Extension Fields

You heavily need to understand Vector Spaces in Abstract Algebra prior to this.

Definition

Simple:

An extension field is exactly what it sounds like, its a bigger field that contains the original field. To put it plainly, let field $E$ be the extension of field $F$. Then both are fields and $F\subset E$. In this case $F$ is the base field and $E$ is the extension field.

Rigorous:

Let $F$ and $E$ be fields. Then, if there exists a homomorphism $\psi :F\to E$ which is injective, then E is an extension field of the base field $F$, and $\psi$ is known as the field extension.

Lets show that $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is an extension of $\mathbb{Q}(\sqrt{2})$

First, the way you read $\mathbb{Q}(\sqrt{2})$ is the smallest field containing both $\mathbb{Q}$ and $(\sqrt{2})$. Similarly, $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is the smallest field containing $\mathbb{Q}$ and $(\sqrt{2}+\sqrt{3})$.

Then, we just have to show that $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ contains $(\sqrt{2})$. Notice $(\sqrt{2}+\sqrt{3}{)}^{-1}=\sqrt{3}-\sqrt{2}$ Then, $\sqrt{2}+\sqrt{3}-(\sqrt{3}-\sqrt{2})=2\sqrt{2}$ Lastly, we know $\frac{1}{2}\in \mathbb{Q}\subset \mathbb{Q}(\sqrt{2}+\sqrt{3})$ So we can show $\frac{1}{2}(2\sqrt{2})=\sqrt{2}$

And now we have shown that $\sqrt{2}\in \mathbb{Q}(\sqrt{2}+\sqrt{3})$ thus this is a proper extension field.

To build on this, we can show $\frac{1}{2}(\sqrt{2}+\sqrt{3})-\frac{1}{2}(\sqrt{3}-\sqrt{2})=\sqrt{2}$ or that $\sqrt{2}$ can be shown as a linear combination of $(\sqrt{2}+\sqrt{3})$ and $(\sqrt{3}-\sqrt{2})$.

Constructing a Field extension of ${\mathbb{Z}}_2$

Notice that $p(x)=x^2+x+1$ is irreducible in ${\mathbb{Z}}_2$ as $p(0)=1$ and $p(1)=1$ (There would need to be a factor of deg 1).

With this, we want to create a field extension that has an element $a$ such that $p(a)=0$.

First, we know ${\mathbb{Z}}_2$ is a field and $p(x)$ is irreducible, and so from Introductory Ring Theory:

$p(x)irreduceable⟺⟨p(x)⟩maximal⟺{\mathbb{Z}}_2/⟨p(x)⟩field$ Thus we can notice that ${\mathbb{Z}}_2[x]/⟨p(x)⟩$ is a field. (Which is needed for it to be an extension field lmao)

Now, lets take a look at this field. Through the division algorithm we know every element in this field is some $f(x)+⟨p(x)⟩$

So $f(x)=p(x)q(x)+r(x)$ where the degree of $r(x)$ is less then $2$. Then, as $p(x)+q(x)$ are in the ideal we know $f(x)+⟨p(x)⟩=r(x)+f(x)⟩$

Finding all remainders for $p(x)=x^2+x+1$ we see they can only be elements under deg 2. In ${\mathbb{Z}}_2[x]$ this leaves the elements of deg 0: $0,1$; and the elements of deg 1: $x,x+1$.

Thus we have the following elements in ${\mathbb{Z}}_2[x]/⟨p(x)⟩$: $0+⟨p(x)⟩$ $1+⟨p(x)⟩$ $x+⟨p(x)⟩$ $x+1+⟨p(x)⟩$

Hence ${\mathbb{Z}}_2[x]⟨p(x)⟩=\{0,1,x,x+1\}$.

Now, we are aware that both $0$ and $x^2+x+1$are in the same ideal, thus we have it so $[x^2+x+1]=[0]$ So we also know that : $[x{]}^2+[x]+[1]=[x^2+x+1]=[0]$ so there must be some $x$ which gives us a zero for $x^2+x+1$ in this equation, as each term only gives us one constant. Call that zero $\alpha$. By the first isomorphism theorem, we know this ring is now isomorphic to ${\mathbb{Z}}_2(\alpha )$ as the kernel of the homomorphism from $\varphi :{\mathbb{Z}}_2[x]\to {\mathbb{Z}}_2(\alpha )$ would be $⟨p(x)⟩$ making ${\mathbb{Z}}_2[x]⟨p(x)⟩\cong \mathbb{Z}(\alpha )$ as evaluation homomorphisms are surjective.

Based on this, we also know that $p(\alpha )={\alpha }^2+\alpha +1=0$.

Kronecker’s theorem

Let $F$ be a fiend and $p(x)$ be a non-constant polynomial of $F[x]$. Then there exists an extension field $E$ of $F$ and also there exists some $\alpha \in E$ where $p(\alpha )=0$

Essentially for any polynomial ring of a Field, every polynomial has a zero in an extension field (or $F[x]$)

I reproved the theorem here: Proof of Kronecker’s Theorem

Proof:

