An element $\alpha\in K$ is algebraic over $F$ if it is the root of some nonzero polynomial in $F[x]$. If $\alpha$ is not algebraic over $F$, then $\alpha$ is said to be transcendental over $F$. The field extension $K/F$ is said to be algebraic if every element of $K$ is algebraic over $F$.
Example 1. The element $\sqrt{2}$ is algebraic over the field $\mathbb{Q}$, and the field extension $\mathbb{Q}(\sqrt{2})\cong \mathbb{Q}[x]/(x^2-2)$ is algebraic. Indeed, every element $\alpha\in \mathbb{Q}(\sqrt{2})$ can be written as $\alpha=a+b\sqrt{2}$ for some $a, b \in \mathbb{Q}$, and it is a root of the polynomial $p(x)=x^2-2ax+(a^2-b^2)$.
One natural question arises here is: if $\alpha$ is algebraic over $F$, then is $F(\alpha)/F$ an algebraic extension? It turns out the answer is yes. In fact, we have the following equivalence:
\[\alpha \text{ is algebraic over } F \iff F(\alpha)/F \text{ is algebraic} \iff [F(\alpha):F]<\infty\]This equivalence suggests that the field extension we constructed in the previous section are all algebraic. Before proving the equivalence, we first prove a simple lemma.
Lemma (Minimal Polynomial). Let $\alpha$ be algebraic over $F$. Then there exists a unique monic irreducible polynomial $m_{\alpha}(x)$ with $\alpha$ as a root. We call $m_{\alpha}(x)$ the minimal polynomial of $\alpha$ over $F$. Moreover, a polynomial $f(x)$ has $\alpha$ as a root if and only if $m_{\alpha}(x)$ divides $f(x)$.
Proof
Let $g(x)$ be the polynomial of least degree with $\alpha$ as a root, and by multiplying by a constant we can assume $g(x)$ is monic. If $g(x)$ is not irreducible, then $g(x)=a(x)b(x)$, so $g(\alpha)=a(\alpha)b(\alpha)=0$, which means that either $a(\alpha)=0$ or $b(\alpha)=0$. This contradicts the minimality of the degree of $g(x)$. Therefore, $g(x)$ must be irreducible and hence is the minimal polynomial of $\alpha$.
Now let $f(x)$ be any polynomial with $\alpha$ as a root. We can write
\[f(x)=g(x)q(x)+r(x) \implies f(\alpha)=r(\alpha)=0\]By minimality of $g(x)$, $r(x)$ must be the zero polynomial. Therefore, $f(x)$ is divisible by $g(x)$.
Theorem. The following are equivalent
\[\alpha \text{ is algebraic over } F \iff F(\alpha)/F \text{ is algebraic} \iff [F(\alpha):F]<\infty\]Proof
We will prove the equivalence by showing
\[\alpha \text{ algebraic over } F\implies [F(\alpha):F]<\infty \implies F(\alpha)/F \text{ algebraic}\implies \alpha \text{ algebraic over } F\](1):
Suppose $\alpha$ is algebraic, then from the above lemma, its minimal polynomial $m_\alpha(x)$ exists. Since $F(\alpha)\cong F[x]/(m_\alpha(x))$, we have
(2):
Suppose $[F(\alpha): F]=n<\infty$, and let $\beta\in F(\alpha)$. Since $F(\alpha)$ is an $n$-dimensional vector space over $F$, the $n+1$ elements
are linearly dependent over $F$. Hence there exist $c_0,c_1,\dots,c_n\in F$, not all zero, such that
\[c_0+c_1\beta+\cdots+c_n\beta^n=0.\]Which implies that $\beta$ is a root of the nonzero polynomial $p(x)=c_0+c_1x+…c_nx^n$.
(3):
Suppose $F(\alpha)/F$ is algebraic. Since $\alpha\in F(\alpha)$, it follows that $\alpha$ is algebraic over $F$.
This concludes the proof.
A useful consequence of the theorem is that finite extensions $K/F$ are automatically algebraic. Indeed, let $\alpha\in K$, then we have an intermediate field
\[F\subseteq F(\alpha)\subseteq K \implies [F(\alpha):F]\le [K:F]<\infty\]Which, by the above theorem, $\alpha$ is algebraic over $F$. And since this holds for any $\alpha\in K$, $K$ is an algebraic extension. However, the reverse implication need not hold, we show a counterexample below.
Example 2. Let $\overline{\mathbb{Q}}$ be the set of all elements of $\mathbb{C}$ that are algebraic over $\mathbb{Q}$. Then $\overline{\mathbb{Q}}/\mathbb{Q}$ is algebraic by definition. This extension is not finite. For each $n\ge 1$, the element $\sqrt[n]{2}$ is algebraic over $\mathbb{Q}$, so
\[[\overline{\mathbb{Q}}:\mathbb{Q}]\ge [\mathbb{Q}(\sqrt[n]{2}):\mathbb{Q}]=n\]Since this holds for every $n$, the degree $[\overline{\mathbb{Q}}:\mathbb{Q}]$ cannot be finite.
The counterexample shows that algebraic extensions need not be finite. The issue is that $\overline{\mathbb{Q}}$ is obtained by adjoining infinitely many algebraic elements. If we rule this out by requiring the extension to be generated by finitely many elements, then the converse becomes true.
We say that $K/F$ is finitely generated if there exist elements $\alpha_1,\dots,\alpha_n\in K$ such that
\[K=F(\alpha_1,\dots,\alpha_n).\]Theorem. Let $K/F$ be a finitely generated field extension. Then
\[K/F \text{ is algebraic} \iff [K:F]<\infty.\]Proof
We already showed that every finite extension is algebraic. So it remains to prove the converse.
Suppose $K/F$ is algebraic and finitely generated. Then
\[K=F(\alpha_1,\dots,\alpha_n)\]for some $\alpha_1,\dots,\alpha_n\in K$. Define a tower of fields
\[F=F_0\subseteq F_1\subseteq \cdots \subseteq F_n=K,\]where
\[F_i=F(\alpha_1,\dots,\alpha_i).\]Since $K/F$ is algebraic, each $\alpha_i$ is algebraic over $F$. Therefore $\alpha_i$ is also algebraic over $F_{i-1}$, because $F[x]\subseteq F_{i-1}[x]$.
Thus each simple extension
\[F_i=F_{i-1}(\alpha_i)\]is finite over $F_{i-1}$. Hence
\[[F_i:F_{i-1}]<\infty\]for every $i$. By the tower law,
\[[K:F]=[F_n:F_0]=\prod_{i=1}^n [F_i:F_{i-1}]<\infty.\]Therefore $K/F$ is finite.