Field Extensions

Definition.
Let $E/F$ be a field extension. Then $E$ is naturally a vector space over $F$. We define the degree of $E$ over $F$ by:

where $\dim_F E$ denotes the dimension of $E$ as an $F$-vector space. We say that $E/F$ is a finite extension if $ [E : F] < \infty $.

Definition / Lemma.
Let $E/F$ be a field extension, and let $\alpha \in E$ be algebraic over $F$ with minimal polynomial $m_{\alpha, F}(x)$ of degree $d$. We define the degree of $\alpha$ over $F$ as $d$, denoted by $\deg (\alpha,F) = d.$ Moreover:

1. \[F(\alpha) = \left\{ c_0 + c_1 \alpha + \cdots + c_{d-1} \alpha^{d-1} : c_i \in F \right\}\]
2. \[[F(\alpha) : F] = d\]

Proof (Sketch).
Since $\alpha$ is algebraic over $F$, we have $F(\alpha) = F[\alpha] = { p(\alpha) : p(x) \in F[x] }$. In particular:

\[F(\alpha) \supset \left\{ c_0 + c_1 \alpha + \cdots + c_{d-1} \alpha^{d-1} : c_i \in F \right\}\]

For the reverse inclusion, we use the fact that $\alpha^d$ can be written as a linear combination of lower powers using the minimal polynomial. Inductively, every $\alpha^k$ for $k \geq d$ lies in the $F$-span of ${1, \alpha, \ldots, \alpha^{d-1}}$.

To prove the second part, we observe that ${1, \alpha, \ldots, \alpha^{d-1}}$ is linearly independent and spans $F(\alpha)$, so it forms an $F$-basis of $F(\alpha)$.

Lemma (Tower Law).
Let $F \subset E \subset L$ be fields such that both $E/F$ and $L/E$ are finite extensions. Then $L/F$ is also a finite extension, and:

Proof (Sketch).
Let ${e_1, \ldots, e_n}$ be an $F$-basis of $E$, and let ${l_1, \ldots, l_m}$ be an $E$-basis of $L$. Then the set:

\[\{ e_i l_j : 1 \leq i \leq n,\ 1 \leq j \leq m \}\]

spans $L$ over $F$ and is linearly independent. Hence it forms an $F$-basis of $L$.

Definition.
A field extension $E/F$ is called algebraic if every element $\alpha \in E$ is algebraic over $F$.

Lemma (Finite Extensions are Algebraic).
If $E/F$ is a finite extension, then it is algebraic.

Proof (Sketch).
Let $\alpha \in E$, and consider the set ${1, \alpha, \alpha^2, \ldots}$. Since $E$ has finite dimension over $F$, this set must be linearly dependent. Then, a nontrivial linear relation of those $\alpha^i$ gives a polynomial in $F[x]$ with $\alpha$ as a root.

Theorem (Transitivity of Algebraic Extensions).
Let $F \subset E \subset L$ be field extensions.

1. Then $L/F$ is algebraic if and only if both $E/F$ and $L/E$ are algebraic.

Proof (Sketch).
Suppose both $E/F$ and $L/E$ are algebraic. Pick $\alpha \in L$. Since $L/E$ is algebraic, $\alpha$ satisfies a polynomial:

\[p(x) = \sum_{i=0}^n e_i x^i,\quad e_i \in E,\quad p(\alpha) = 0\]

Since $E/F$ is algebraic, each $e_i$ is algebraic over $F$. Then the field $F(e_0, \ldots, e_n)$ is a finite extension of $F$, and $\alpha$ is algebraic over it.

Therefore:

\[[F(e_0, \ldots, e_n, \alpha) : F] < \infty\]

and $\alpha$ is algebraic over $F$.

Conversely, if $L/F$ is algebraic, then in particular every element of $E \subset L$ is algebraic over $F$, so $E/F$ is algebraic. Also, for $\alpha \in L$, if it satisfies a polynomial in $F[x] \subset E[x]$, then $\alpha$ is algebraic over $E$ as well. So $L/E$ is algebraic.