# Separable Polynomial

In the last post I’ve introduced the term of a separable polynomial. In this post I want to dive deeper into the defintion and persent some properties and criteria of separable polynomials.

## Characteristic of a field

My major plan for this post is to present and prove a criteria that will determine wether or not a polynomial is separable. To do so, we need to define a new property of a field. It’s Characteristic.

For every field $F$ there is a natural homomorphism:

i:\mathbb{Z}\to F

Where:

i(1)=1_F

And that’s it. every other $n\in \mathbb{Z}$ is determined by the value of $1$:

n\geq 0:\ \ i(n)=i(\overbrace{1+1+\dots+1}^{n\text{ times}})=\overbrace{i(1)+i(1)+\dots+i(1)}^{n\text{ times}}
=\overbrace{1_F+1_F+\dots+1_F}^{n\text{ times}}=n_F

And if $n< 0$ then:

i(-n)=-i(n)=-n_F

That’s a homomorphism, thus, it has a kernel $K=\ker i$.

If $K=(0)$ then we say that F has Characteristic 0.

If that’s not the case, by the first isomorphism theorem we have:

\mathbb{Z}/K\cong\text{Im}(i)\sub F

Since $F$ is a field, then $\text{Im}(i)$ is an integral domain – as a subring of one. Therefore, $\mathbb{Z}/K$ is an integral domain as well. Thus, $K$ is a prime ideal (I proved it in this post). But we know what prime ideals of $\mathbb{Z}$ look like – they are of the form $\langle p \rangle$ where $p\in\mathbb{Z}$ is prime.

In that case, we say that $F$ has Characteristic $p$.

We denote Characteristic of a field as $\text{char}(F)$

Notice that in such a field: $p=0$.

## Cool property in fields with charactetistic $p$$p$

If you recall, the Binomial theorem shows us how to simplify expressions of the form $(a+b)^n$ (I am actually planning on writing a post about it in the next week). According to it:

(a+b)^p=\sum_{k=0}^p {p\choose k}a^{p-k}b^{p}

Let’s see what’s happening when we are working with field with Characteristic $p$:

If $k=0,p$ then:

{p\choose0}={p\choose p}=1

If $0 then:

{p\choose k}=\frac{p!}{k!(p-k)!}=\frac{p!}{1\cdot2\cdots k\cdot1\ \cdot\ 2\cdots(p-k)}

Each one of the factor in the denominator is a number that is smaller than $p$. Thus $p$ doesn’t divide any of the factors. Since $p$ is prime – we know that it won’t devide the product as well.

However, $p$ divides the numerator – $p!$. Thus, $p$ divides the binomial coefficient! Then there is some $m\in\mathbb{Z}$ such that

{p\choose k}=m\cdot p=m\cdot 0_F=0_F

That’s cool, from that property we conclude that:

(a+b)^p=\sum_{k=0}^p {p\choose k}a^{p-k}b^{p}=a^p+{p\choose 1}a^{p-1}b+\dots+{p\choose p-1}a^{}b^{p-1}+b^p
a^p+0\cdot a^{p-1}b+\dots+0\cdot a^{}b^{p-1}+b^p=a^p+b^p

That’s really cool. We got that if we are working with a field with Characteristic $p$ then:

(a+b)^p=a^p+b^p

## Formal derivative

The last thing I want to present before the theorem is the formal derivative of a polynomial. If $f=\sum_{k=0}^na_kx^k$ then the formal derivative is defined as:

f^\prime=\sum_{k=0}^nka_kx^{k-1}

This is just like the “regular” derivative but this is a formal definition – we didn’t derive it from a limit calculation and we don’t care about tangents and other terms from analysis.

It’s not hard to verify that:

• $(f+g)^\prime=f^\prime+g^\prime$
• $(c\cdot f)^\prime=c\cdot f^\prime$
• $(fg)^\prime=f^\prime g+ f g^\prime$

And all the other properties we are familiar with (such as ‘chain rule’…)

The derivative is going to play a part in the theorem.

## The Criteria

Suppose that $f\in F[x]$ is an irreducible polynomial. Then the following are equivalent

1. $f$ is not separable.
2. $f^\prime$ = 0.
3. $\text{char}(F)=p>0$ and $f(x)=g(x^p)$ for some $g\in F[x]$.

Some notes on the criteria:

• Notice that the criteria guarantees us that if we are working with a field with charecteristic 0 then any irreducible polynomial is separable.
• We also see that the separabilty of $f$ is determined only by it’s formal derivative. Thus, being a separable polynomial is not a property that depends on the splitting field.

