Finding

A problem like this can be tackled by listing out the initial expansions of a power in order to determine a pattern. Then, by proving that this pattern holds (often by induction), the last decimal digit of a power can be solved completely. For example, the last digit of various powers of 1 to 9 are:

From this we can determine that:

• the last digit of powers of 1 is always 1;
• the last digits of powers of 2 repeat in a cycle of 4, 8, 6, 2;
• the last digits of powers of 3 repeat in a cycle of 9, 7, 1, 3;
• the last digits of powers of 4 alternate between 6 and 4;
• the last digit of powers of 5 is always 5;
• the last digit of powers of 6 is always 6;
• the last digits of powers of 7 repeat in a cycle of 9, 3, 1, 7;
• the last digits of powers of 8 repeat in a cycle of 4, 2, 6, 8;
• the last digits of powers of 9 alternate between 1 and 9.

Example

Find the last digit of 2^2016.

The last digits of the powers of two repeat in a cycle of 2, 4, 8, 6. Dividing 2016 by 4 we get 504 (with no remainder). Hence the sequence of digits repeats 504 times with no extra entries, so the last digit should be 6.

### Modular Arithmetic

The above idea can be expressed more elegantly with modular arithmetic. Finding the last digit of a number is the same as finding the remainder when this number is divided by 10. In general, the last digit of a power in base n is its remainder upon division by n. So, for decimal numbers, we compute mod 10 to find the last digit, mod 100 to find the last two digits, etc.

For example,

### Chinese Remainder Theorem

The Chinese Remainder Theorem states that

if one knows the remainders of the division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime

– Wikipedia

Applying this theorem, we can find a number mod 2^n and mod 5^n and then combine those results to find that number mod 10^n.

Example

Find the last two digits of 74^540.

Observe that 100 = 4 x 25 and gcd(4,25) = 1. So we can compute 74^540 mod 4 and mod 25, then combine these to find it mod 100.

74^540 ≡ 2^540 x 37^540 ≡ 0 (mod 4)

and

7^540 ≡ (-1)^540 ≡ 1 (mod 25)

The unique solution mod 100 to x ≡ 0 (mod 4), x ≡ 1 (mod 25) is 76, so this is the answer.

### Euler’s Theorem

Euler’s Theorem states that

“if n and a are coprime positive integers, then

where  is Euler’s totient function.”

-Wikipedia

So an exponent b can be reduced modulo φ(n) to a smaller exponent without changing the value of a^b (mod n).

Example

Find the last two digits of 33^42.

φ(100) = 40, so 33^40≡ 1 (mod 100). So, 33^42 ≡ 33^2 (mod 100). This is easier to compute: 33^2 = 1089, so answer is 89.

### Binomial Expansion

Another method is to expand the power using the Binomial theorem in such a way that many of the terms vanish mod 10. This is generally only useful when the base of the exponent is of the form 10k +/- 1.

Example

Find the last two digits of 31^25.

After the second term, every term contains at least 2 zeros, and so they vanish mod 100. The first two terms = 751 ≡ 51 (mod 100) and so the last two digits of 31^25 are 51.

Hope you enjoyed this slightly longer post! M x

## Noether’s Theorem

Today I thought I’d write a blog post about an interesting theorem I learnt whilst studying my Variational Principles module – Noether’s Theorem.

To understand Noether’s Theorem, we must first understand what is meant by a symmetry of a functional.

Given

suppose we change the variables by the transformation t –> t*(t) and x –> x*(t) to obtain a new independent variable and a new function. This gives

where α* = t*(α) and β* = t*(β).

If F*[x*] = F[x] for all x, α and β, then this transformation * is called a symmetry.

What is a continuous symmetry?

Intuitively, a continuous symmetry is a symmetry that we can do a bit ofFor example, a rotation is a continuous symmetry, but a reflection is not.

### Noether’s Theorem

Noether’s Theorem – proven by mathematician Emmy Noether in 1915 and published in 1918 – states that every continuous symmetry of F[x] the solutions (i.e. the stationary points of F[x]) will have a corresponding conserved quantity.

### Why?

Consider symmetries that involve only the x variable. Then, up to first order, the symmetry can be written as:

t –> t, x(t) –> x(t) + εh(t)

where h(t) represents the symmetry transformation. As the transformation is a symmetry, we can pick ε to be any small constant number and F[x] does not change, i.e. δF = 0. Also, since x(t) is a stationary point of F[x], we know that if ε is any non-constant, but vanishes at the end-points, then we have δF = 0 again. Combining these two pieces of information, we can show that there is a conserved quantity in the system.

For now, do not make any assumptions about ε. Under the transformation, the change in F[x] is given by

Firstly, consider the case where ε is constant. Then the second integral vanishes and we obtain

So we know that

Now, consider a variable ε that is not constant, but vanishes at the endpoints. Then, as is a solution, we must have that δF = 0. Therefore,

If we integrate the above expression by parts, we get that

Hence the conserved quantity is:

Not all symmetries involve just the x variable, for example we may have a time translation, but we can encode this as a transformation of the x variable only.

