Notes - Galois Theory HT25, Separability

Created: October 28, 2024
Updated: January 21, 2025

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, or
  • 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, or
  • If $f$ 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 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:

Suppose that $\langle f, D(f) \rangle = \langle f \rangle$. 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). Hence, $\langle f, D(f) \rangle = F[t]$.

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.

Give an @example of a non-zero polynomial $f$ over a field of positive characteristic which is irreducible but not separable (in our definition of separability, every non-zero polynomial is separable if it’s over a field of characteristic zero).

Consider $F := \mathbb F _ p(t)$ be the field of fractions over the finite field of size $p$, and let $K$ be a field extension of $F$ containing a root $\alpha$ of $f(x) := x^p - t$.

Then $f$ is irreducible over $F$ (@todo, once I’ve finished sheet 1), but it is not separable since its formal derivative is zero.



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