fields – Some Mathematical Farts

Algebras and why modules are a thing

March 16, 2018March 16, 2018 YatimaLeave a comment

A $k$ -algebra $A$ where $k$ is a field, is a ring with the added structure of being a vector space over $k$ (with respect to the same addition) where scalar multiplication behaves nicely with the ring multiplication. $\mu(ab) = (\mu a)b = a(\mu b)$ for $\mu \in k$ and $a, b \in A$ . The atypical example is a matrix algebra of dimension $n$ with matrix elements in $k$ denoted $M_{n}(k)$ . There is a notion of algebra morphism which is not too hard to figure out with the above or a quick google, so we have a notion of isomorphism.

Matrix algebras are very special, they have vectors in which they can act on. For an $n$ -dimensional matrix algebra these objects are the tuples $(x_1, \cdots, x_n)$ where $x_i \in k$ . As a collection, they are denoted $k^n$ . These form a vector space and as algebras $M_{n}(k) \cong \text{Aut}(k^n)$ . Where the algebra operation in $M_n(k)$ is matrix multiplication and $\text{Aut}(k^n)$ is the algebra of vector space automorphisms $k^n \longrightarrow k^n$ with the algebra operation being composition.

The definition of an $A$ -module can be found elsewhere but to motivate its origins, it is precisely the articulation of assigning some ‘vectors’ in some vector space $V$ to a general $k$ -algebra $A$ which $A$ can act on $V$ similarly to matrices on tuples. If you struggle to remember the definition of an $A$ -module just think like this and you should be able to reconstruct it.

Modules can even be defined for a ring $R$ , with the relaxation of the structure of a vector space. And so we can assign ‘vectors’ to rings by looking at $R$ -modules. So we can ask the question, what are $\mathbb{Z}$ -modules? These are in fact abelian groups (why?) and so the collection of possible ‘vectors’ for $\mathbb{Z}$ can be thought of as just abelian groups.

I think that’s pretty nice.

Sums of powers in Finite Fields

March 13, 2017March 19, 2017 YatimaLeave a comment

I like finite fields. This isn’t hard to see as two of my posts (post 1 & post 2) have been about their structure but sticking with this ‘theme’ I’ve forced upon myself lets investigate them some more! Unlike previous posts I’m just going to jump straight into the maths.

Theorem: We let $K =\mathbb{F}_{p^q}$ be a finite field with $p^q$ elements where $p$ is prime and $q \in \mathbb{Z}_{>0}$ . If we let $u \in \mathbb{Z}$ the sum $S(x^u) := \sum_{x \in K}x^u$ is equal to $-1$ if $u \geq 1$ and is divsible by $p^q-1$ or $0$ otherwise.

Proof of Theorem: If we have $u = 0$ then $S(x^u) = |K| = p^q = 0$ as we have $Char K = p$ . Now we consider when $u \geq 1$ and when $p^q -1$ divides $u$ . Trivially $0^u = 0$ and if $x \in K \backslash \{0\}$ we have $x^u = x^{k(p^q-1)} = 1$ for some $k \in \mathbb{Z}$ . This follows directly from Lagrange’s theorem applied to the multiplicative group $K^*$ of units and that $Char K = p$ . This gives us $S(x^u) = (p^q - 1) = -1$ by again using the characteristic of $K$ .

The last case, when $u$ is not divisible by $p^q -1$ , is not as direct as the others (hence the new paragraph) and requires knowledge of the structure of $K^*$ the units of $K$ . In fact $K^* \cong \mathbb{Z}_{p^q - 1}$ which if you’re not familiar with is shown in ‘post 2’. This means that there exists an element $y \in K^*$ such that $y^u \neq 1$ . We can simply take $y$ to be the generator of $K^*$ . As we are in a field we can say $S(x^u) = \sum_{x \in K}x^u = \sum_{x \in K}y^ux^u$ $= y^u S(x^u)$ so we have $(1- y^u) S(x^u) = 0$ and due to $y^u \neq 1$ we must conclude that $S(x^u) = 0$ . We are done!

