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 $K$.
9.27.1.1
\begin{equation} \label{fields-equation-inside-omega} \vcenter { \xymatrix{ L \ar[r] & \Omega \\ k \ar[r] \ar[u] & K \ar[u] } } \end{equation}
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.
\[ K \otimes _ k L \longrightarrow KL,\quad \sum x_ i \otimes y_ i \longmapsto \sum x_ i y_ i \]
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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)