Galois Theory HT25, Separability

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.

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This is also equivalent to the condition that $\text{gcd}(f, f’) = 1$.

@important~

$m _ {F, \alpha}$ is a separable polynomial.

@important~

It is separable.

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.

$f$ has exactly $\deg f$ distinct roots in $K$, i.e. $f$ has no repeated roots in any splitting field.

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.

\[f(x) = g(x^p)\]

for some $g \in K[x]$.

@exam~

Suppose $\alpha$ is separable over $F$. Let $m := m _ {F, \alpha}$. Then $D(m) \ne 0$ as $\alpha$ is separable.

Then $\alpha$ is a root of $t^p - a^p \in F(\alpha^p)[t]$. Hence $m \mid t^p - a^p$. Viewing this divisibility in the larger ring,

\[m \mid (t - \alpha)^p\]

As $m$ has no repeated roots in $F(\alpha)$ (by separability), this implies $t - \alpha \in F(\alpha^p)[t]$ and $\alpha \in F(\alpha^p)$. Hence $F(\alpha) = F(\alpha^p)$.

Conversely, suppose $\alpha$ is not separable over $F$. Since $m$ is irreducible, $m \in F[t^p]$ and hence there exists $g \in F[t]$ such that $m = g(t^p)$. Then $\deg m = p \deg g$, so $\deg g < \deg m$.

Since $\alpha^p$ is a root of $g$, it follows

\[\begin{aligned} \text{}[F(\alpha^p) : F] &\le \deg g \\\\ &< \deg m \\\\ &= [F(\alpha) : F] \end{aligned}\]

Hence $F(\alpha^p) \ne F(\alpha)$, as required.

@exam~

Consider the extensions:

• $F \subseteq F(\alpha + \beta) \subseteq F(\alpha + \beta, \beta)$
• $F \subseteq F(\alpha^p + \beta^p) \subseteq F(\alpha^p + \beta^p, \beta^p)$

Since $\alpha, \beta$ are separable, by the result we have

\[F(\alpha^p + \beta^p, \beta^p) = F(\alpha^p, \beta^p) = F(\alpha, \beta) = F(\alpha + \beta, \beta)\]

Hence (?) by the Tower Law,

\[[F(\alpha + \beta) : F] \cdot [F(\alpha + \beta, \beta) : F(\alpha + \beta)] = [F(\alpha^p + \beta^p) : F] \cdot [F(\alpha^p + \beta^p, \beta^p) : F(\alpha^p + \beta^p)]\]

But then $F(\alpha^p + \beta^p) = F((\alpha + \beta)^p) \subseteq F(\alpha + \beta)$. Hence

\[[F(\alpha^p + \beta^p) : F] \le [F(\alpha + \beta) : F]\]

If $g$ is the minimal polynomial of $\beta$ over $F(\alpha + \beta)$, then $0 = g(\beta)^p = h(\beta^p)$ for some monic

\[h \in F((\alpha + \beta)^p)[t] = F(\alpha^p + \beta^p)[t]\]

with $\deg h = \deg g$, as $h$ is obtained by raising each coefficient to the power of $p$. Hence

\[[F(\alpha^p + \beta^p, \beta^p) : F(\alpha^p + \beta^p)] \le \deg h = \deg h = [F(\alpha + \beta, \beta) : F(\alpha + \beta)]\]

Then we have equalities in the above, so

\[F(\alpha + \beta) = F((\alpha + \beta)^p)\]

and so $\alpha + \beta$ is separable over $F$ by the result.

@exam~

We first show that if $K$ is not perfect, then $K^p \ne K$ (where $K^p = \lbrace k^p \mid k \in K \rbrace$, and this is actually an equivalence).

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Suppose $K^p = K$ and $F/K$ is a finite extension. Pick $\gamma \in F$. We have to show that the minimal polynomial $m _ {K, \gamma}$ is separable.

If $m _ {K, \gamma}’(x) \ne 0$, then $(m _ {K, \gamma}(x), m _ {K, \gamma}’(x)) = 1$, since $m _ {K, \gamma}$ is irreducible and $\deg m _ {K, \gamma}’ < \deg m _ {K, \gamma}$.

Thus if $m _ {K, \gamma}’ \ne 0$, then $m _ {K, \gamma}$ has no multiple roots and thus $m _ {K, \gamma}$ is separable (recall it suffices to just check the minimal polynomials of the generators).

Now consider the case where $m _ {K, \gamma}’(x) = 0$. Then $m _ {K, \gamma}(x) = f(x^p)$ for some $f \in K[x]$ (by results about fields of characteristic $p$). Write

\[f(x) = \sum b_d x^d\]

and define

\[f^{1/p}(x) := \sum b_d^{1/p} x^d\]

where $b _ d^{1/p}$ is the $p$-th root of $b _ d$ in $K$. Then

\[m_{K, \gamma}(x) = f(x^p) = (f^{1/p}(x))^p\]

so that $m _ {K, \gamma}$ is not irreducible, a contradiction.

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Hence we now know that $K^p \ne K$. Let $\phi _ L$ be the Frobenius endomorphism of $L$. Then $\varphi _ L(K) \subsetneq K$, but $\varphi _ L (L) = L$. Hence $\varphi _ L : L \to L$ is a field automorphism.

Applying $\varphi _ L^{-1}$ repeatedly to the proper inclusion $\varphi _ L(K) \subsetneq K$, we obtain the infinite sequence of proper inclusions

\[\varphi_L(K) \subsetneq K \subsetneq \varphi_L^{-1}(K) \subsetneq \varphi_L^{-2}(K) \subsetneq \cdots\]

which contradicts the fact $L/K$ is finite, since it implies that the degree is infinite.