1. First, we only need to check irreducible polynomials as its clear if the polynomial was reducible it would have a zero in the given field.
2. Let $p(x)\in F[x]$ be such a polynomial. Then we have it so $⟨p(x)⟩$ is a maximal ideal in $F[x]$ and thus the quotient ring $F[x]/⟨p(x)⟩$ is a field. We will prove this is the extension field which contains the root of $p(x)$.
3. Now, lets first show that $F[x]/⟨p(x)⟩$ is actually an extension field of $F$ by defining an injective ring homomorphism from $\psi :F\to F[x]/⟨p(x)⟩$ defined as $\psi (f)=f+⟨p(x)⟩$.
4. We can check this is a ring homomorphism as, letting $a,b\in F$ $\psi (a+b)=((a+b)+⟨p(x)⟩)=(a+⟨p(x)⟩)+(b+⟨p(x)⟩)=\psi (a)+\psi (b)$ $\psi (ab)=(ab+⟨p(x)⟩)=(a+⟨p(x)⟩)(b+⟨p(x)⟩)=\psi (a)\psi (b)$
5. Lets also prove injectivity:
   1. Assume $\psi (a)=\psi (b)$
   2. then $a+⟨p(x)⟩=b+⟨p(x)⟩$.
   3. Then we have it so $(a-b)+⟨p(x)⟩=0+⟨p(x)⟩$
   4. Notice that this means $a-b\in ⟨p(x)⟩$
   5. The only constant polynomial (and therefor member of $F$) in $⟨p(x)⟩$ is 0 as $⟨p(x)⟩$ is a polynomial with at least deg 1, and as $a,b\in F$ it is clear $a-b=0$ or $a=b$
6. Now lets define F as a subfield of E such that $\{f+⟨p(x)⟩\mid \forall f\in F\}$ It is important to understand this holds as every coset only has one polynomial at max, so these are all different cosets.
7. review

Example:

Let $p(x)=x^5+x^4+1\in {\mathbb{Z}}_2[x]$. This polynomial has irreducible factors $x^3+x+1$ and $x^2+x+1$. Given this, find the extension fields on which zeros exist, define the base fields, and the elements of the extension field.

Zeros for this polynomial exist in the extension fields ${\mathbb{Z}}_2[x]/⟨x^3+x+1⟩$ ${\mathbb{Z}}_2[x]/⟨x^2+x+1⟩$ These are both extensions of the base field ${\mathbb{Z}}_2$.

Elements in ${\mathbb{Z}}_2[x]/⟨x^3+x+1⟩$ are the remainders when you modulo $x^3+x+1$. By the remainder theorem these are all elements degree 2 or lower, which are precisely: $0$ $1$ $x$ $x+1$ $x^2$ $x^2+x$ $x^2+x+1$ $x^2+1$

Algebraic Elements

Definitions

Algebraic:

An element in an extension field $E$ is algebraic over the base field $F$ if it serves as a root for an element in $F[x]$

If it is not a root for any element in the base field, it is transcendental over the base field.

An extension field where every element is algebraic is called an algebraic extension of F.

We can denote $F(a_1,a_2,....,a_n)$ as the smallest group containing $F$ and $a_1,a_2,....,a_n$ which is an extension.

If $E$ is generated by $F$ and one other element it is called a simple extension, ie $E=F(a)$

Examples:

As $(x^2-2)$ exists in $\mathbb{Q}[x]$ we know $\sqrt{2}$ is transcendental over $\mathbb{Q}$. Similar with $i$. $\pi$ and $e$ are not algebraic over $\mathbb{Q}$ (not trivial).

Most real numbers are transcendental over $\mathbb{Q}$

Noting that $\mathbb{Q}$ is a subfield of $\mathbb{C}$, imaginary numbers that are transcendental or algebraic over $\mathbb{Q}$ are called transcendental numbers or algebraic numbers respectfully.

Lets show that $\sqrt{2+\sqrt{3}}$ is algebraic over $\mathbb{Q}$. We can do this easily by just tracking changes to as we set it equal to 0: $\sqrt{2+\sqrt{3}}=x$ $2+\sqrt{3}=x^2$ $\sqrt{3}=x^2-2$ $3=(x^2-2{)}^2$ $0=(x^2-2{)}^2-3=x^4-4x^2+4-3=x^4-4x^2+1$ Thus $\sqrt{2+\sqrt{3}}$ is the zero for the polynomial $x^4-4x^2+1$ in $\mathbb{Q}$ and therefore algebraic.

If $a\in E$ is algebraic over $F$ then there exists a unique monic (no coefficients) polynomial of smallest degree where $p(a)=0$ (a is a root). If a is a root for any other elements then the $p(x)$ divides that element. (like how x-a divides all elements of which a is a root).

The unique monic polynomial $p(x)$ is called the minimal polynomial, and the order of that polynomial is called the degree of $a$ over $F$

For $p(x)$ being the minimal polynomial of some $\alpha \in E$, we can define a isomorphism $F(\alpha )\cong F[x]/⟨p(x)⟩$

This clearly works through the first isomorphism theorem because p(x) would be a factor of any element of which $\alpha$ is a root. Thus, the kernel of the evaluation homomorphism from $F[x]\to F(\alpha )$ is $⟨p(x)⟩$, which clearly means $F(\alpha )\cong F[x]/⟨p(x)⟩$ through the first isomorphism theorem.

For a basic extension $E=F(\alpha )$ where $\alpha$ is algebraic over $F$ (root of a n element) and degree of $\alpha$ is $n$ then every element of $E$ can be expressed as a ‘linear combination’ with n elements of F:

$f_0+f_1\alpha +f_2{\alpha }^2.....f_{n-1}{\alpha }^{n-1}$ for $f_i\in F$

If $E$ is a vector space over $F$ of dimension $n$, then we say $[E:F]=n$. This is noted the finite extension field degree $n$ over $F$

Remember, from Vector Spaces in Abstract Algebra dimension is just a length of the basis.

Every finite extension field $E$ over $F$ is algebraic.

Means all values in $E$ are the root for some value in $F[x]$

Proof:

1. Let $[E:F]=n$. Then we know at most there are $n$ independent vectors.
2. If we create

For any we can chain the index notation. ie if K is a finite extension of E which is a finite extension of F

$[K:F]=[K:E][E:F]$ This can be chained indefinitely.

for $[E(\alpha ):F]=degp(x)$ where p(x) is the minimal polynomial for \alpha in F[x]