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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!

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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.

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