Finite Field Question: Which of the followings are true? I have the following True or False question that I am having trouble getting it correct. I've written down my thoughts on each choice. If anyone could verify my thoughts or tell me where I made a mistake, I would really appreciate.

Question: Which of the followings are true?

*

*It is possible for an irreducible polynomial in $\mathbb{Q}[X]$ to be inseparable.

*Every irreducible polynomial over a finite field $K$ is separable.

*The polynomial $X^p - t\in\mathbb{F}_p (t)[X]$ over the finite field of rational functions $\mathbb{F}_p (t)$ is separable.

*The field $\mathbb{F}_p(t^{1/p})$ for the polynomial $X^p - t$ has degree $p$ over $\mathbb{F}_p(t)$ and is a splitting field for $X^p - t$.

*Let $K$ be the algebraic closure of $\mathbb{F}_p (t)$. Let $\mathbb{F}_p (t^{1/p})$ be a stem field of $X^p - t$. Then there are $p$ $\mathbb{F}_p (t)$-homomorphisms from $\mathbb{F}_p(t^{1/p})$ to $K$.


My thoughts:

*

*False. Let $p\in\mathbb{Q}[X]$ be irreducible. Since $char(\mathbb{Q})=0$, we must have $\deg(\gcd(p,p'))\le\deg(p')<\deg(p)$. This implies that $\gcd(p,p')=1$, which implies that $p$ is separable.


*My "guess" is False. We can take a polynomial $p$ such that $p'$ vanishes. This sounds plausible since we are in a finite field, and we can take polynomials in form of $X^{p^r}$. Then $p$ is inseparable. But I am not sure if we can get irreducible polynomials at the same time. I remember that $X^p - 1$ is irreducible but I forgot the proof.


*False. Let $f = X^p -t$, then $f' = 0$ since $char(\mathbb{F}_p(t))=p$. Then $\gcd(f,f')=f\ne 1$. So $f$ has to be inseparable.


*True. $t^{1/p}$ is a root of $f = X^p - t$ and $f$ is irreducible. So degree of extension is $p$. $\mathbb{F}_p(t^{1/p})$ is a splitting field because $X^p - t=(X-t^{1/p})^p$ due to Frobenius map.


*I have no idea but I would guess "True" if I have to guess :(
 A: Corrections/comments/whatnot:


*

*Your answer to part 2 is incorrect. The key is that when $K$ is a finite field of characteristic $p$, every one of its elements is also a $p$th power. That is to each $z\in K$ there exists a $y\in K$ such that $y^p=z$. If the derivative of $f(x)\in K[x]$ vanishes, then $f(x)=a_0+a_1x^p+a_2x^{2p}+\cdots+a_nx^{np}$ for some natural number $n$ and some elements $a_i\in K$. Selecting elements $b_i\in K$ such that $b_i^p=a_i$ we see, by the characteristic $p$ power law $(u+v)^p=u^p+v^p$, that
$$
f(x)=(b_0+b_1x+b_2x^2+\cdots+b_nx^n)^p.
$$
So if $f'=0$ then $f$ cannot be irreducible in $K[x]$.

*Your answers to parts 1,3, and 4 are correct.

*Your guess to part 5 is, however, incorrect. There is only a single such homomorphism. This is because the usual logic shows that such a homomorphism $\phi$ is fully determined, if we know $y=\phi(t^{1/p})\in K$. This element must satisfy the equation $y^p=t$ But the result of part 4 shows that there is only one such element $t\in K$.

