# How to show x^p-x+a is irreducible in F_p?

#### watson

How can I show that $$\displaystyle x^p-x+a$$ is irreducible in a finite field $$\displaystyle F_p$$? Thanks!

#### Turgul

Well, what is your toolbox? Are you familiar with field extensions? Galois theory?

What have you tried?

#### watson

Hi,

Thank you for your response. I am familiar with field extensions and Galois theory. First I want to say that for the question, I am assuming that $$\displaystyle a\not=0$$, else $$\displaystyle x^p-x+a$$ is not irreducible (note that $$\displaystyle x^p-x+0=x^p-x=x(x^{p-1}-1)$$).

What I tried: Suppose $$\displaystyle x^p-x+a$$ is not irreducible. Then we can write $$\displaystyle x^p-x+a=g(x)h(x)$$, where the degree of g and h are less than the degree of our polynomial, with positive degrees. Also, note that $$\displaystyle d/dx(x^p-x+a)=px^{p-1}-1=-1$$. We then have the following:

$$\displaystyle -1=d/dx(x^p-x+a)=g'(x)h(x)+g(x)h'(x)$$..

I'm just not really sure if I'm even going in a direction that might be helpful at this point.

#### Turgul

The primary thing we gain by looking at the derivative is that since the derivative is constant, our polynomial cannot have repeated roots, so none of the factors are repeated. It does not give us information on how many factors there are.

As you point out, the polynomial factors if $$\displaystyle a=0$$. But you can do better than that: by Fermat, we know $$\displaystyle \alpha^p = \alpha$$ for all $$\displaystyle \alpha \in F_p$$, hence every $$\displaystyle \alpha$$ is a root, hence $$\displaystyle x^p - x = (x-0)(x-1) \cdots (x-(p-1)) = x(x-1) \cdots (x-(p-1))$$.

Now let us look at the general polynomial $$\displaystyle x^p - x + a$$. We know this polynomial has a root in some extension of $$\displaystyle F_p$$ (it might just be $$\displaystyle F_p$$ itself, as in the case $$\displaystyle a = 0$$). Let $$\displaystyle \alpha$$ be such a root in this (minimal) extension. Then what can we say about $$\displaystyle \alpha + 1$$?

I claim that if a finite field of characteristic $$\displaystyle p$$ has one root of $$\displaystyle x^p - x +a$$, then it has all of them, hence $$\displaystyle x^p - x +a$$ splits completely over any extension of $$\displaystyle F_p$$ in which it has a root. Think about why this is true. Then think about how you can use Galois theory to show that $$\displaystyle x^p - x + a$$ is either irreducible over a field of characterstic $$\displaystyle p$$ or factors completely. Then all that remains to show is why $$\displaystyle x^p - x + a$$ does not have a root if $$\displaystyle a \neq 0$$.

#### watson

So if $$\displaystyle \alpha$$ is a root of $$\displaystyle x^p-x+a$$, then we have $$\displaystyle \alpha^p-\alpha+a=0$$. So then:

$$\displaystyle (\alpha+1)^p-(\alpha+1)+a=\alpha^p+1^p-\alpha-1+a=\alpha^p-\alpha+a=0,$$

so $$\displaystyle \alpha+1$$ is also a root if $$\displaystyle \alpha$$ is a root. I can't quite see how to use this though..

Also I previously had showed that $$\displaystyle x^p-x+a$$ was separable, so then if I could show that some extension field $$\displaystyle K$$ is the splitting field over $$\displaystyle F_p$$ for $$\displaystyle x^p-x+a$$, I would have that $$\displaystyle K/F$$ is Galois. But I suppose this is overkill, since isn't every finite field extension of a finite group a Galois extension?

I'll try some more things though..

#### Turgul

Well now you know that if $$\displaystyle \alpha$$ is a root, so is $$\displaystyle \alpha + 1$$. What about $$\displaystyle \alpha + 2$$? $$\displaystyle \alpha + 3$$?
If a field of characteristic $$\displaystyle p$$ has a root of the polynomial, how many (distinct) roots must the field have (at least)?

Indeed, you are correct that any finite extension of a finite field is Galois. In particular, if you believe $$\displaystyle f(x) = x^p - x + a$$ is irreducible, then it must be that the extension of $$\displaystyle F_p$$ which contains a single root $$\displaystyle \alpha$$ of $$\displaystyle f$$, it must contain all roots of $$\displaystyle f$$ and there must be automorphisms of the extension, fixing $$\displaystyle F_p$$, which transitively permute the roots.

Can you say what the Frobenius Automorphism does to the roots of $$\displaystyle f$$?