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## MATHS BITE: Shoelace Theorem

The Shoelace theorem is a useful formula for finding the area of a polygon when we know the coordinates of its vertices. The formula was described by Meister in 1769, and then by Gauss in 1795.

### Formula

Let’s suppose that a polygon P has vertices (a1, b1), (a2, b2), …, (an, bn), in clockwise order. Then the area of P is given by

$$\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|$$

The name of this theorem comes from the fact that if you were to list the coordinates in a column and mark the pairs to be multiplied, then the image looks like laced-up shoes.

### Proof

(Note: this proof is taken from artofproblemsolving.)

Let $\Omega$ be the set of points that belong to the polygon. Then

$$A=\int_{\Omega}\alpha,$$

where $\alpha=dx\wedge dy$.

Note that the volume form $\alpha$ is an exact form since $d\omega=\alpha$, where

$$\omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega}$$

Substitute this in to give us

$$\int_{\Omega}\alpha=\int_{\Omega}d\omega.$$

and then use Stokes’ theorem (a key theorem in vector calculus) to obtain

$$\int_{\Omega}d\omega=\int_{\partial\Omega}\omega.$$

where

$\partial \Omega=\bigcup A(i)$

and $A(i)$ is the line segment from $(x_i,y_i)$ to $(x_{i+1},y_{i+1})$, i.e.  is the boundary of the polygon.

Next we substitute for $\omega$:

$$\sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}.$$

Parameterising this expression gives us

$$\frac{1}{2}\sum_{i=1}^n\int_0^1{(x_i+(x_{i+1}-x_i)t)(y_{i+1}-y_i)}-{(y_i+(y_{i+1}-y_i)t)(x_{i+1}-x_i)\,dt}.$$

Then, by integrating this we obtain

$$\frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)- (y_{i}+y_{i+1})(x_{i+1}-x_i)].$$

This then yields, after further manipulation, the shoelace formula:

$$\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).$$

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## New Podcast!

Today is a quick post to let you know that the writer of one of my favourite blogs Roots of Unity, Evelyn Lamb, and the mathematician Kevin Knudson from the University of Florida have created a new podcast called ‘My Favourite Theorem‘!

Lamb says:

In each episode, logically enough, we invite a mathematician on to tell us about their favourite theorem. Because the best things in life are better together, we also ask our guests to pair their theorem with, well, anything: wine, beer, coffee, tea, ice cream flavours, cheese, favourite pieces of music, you name it. We hope you’ll enjoy learning the perfect pairings for some beautiful pieces of math.

Click here to listen to the first episode, which features Lamb and Knudson telling us about their favourite theorems!

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## Goodstein Theorem

On reading a magazine on Gödel’s Incompleteness Theorems, I came across a family of sequences of non-negative integers called Goodstein sequences and the Goodstein Theorem involving these sequences.

Goodstein’s thoerem is a statement about the natural numbers, proved by Reuben Goodstein in 1944, which states that every Goodstein sequence converges to 0.

### What is a Goodstein Sequence?

To understand what a Goodstein sequence, first we must understand what hereditary base-n notation is. Whilst this notation is very similar to the usual base-n notation, base-n notation is not sufficient for Goodstein’s theorem.

To convert a base-n representation to a hereditary base-n notation, first rewrite all of the exponents in base-n notation. Then rewrite any exponents inside the exponents, and continue this way until every number in the expression has been converted to base-n notation.

For example, 35 = 25 + 2 + 1 in ordinary base-2 notation but  in hereditary base-2 notation.

Now we can define the Goodstein sequence G(m) of a natural number m. In general the (n+1)-st term G(m)(n+1) of the Goodstein sequence of m is given as follows (taken from Wikipedia):

• Take the hereditary base-n + 1 representation of G(m)(n).
• Replace each occurrence of the base-n + 1 with n + 2.
• Subtract one. (Note that the next term depends both on the previous term and on the index n.)
• Continue until the result is zero, at which point the sequence terminates.

In spite of the rapid growth of the terms in the sequence, Goodstein’s theorem states that every Goodstein sequence eventually terminates at 0, no matter what the starting value is.

Mathematicians Laurie Kirby and Jeff Paris proved in 1982 that Goodstein’s theorem is not provable in ordinary Peano arithmetic. In other words, this is the type of theorem described in 1931 by Gödel’s incompleteness theorem.

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The Banach-Tarski Paradox is a theorem in geometry which states that:

“It is possible to decompose a ball into five pieces which can be reassembled by rigid motions to form two balls of the same size as the original.”

It was first stated in 1924, and is called a paradox as it contradicts basic geometric intuition.

An alternate version of this theorem tells us that:

“It is possible to take a solid ball the size of a pea, and by cutting it into a finite number of pieces, reassemble it to form a solid ball the size of the sun.”

Below is an awesome video explaining how this paradox works:

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