What can this theorem tell us? A quick study of the result gives us that if $u \in \mathbb{Z}$ is odd and the prime $p \neq 2$ we must get $S(x^u)= 0$ as $p^q -1$ is always even so it cannot divide $u$ . This is quite remarkable as the result implies there are huge amounts of cancellation in the sum which a prior has no reason to occur.

The Units of finite fields

March 9, 2017April 4, 2017 YatimaLeave a comment

Given a field $K$ a unit is an element with a multiplicative inverse and the collection of all of them has a group structure induced from the field and is called the multiplicative group of units denoted $K^*$ . All elements in a field, apart from $0$ , have a multiplicative inverse so we can say $K^* = K \backslash \{0\}$ . When given a field, one can ask what the structure of the group of units is and in this post we are going to look at this question for finite fields.

Theorem: The multiplicative group of units for a finite field is cyclic more precisely, given $p$ prime and $n \in \mathbb{N}$ we have $\mathbb{F}_{p^n}^* \cong \mathbb{Z}_{p^n-1}$ .

Before proving this we need to introduce a nice function known as Euler’s $\varphi$ -Function which is defined as $\varphi : \mathbb{N} \longrightarrow \mathbb{N}$ such that $\varphi(n) = |\{ a \in \mathbb{N} : a \leq n, gcd(a,n) = 1 \} |$ where $gcd(a,n)$ is the greatest common divisor of $a$ and $n$ . One way of thinking about the $\varphi$ -function is that it counts the number of generators in a cyclic group or equivalently $\varphi(n) = |\{a \in \mathbb{Z}_n : \mathbb{Z}_n = \langle a \rangle \}|$ thanks to the fact all cyclic groups are ‘integers’. This new interpretation follows from that for $a \in \mathbb{Z}_n$ we have $|a| = \frac{n}{gcd(a,n)}$ which is left for you to check.

Let us use this new perspective of the $\varphi$ -function to establish some of its properties. One such property is that for any $n \in \mathbb{N}$ we have $n = \sum_{d | n }\varphi(d)$ . To show this we look at the cyclic group $\mathbb{Z}_n$ and notice for each $d | n$ there is a unique $x \in \mathbb{Z}_n$ such that $|x| = n$ (where $|.|$ is the order of an element). This is a property of cyclic groups and is left for you to prove… That is if you don’t believe me. Let $C_d$ be the collection of generators in the unique subgroup, $\langle x \rangle$ , with $d$ elements. Since all the elements of $\mathbb{Z}_n$ are contained in a $C_d$ for some $d$ we have $n = |\mathbb{Z}_n| =\sum_{d |n}|C_d| = \sum_{d | n}\varphi(d)$ .

Now we have all the tools to prove our theorem. Lets get to it.

Proof of Theorem: We are looking at the finite field $K = \mathbb{F}_{p^n}$ with $p^n$ elements and we first notice that for $d \in \mathbb{N}$ such that $d \;| \;(p^n - 1) = |K^*|$ the set $A_d := \{x \in K^* : x^d = 1 \}$ is such that $|A_d| \leq d$ . This is because the equation $x^d = 1$ in $K$ has at most $d$ solutions due to the polynomial $x^d -1$ having degree $d$ . We shall define the set $\overline{A}_d := \{x \in K^* : |x| = d \}$ and look at its cardinality. If $\overline{A}_d \neq \emptyset$ then for $x \in \overline{A}_d$ we have $\langle x\rangle \cong \mathbb{Z}_d$ and so by Lagrange’s Theorem $\langle x \rangle \subseteq A_d$ but with the fact $|A_d| \leq d = |\langle x \rangle|$ we get $A_d = \langle x \rangle$ hence we must get $|\overline{A}_d | = \varphi(d)$ . To summarise either the cardinality of $\overline{A}_d$ is $0$ or $\varphi(d)$ . Suppose that $\overline{A}_d= \emptyset$ for some $d$ using the properties of the $\varphi$ -function we have $p^n - 1 = \sum_{f | (p^n - 1)}|\overline{A}_f | = \sum_{f \neq d | (p^n -1) }\varphi(f)$ $< p^n -1$ which is a contradiction so we must always be in the second case. Taking $d = p^n - 1 = |K^*|$ we get $\overline{A}_{|K^*|} \neq \emptyset$ so a generator of $K^*$ must exist so it is cyclic with $p^n -1$ elements and so $K^* \cong \mathbb{Z}_{p^n - 1}$ due to the classification of cyclic groups.

