Notes - Galois Theory HT25, Separability
This is not the same definition of separable used elsewhere. In the definition given in this course, a polynomial like $f(x) = (x-1)^2$ is still separable since each of it’s irreducible factors is still separable.
Flashcards
Suppose:
- $f \in F[t]$
@Define what it means for $f$ to be separable.
There are two cases:
- If $f$ is irreducible, it is separable if $D(f) \ne 0$ where $D(f)$ is the formal derivative
- If $f$ is not irreducible, it is separable if each of its irreducible factors in $F[t]$ is separable.
A polynomial $f \in F[t]$ is said to be separable if:
- $f$ is irreducible, and $D(f) \ne 0$ where $D(f)$ is the formal derivative
- $f$ is not irreducible, and each of its irreducible factors in $F[t]$ is separable.
What does this particular definition of separability imply about every (non-constant? @todo) polynomial over a field of characteristic zero?
It is separable.
Suppose:
- $f \in F[t]$ is an irreducible, separable polynomial
@State and quickly @prove a useful lemma about the relationship between $f$ and its derivative, $D(f)$.
There exist $p, q \in F[t]$ such that $p f + q D(f) = 1$.
Proof: Since $f$ is separable, $D(f)$ is a non-zero polynomial of stricly smaller degree than $f$.
Since $f$ is irreducible, $\langle f \rangle$ is maximal. There are two cases:
Case 1: $\langle f, D(f) \rangle = \langle f \rangle$. Then $D(f) \in \langle f \rangle$, so there exists some $g \in F[t]$ such that $fg = D(f)$. But since $D(f)$ is nonzero, $g$ must also be nonzero. But then $\deg D(f) \ge \deg f$, which is impossible (by the definition of derivative). So this case can’t actually happen.
Case 2: $\langle f, D(f) \rangle$. In this case, there exists some $p, q$ such that $pf + qD(f) = 1$ as claimed.
Suppose:
- $f \in F[t]$ is an irreducible, separable polynomial
- $K$ is any field extenion of $F$ over which $f$ splits completely
@State a result that links these conditions to the number of roots of $f$.
$f$ has exactly $\deg f$ distinct roots in $K$.
Suppose:
- $f \in F[t]$ is an irreducible, separable polynomial
- $K$ is any field extenion of $F$ over which $f$ splits completely
@Prove that then $f$ has exactly $\deg f$ distinct roots in $K$, appleaing to a useful lemma that you do not need to prove.
The useful lemma in question is that there exists $p, q \in F[t]$ such that $pf + qD(f) = 1$.
Suppose that $\alpha$ is a repeated root of $f$. Then $f = (x - \alpha)^2 g$ for some $g \in K[t]$. Hence
\[D(f) = 2(x - \alpha) g + (t - \alpha)^2 D(g)\]So $D(f)(\alpha) = 0$ also. But then letting $x = \alpha$ in $pf + qD(f) = 1$ gives $0 = 1$, a contradiction, so no such repeated root can exist.