Compositions only of algebraic extensions are algebraic

Let $𝐹 : 𝐿 : 𝐾$ be a tower of field extensions. Then $𝐹 : 𝐾$ is algebraic iff both $𝐿 : 𝐾$ and $𝐹 : 𝐿$ are algebraic.¹ field

Proof

If $𝐹 : 𝐾$ is algebraic, then every element of $𝐹$ is algebraic over $𝐾$ ; therefore $𝐿 : 𝐾$ and $𝐹 : 𝐿$ are algebraic.

Conversely, suppose $𝐿 : 𝐾$ and $𝐹 : 𝐿$ are algebraic, and let $𝛼 \in 𝐹$ . Then there exists a polynomial
$𝑓 (𝑥) = 𝑛 \sum 𝑖 = 0 𝑎 𝑖 𝑥 𝑖 \in 𝐿 [𝑥]$
such that $𝑓 (𝛼) = 0$ , whence $𝛼$ is algebraic over the subfield
$𝐾 (𝑎 0, \dots, 𝑎 𝑛) \leq 𝐿$
so
$𝐾 (𝑎 0, \dots, 𝑎 𝑛, 𝛼) : 𝐾 (𝑎 0, \dots, 𝑎 𝑛)$
is finite. Now
$𝐾 (𝑎 0, \dots, 𝑎 𝑛) : 𝐾$
is finite by ^P1 since all the $𝑎 𝑖$ are algebraic over $𝐾$ by construction. Thus by the basic property of an Intermediate field extension,
$𝐾 (𝑎 0, \dots, 𝑎 𝑛) : 𝐾$
is finite. To summarize, we have the tower
where squiggly lines are algebraic and dashed lines are finite. Finally we see that
$𝐾 (𝑎 0, \dots, 𝑎 𝑛, 𝛼) : 𝐾$
must be finite and thus $𝛼$ is algebraic over $𝐾$ .

tidy | en | SemBr

2009. Algebra: Chapter 0, §VII.1.3, p. 395 ↩

Compositions only of algebraic extensions are algebraic

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Compositions only of algebraic extensions are algebraic

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