What do Finite Fields look like?

January 10, 2017March 19, 2017 YatimaLeave a comment

A field normally denoted $K$ is an abstract structure that satisfies a list of axioms, these axioms are motivated by the algebraic properties of the rational numbers $\mathbb{Q}$ and other number systems such as the real numbers $\mathbb{R}$ and the complex numbers $\mathbb{C}$ . With the privilege of being alive now rather than in the 1800s abstract mathematics, particularly abstract algebra, is widely taught and fields can just be considered as a set with a nice structure (I’m not trying to dumb down the concept but to emphasis that fields aren’t illusive and magical). In this post I will be talking a bit about Field extensions, Rings, Integral Domains and the structure of the integers as a ring. A quick warning: this post is rather long.

An extremely important concept in field theory, more generally ring theory, is the characteristic of such a structure denoted $\mbox{Char} \; K$ . The characteristic is the first step in the direction to classify fields/rings as, informally, it tells you what the smallest subfield/ring our structure contains. More precisely we can map $\mathbb{Z}$ into $K$ by $\varphi : \mathbb{Z} \longrightarrow K$ defined by $\varphi(n) = \sum_{k=1}^{n} 1$ , $\varphi(-n) = \sum_{k=1}^{n} - 1$ for $n \in \mathbb{Z}^+$ and $\varphi(0) = 0$ . This is a ring homomorphism and the kernel is an integral domain of $\mathbb{Z}$ so it will be of the form $n\mathbb{Z} = \{nx : x \in \mathbb{Z} \}$ where $n \in \mathbb{N}$ and using the first isomorphism theorem $\mathbb{Z}/ n \mathbb{Z} \cong \varphi(\mathbb{Z})$ and as the image is an integral domain $n = p$ where $p$ is prime or $n = 0$ . We define the characteristic as this value so either $\mbox{Char} K = p$ where $p$ is prime or $0$ . Another way to think about the characteristic, is that it’s the smallest number of times you have to continuously add the multiplicative identity to get the additive identity.

Throughout the rest of the post $K$ will be a finite field. With $K$ being finite it’s not hard to show it must have $\mbox{Char} K = p$ where $p \neq 0$ is prime. This follows by the short argument that if $\mbox{Char} K = 0$ the field must contain $\mathbb{Z}$ as a subring, this cant be so as $\mathbb{Z}$ is infinite and hence so would $K$ which gives us a contradiction. This tells us that $K$ must contain the field $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ if $\mbox{Char} K = p$ and it’s the smallest field $K$ contains. Looking at the degree of the extension $[K : \mathbb{F}_P] = n$ (if you’re unfamiliar with this concept click here) we have $|K| = p^n$ so we get a bit more information, the size of a finite field must be a power of a prime.

We’ll assume our finite field $K$ has $\mbox{Char} K = p$ from now on and in doing so we have an endomorphism $\sigma : K \longrightarrow K$ defined by $\sigma(x) = x^p$ for each $x \in K$ . This map is given a special name, the Frobenius endomorphism, and as all things with a name it’s important.