Ok, let’s prove it:

#### $1\Rightarrow 2$$1\Rightarrow 2$

Suppose that $f$ is not separable. Then in a splitting field we can represent $f$ as:

f(x)=(x-\alpha)^2\cdot h(x)\ \ , \ \ h\in E[x]

Let’s calculate the derivative:

f^\prime(x)=2(x-\alpha)h(x)+(x-\alpha)^2h^\prime(x)

Notice that $f^\prime(\alpha)=0$. i.e. $\alpha$ is a root of $f^\prime$. Since $f$ is irreducible over $F$ (thus, it is the minimal polynomial of $\alpha$), we get that:

f|f^\prime

However, $\deg(f)=\deg(f^\prime)+1>\deg(f)^prime$. Since $f$ divides $f^\prime$ we get that $\deg(f)<\deg(f^\prime)$ or $\deg(f^\prime) = 0$. But if $\deg(f)<\deg(f^\prime)$ we will get a contradiction to $f$ being the minimal polynomial.

Therefore, $f^\prime$ is constant. But we’ve seen that $f^\prime(\alpha)=0$. So we can conclude that $f^\prime = 0$.

#### $2\Rightarrow 3$$2\Rightarrow 3$

If $f=\sum_{k=0}^na_kx^k$ then $f^\prime=\sum_{k=0}^nka_kx^{k-1}$ which is 0 by our assumption.

Therefore, all the coefficients are 0:

ka_k=0\text{ for }0\leq k\leq\deg(f)

However, $f$ itself is not the zero polynomial, therefore there are coefficients that are not zero. Let’s pick one:

ka_k=0,a_k\neq0

Therefore $k=0$. Thus $p$ divides $k$ and we can write $k=m_k\cdot p$. Thus:

f(x)=\sum_{k:a_k\neq 0}a_kx^k=\sum_{k:a_k\neq 0}a_{m_k\cdot p}x^{n_k\cdot p}=\sum_{k:a_k\neq 0}a_{m_k\cdot p}(x^{n_k})^{p}

Then we can define:

g(x)=\sum_{k:a_k\neq 0}a_{m_k\cdot p}x^{n_k}

And get that $g(x^p)=f(x)$.

#### $3\Rightarrow 1$$3\Rightarrow 1$

By our assumption we have such $g$. Then we can split $g$ in some splitting field $E$ to get:

g(x)=\prod_{i=1}^n(x-\alpha_i)

Thus:

f(x)=g(x^p)=\prod_{i=1}^n(x^p-\alpha_i)

Now, over the field $E[\sqrt[p]{\alpha_1},\dots,\sqrt[p]{\alpha_n}]$ we know that:

x^p-\alpha_i=(x-\sqrt[p]{\alpha_i})^p

By the fact that we are working with a field that extends a field with charecteristic $p>0$ (then the extension field also has charecteristic $p>0$) and by the cool property I proved. Thus, over the field $E[\sqrt[p]{\alpha_1},\dots,\sqrt[p]{\alpha_n}]$ we have:

f(x)=\prod_{i=1}^n(x-\sqrt[p]{\alpha_i})^p

And that’s indeed not a separable polynomial.

### Irreducible and not separable polynomial

This theorem shows us that being an irreducible and not separable at the same time – is kind of a strong property. I am going to give now one of the most famous example of such a polynomial, and as you will – It is not a trivial example at all.

The field we are goung to deal with is

E=\mathbb{F}_p(t)

This is the field of the rational functions over the field $\mathbb{F}_p$ where:

\mathbb{F}_p=\{0,1,2\dots,p-1\}

with addition and multiplication modulo $p$.

Here are some examples of elements in this field when $p=3$:

• $t^2+3^+1$
• $t^4+2t+1$

Ok, now I am going to define a sub-field:

F=\mathbb{F}_p(t^p)

Over $F$, the polynomial

f(x)=x^p-t^p

is irreducible. Why?

Over $E$, $t$ is a root of $f$. moreover, clearly the $\text{char}(F)=p$, thus, over $E$:

f(x)=x^p-t^p=(x-t)^p

Again, this follows from the cool property. therefore, any polynomial that divides $f$ is of the form:

(x-t)^i

It’s constant term is $t^i$ and if $i then $t^i\notin F$. Therefore, none of it’s divisors is in $F[x]$, thus, it is irreducible.

In addition, we’ve seen through the explanation that $f$ is not separable.

## Summary

We now have a tool to check wether a polynomial is separable or not. This is going to be useful in the future.

In the next post we are finally going to meet a group – The Automorphism group. And that’s where galois theory actually begins!