# Proving that polynomial f must be of this form if the polynomial is given to be separable

If $$f \in K[x]$$ is monic irreducible, $$\deg (f) \geq 2$$, and has all its roots equal (in a splitting field), then $$\text{char }K = p \neq 0$$, and $$f = x^{p^n} - a$$ for some $$n\geq 1$$ and $$a\in K$$ .

Characteristic of Field $$F$$ is always a prime( Characteristic of Integral Domain is always prime) and if it is $$0$$ then the field is $$0$$ field and so $$f$$ is zero polynomial.

So, the characteristic of the field must be prime. Now, if all roots of $$f$$ are equal in some splitting field then it will be given in that splitting field by $$(x-u)^m$$ where $$m$$ is the degree of $$f$$ and $$u$$ is the root that is given equal.

But I don't get it how can I prove f to be equal to $$f = x^{p^n} - a$$ with the information given in the question. Can you please give some direction?

Thanks!

First of all, $$\text{char}K=0$$ means (by convention) that the unique ring homomorphism $$\mathbb Z\rightarrow K$$ is injective, e.g. when $$K=\mathbb Q$$, not if the ring is the zero ring. However if $$\text{char}K=0$$ then all irreducible polynomials are separable, which $$f$$ isn't. Therefore $$\text{char}K\neq0$$. Suppose $$\text{char} K=p$$.
By considering $$\gcd(f,f')=1$$ and $$f$$ being irreducible and inseparable, you'd realise $$f(x)=g(x^p)$$ for some $$g\in K[x]$$. Now notice $$g$$ must be irreducible easily, and it has only one root, because the roots of $$g$$ are simply $$p$$th powers of the only root of $$f$$. Therefore, either $$g$$ is inseparable or $$\deg g=1$$. Repeat the argument with $$g$$ in place of $$f$$ would eventually get you the result.