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

Question

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) or 0.

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!

Answer 1

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.

Answer 2

Claim 1: "Suppose we have a field $K$, $f \in K[x]$ is an irreducible polynomial and $0 \neq g \in K[x]$, prove that either $\gcd(f, g) = 1$ or $f$ divides $g$."

An elementary proof: Hint: $k[x]/(f(x))$ is a field. Assume $f(x)\in k[x]$ is (non-constant) irreducible and assume $g(x)\neq 0$ and let $k(\alpha):=k[x]/(f(x))$ where by definition $\alpha:=\overline{x}$.

Assume $g\notin (f(x))$. It follows $g(\alpha)\neq 0$ and since $k(\alpha)$ is a field, there is an element $h(\alpha)$ with $h(\alpha)g(\alpha)=1$ in $k(\alpha)$. It follows there is a polynomial $p(x)\in k[x]$ with

$$h(x)g(x)=1+p(x)f(x)$$

hence

$$h(x)g(x)-p(x)f(x)=1$$

and it follows there is an equality of ideals $(g(x),f(x))=1$. Conversely if $(g,f)=1$ it follows there are polynomials $h,p$ with $hg+pf=1$. If $g\in (f(x))$ you would get $g=\tilde{g}f$ and

$$ h\tilde{g}f+pf=(h\tilde{g}+p)f=1$$

a contradiction since $f(x)$ is non-constant. Hence $(g,f)=1$ iff $g\notin (f(x))$. Hence two situations can occur $g \in (f(x))$ or $g\notin (f(x))$ or equivalently $g\in (f(x))$ or $(g,f)=1$.

Claim 2: A polynomial $f(x) \in K[x]$ is inseparable iff $(f(x),f'(x))\neq (1)$.

Proof: If $f(x)=(x-a)^2h(x)$ is inseparable $a\in k$ it follows $(f(x),f'(x)) \neq (1)$ since $(x-a)$ is a factor of $f(x)$ and $f'(x)$. Conversely if $(f(x),f'(x))=(d(x))$ where $d(x)$ is non-constant it follows $f(x)=f_1(x)d(x)$ and $f'(x)=F_1(x)d(x)$ and any root $a$ of $d(x)$ is a root of $f(x)$ and $f'(x)$ hence $f(x)$ is inseparable.

Question: "Li "By considering $gcd(f,f′)=1$ and $f$ being irreducible and inseparable, you'd realise $f(x)=g(x^p)$ for some $g∈K[x]$" Can you please give a proof of this? – No -One 33 mins ago"

Answer: Hence if $f(x)\in K[x]$ is irreducible and inseparable it follows $(f(x),f'(x))\neq 1$ and hence $f(x)$ divides $f'(x)$ but since $deg(f'(x))< deg(f(x))$ this is possible iff $f'(x)=0$ which is iff $char(k)=p>0$ and $f(x)=G(x^p)$ for some $G(t)\in K[t]$.