Let $k$ be a field, $K$ and $L$ field extensions of $k$. Suppose also that $K$ and $L$ are embedded in some larger field $\Omega $.
Definition 9.27.1. Consider a diagram
of field extensions. The compositum of $K$ and $L$ in $\Omega $ written $KL$ is the smallest subfield of $\Omega $ containing both $L$ and $K$.
It is clear that $KL$ is generated by the set $K \cup L$ over $k$, generated by the set $K$ over $L$, and generated by the set $L$ over $K$.
Warning: The (isomorphism class of the) composition depends on the choice of the embeddings of $K$ and $L$ into $\Omega $. For example consider the number fields $K = \mathbf{Q}(2^{1/8}) \subset \mathbf{R}$ and $L = \mathbf{Q}(2^{1/12}) \subset \mathbf{R}$. The compositum inside $\mathbf{R}$ is the field $\mathbf{Q}(2^{1/24})$ of degree $24$ over $\mathbf{Q}$. However, if we embed $K = \mathbf{Q}[x]/(x^8 - 2)$ into $\mathbf{C}$ by mapping $x$ to $2^{1/8}e^{2\pi i/8}$, then the compositum $\mathbf{Q}(2^{1/12}, 2^{1/8}e^{2\pi i/8})$ contains $i = e^{2\pi i/4}$ and has degree $48$ over $\mathbf{Q}$ (we omit showing the degree is $48$, but the existence of $i$ certainly proves the two composita are not isomorphic).
Definition 9.27.2. Consider a diagram of fields as in (9.27.1.1). We say that $K$ and $L$ are linearly disjoint over $k$ in $\Omega $ if the map
is injective.
The following lemma does not seem to fit anywhere else.
Lemma 9.27.3. Let $E/F$ be a normal algebraic field extension. There exist subextensions $E / E_{sep} /F$ and $E / E_{insep} / F$ such that
$F \subset E_{sep}$ is Galois and $E_{sep} \subset E$ is purely inseparable,
$F \subset E_{insep}$ is purely inseparable and $E_{insep} \subset E$ is Galois,
$E = E_{sep} \otimes _ F E_{insep}$.
Proof. We found the subfield $E_{sep}$ in Lemma 9.14.6. We set $E_{insep} = E^{\text{Aut}(E/F)}$. Details omitted. $\square$
Comments (0)