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To show it is irreducible I just show that $\sqrt[p]\alpha \notin F$, since it is a zero of the polynomial in question and, if it is not, then it cannot be reducible. (Correct me, if I am wrong here. I am not entirely sure if that's true.) $\sqrt[p]\alpha \notin F$ obviously holds, because otherwise $\alpha \in F$ and a minimal polynomial $x-\alpha$ is enough to show that $\alpha$ isn't transcendental. Which is a contradiction and therefore $f$ is irreducible.

To show that is inseparable we look at $f' = (x^p)' - (\alpha)' = px^{p-1}-0 = px^{p-1}$

Now since $char(F)=p$ it holds that $py=0, \forall y \in F$ (This is the second part I am not sure about). And so $f' = 0$, which proves by definition that it is inseparable.

Could you please tell me if this is a viable solution and if not then I could use hint.

Simon Tura
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  • For your first "not entirely sure" see this post. – Dietrich Burde Dec 15 '22 at 15:35
  • @DietrichBurde They talk about no zeroes in $F$. Does it also hold for when just one zero is not in $F$? – Simon Tura Dec 15 '22 at 15:41
  • For $g(x) \in K[x]$, if $\deg g = 2$ or $3$ then the lack of a root of $g(x)$ in $K$ implies it is irreducible, but that is generally not true when $\deg g \geq 4$. Consider $(x^2-2)(x^2-3)$ in $\mathbf Q[x]$: it has no rational root but is reducible! So the fact that for $x^p - a$ in characteristic $p$, a lack of roots implies irreducibility is an unusual example because ordinarily the lack of roots above degree $3$ doesn't imply irreducibility. So proving that requires a careful argument. – KCd Dec 15 '22 at 16:22
  • @KCd Ok, I am getting very confused now. Here it says that it is irreducible if no zeros are in the field. Isn't it irreducible if it doesnt have linear factors? – Simon Tura Dec 15 '22 at 16:44
  • For the specific case of $x^p-a$ in characteristic $p$, a lack of roots implies irreducibility. But that is not because there is general result in math that a polynomial having no roots is irreducible!! In degree 2 or 3 over all fields, it is true that a lack of roots implies irreducibility, but for degree $4$ and higher it is typically false that a lack of roots implies irreducibility. So the fact that for $x^p-a$ in characteristic $p$ a lack of roots implies irreducibility is quite unusual and needs a careful argument. I feel like I am repeating what I already wrote before. – KCd Dec 15 '22 at 16:55

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