As we’ve assumed $K$ is finite we can in-fact conclude $\sigma$ is an automorphism (This is not true for a general field). let $x,y \in K$ we calculate $\sigma(xy) = (xy)^p = {x^p}{y^p} = \sigma(x)\sigma(y)$ and by noticing ${{p}\choose{k}} \equiv 0 \mbox{(mod)}p$ with $1 \leq k \leq p -1$ , using the binomial theorem and that our characteristic is $p$ we calculate $\sigma(x + y) = \sum_{ k = 0}^{k = p} {{p}\choose{k}} x^{k}y^{p-k} = x^p + y^p = \sigma(x) + \sigma(y)$ so $\sigma$ is a homomorphism and as the name suggests an endomorphism as $K$ is closed under multiplication. As $K$ is finite to show $\sigma$ is bijective we only need to show the map is injective or surjective and it’s much easier to prove injectivity. Suppose $\sigma(x) = \sigma(y)$ then $x^p - y^p = 0$ so by the fact we have a homomorphism $(x-y)^p = 0$ and as we are in an integral domain, in particular a finite field, we have $x = y$ .

With all this setup we are in good shape to show what finite fields must look like and surprisingly there is only one such possibility. The result is stated in the theorem bellow and was first proved by E.H Moore in 1893.

Theorem: Let $q = p^n$ where $p$ is prime, let $\Omega$ be an algebraically closed field with $\mbox{Char} \Omega = p$ . There exists a unique subfield $\mathbb{F}_q$ of $\Omega$ with $q$ elements. This is the set of roots of the the polynomial $x^q - x$ more explicit $\mathbb{F}_q = \{x \in \Omega : x^q - x = 0 \}$ .

Proof of Theorem: We have $\Omega$ being algebraically closed with characteristic $p$ so we must have the map $\psi : \Omega \longrightarrow \Omega$ defined by $\psi(x) = x^q$ being an injective homomorphism as $\psi(x) = \sigma(x)^n$ , think of $\psi$ as the concatenation of $\sigma$ $n$ times. Note that it isn’t necessarily an isomorphism as $\Omega$ might not be finite. This means the fixed points of $\psi$ , the set $\{x \in \Omega : x^ q = x \}$ forms a subfield of $\Omega$ which is precisely $\mathbb{F}_q$ .

If we show that our set of fixed points has size $q$ we are done as we’ve found our $\mathbb{F}_q$ of size $q$ . This is equivalent to showing $x^q - x$ has $q$ distinct roots in $\Omega$ . We can easily argue that if $x^q - x$ has roots they must be distinct as we observe if $f \in \Omega[x]$ has a ‘double’ root say $\alpha$ then $(x -\alpha)^2$ divides $f$ so the derivative of $f$ with respect to $x$ , $f'$ is non-constant but $(x^q - x)' = qx^{q-1} - 1 = p^nx^{q-1 } - 1 = -1$ as $\mbox{Char} \Omega = p$ . This means all the roots are distinct and as we have assumed $\Omega$ is algebraically closed so all $q$ of them must exist.

To show uniqueness we assume $K$ is a subfield of $\Omega$ with exactly $q$ elements. Looking at the multiplicative group of units $K^{*} = \{ x\in K : \exists y \in K \mbox{ such that } xy = 1 \} = K \backslash \{0\}$ by definition there are $q -1$ of them, by Lagrange’s Theorem for groups we have $x^{q-1} = 1$ for $x \in K^*$ so $x^q - x = 0$ hence $K^* \subset \mathbb{F}_q$ . Trivially we have $0 \in \mathbb{F}_q$ so $K \subset \mathbb{F}_p$ and as they both contain $q$ elements we have $K = \mathbb{F}_q$ .

It follows directly from this theorem that all finite fields $K$ with $q = p^n$ elements where $p$ is prime is isomorphic to the $\mathbb{F}_q$ defined above. This follows from Zorn’s lemma allowing us to embed $K$ into an algebraically closed field $\Omega$ .

So there we have it.