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



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:


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

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<k<p 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:


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:


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

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:


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:


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

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:


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

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




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


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:


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


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


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:


Over F, the polynomial


is irreducible. Why?

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


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


It’s constant term is t^i and if i<p 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.


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!

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