Separable Degree of Field Extensions is Multiplicative

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Theorem

Let $E / F / k$ be a tower of fields. Then

$\index E k_s = \index E F_s \index F k_s$

where $\index E F_s$ denotes the separable degree of $E / F$.

Proof

If $E / k$ is infinite, both sides are infinite.

Now assume $E / k$ is finite.

Let $L$ be the algebraic closure of $k$.

Let $\set{\sigma_i}$ be the family of distinct embedding of $F$ to $L$ fixing $k$,

By definition 2 of separable degree, the set $\set{\sigma_i}$ has $\index F k_s$ elements.

For each $i$, let $\set{\tau_{i j}}$ be the family of distinct extensions of $\sigma_i$ to $E$.

We can view $L$ as an algebraic closure of $F$. By definition 2 of separable degree, each $\sigma_i$ has precisely

$\index E F_s$

extensions to embeddings of $E$ in $L$.

The set of embeddings $\set{\tau_{i j}}$ contains precisely

$\index E F_s \index F k_s$

elements.

Any embedding of $E$ into $L$ fixing $k$ must be one of the $\tau_{i j}$, and thus we have

$\index E k_s = \index E F_s \index F k_s$

$\blacksquare$

Sources

• 2002: Serge Lang: Algebra (Revised 3rd ed.): Chapter $\text V$: $\S4$: Separable Extensions: Theorem $4.1$