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 there is a natural homomorphism:

i:\mathbb{Z}\to F

Where:

i(1)=1_F

And that’s it. every other is determined by the value of :

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 then:

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

That’s a homomorphism, thus, it has a **kernel** .

If 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 is a field, then is an integral domain – as a subring of one. Therefore, is an integral domain as well. Thus, is a **prime ideal** (I proved it in this post). But we know what prime ideals of look like – they are of the form where is **prime**.

In that case, we say that has **Characteristic** .

We denote Characteristic of a field as

Notice that in such a field: .

## Cool property in fields with charactetistic

If you recall, the Binomial theorem shows us how to simplify expressions of the form (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 :

If then:

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

If 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 . Thus doesn’t divide any of the factors. Since is prime – we know that it won’t devide the product as well.

However, divides the numerator – . Thus, divides the binomial coefficient! Then there is some 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 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 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:

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 is an **irreducible** polynomial. Then the following are equivalent

- is
**not**separable. - = 0.
- and for some .

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 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:

Suppose that is not separable. Then in a splitting field we can represent 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 . i.e. is a root of . Since is irreducible over (thus, it is the minimal polynomial of ), we get that:

f|f^\prime

However, . Since divides we get that or . But if we will get a contradiction to being the minimal polynomial.

Therefore, is constant. But we’ve seen that . So we can conclude that .

If then which is **0** by our assumption.

Therefore, all the coefficients are 0:

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

However, 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 . Thus divides and we can write . 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 .

By our assumption we have such . Then we can split in some splitting field 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 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 (then the extension field also has charecteristic ) and by the cool property I proved. Thus, over the field 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 where:

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

with addition and multiplication modulo .

Here are some examples of elements in this field when :

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

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

Over , the polynomial

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

is irreducible. Why?

Over , is a root of . moreover, clearly the , thus, over :

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

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

(x-t)^i

It’s constant term is and if then . Therefore, none of it’s divisors is in , thus, it is irreducible.

In addition, we’ve seen through the explanation that 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!