Separability of polynomials From my textbook (Robert Ash's Basic Abstract Algebra, section 3.4):

3.4.2 Proposition
If $$f(X)=a_0 + a_1X + \dots + a_nX^n \in F[X],$$ let $f'$ be the derivative of $f$, defined by $$f'(X) = a_1 + 2a_2X + \dots + na_nX^{n-1}.$$
If $g$ is the greatest common divisor of $f$ and $f'$, then $f$ has a repeated root in a splitting field if and only if the degree of $g$ is at least $1$.

There is an example in the textbook that says that $f(X)=(X-1)^2(X-3)$ over $\Bbb{Q}$ is separable, because "the irreducible factors $(X-1)$ and $(X-3)$ do not have repeated roots." However, if we try to apply the proposition, we get a gcd of degree $1$, right? So I was just wondering if the proposition had to assume that $f$ is irreducible or if I was missing something.

3.4.3 Corollary
(1) Over a field of characteristic zero, every polynomial is separable.
(2) Over a field $F$ of prime characteristic $p$, the irreducible polynomial $f$ is inseparable if and only if $f'$ is the zero polynomial. Equivalently, $f$ is a polynomial in $X^p$; we abbreviate this as $f \in F[X^p]$.

Later on, the textbook states "By (3.4.3), part (1), every polynomial over the rationals (or the reals or the complex numbers) is separable. This pleasant property is shared by finite fields as well." However, doesn't 3.4.3 part (2) tell us that $f$ is inseparable if it is a polynomial in $X^p$? How is every polynomial in a finite field separable then?

By (3.4.3) and (3.4.5), every algebraic extension of a field of characteristic zero or a finite field is separable.

This is true because an algebraic extension of a finite field must also be finite, right?
Thank you in advance
 A: 1st question: An irreducible polynomial is separable if it does not have repeated roots in its splitting field. A polynomial is separable if its irreducible factors are. So, there is no contradiction, and Prop. 3.4.2 is ok as stated: the fact is that in your example there is a repeated root even though the polynomial is separable, but this is no contradiction since the polynomial is not squarefree.
2nd question: over a finite field, no polynomial in $x^p$ is irreducible (this is related to the fact that finite fields are perfect, so you can take $p$th roots of all coefficients) so there is, again, no contradiction. The quintessential example of a non-separable polynomial is 
$$f(x)=x^p-t$$ where $t$ has no $p$th root in $F$.
A: In most references, a separable polynomial is one that has no repeated roots in an algebraic closure. This is not so in your case, and maybe this is a cause of some confusion.
For your first question, there is no contradiction because "separable" in your book's sense does not mean that no repeated roots exist.
For your second question: no irreducible polynomial over a finite field $F_q$, $q = p^e$ can be a polynomial in $X^p$, because:
Since $F_q$ is finite, the Frobenius map (injective, because it is a homomorphism of fields) is also surjective, so for any $a_k \in F_q$ you find $b_k \in F_q$ with $b_k^p = a_k$. So $$\sum_{k=0}^r a_k (X^p)^k = \Big(\sum_{k=0}^r b_k X^k\Big)^p$$ is not irreducible.
For your third question: no, algebraic extensions of a finite field need not be finite. However, each element of such an algebraic extension lies in a finite field extension and you can use that.
