Degrees of Field Extensions
Theorem 1. Let $E/F$ be a field extension, and let $a$ be algebraic over $F$. Then
$$
[F(a):F]=\deg f(x),
$$
where $f(x)$ is the minimal polynomial of $a$ over $F$.
Proposition 1. Let $E/F$ be a field extension, let $a$ be algebraic over $F$, and let $f(x)$ be a monic polynomial over $F$. Then the following are equivalent:
$f(x)$ is the minimal polynomial of $a$ over $F$.
$f(a)=0$ and $f(x)$ is irreducible over $F$.
$f(x)$ is the polynomial of smallest degree over $F$ having $a$ as a root.
If $g(x)$ is any polynomial over $F$ having $a$ as a root, then
$$
f(x)\mid g(x).
$$
Proof: The verification is straightforward and is left to the reader.
Proposition 2. Let $E/F$ be a field extension, and let $a$ be algebraic over $F$. Then the minimal polynomial of $a$ over $E$ divides the minimal polynomial of $a$ over $F$.
Proof: Let $f(x)$ be the minimal polynomial of $a$ over $E$, and let $g(x)$ be the minimal polynomial of $a$ over $F$.
Viewed over $E$, the polynomial $g(x)$ is also a polynomial in $E[x]$. By the division algorithm in $E[x]$, there exist $q(x),r(x)\in E[x]$ such that
$$
g(x)=f(x)q(x)+r(x),
$$
where either $r(x)=0$, or
$$
\deg r(x)<\deg f(x).
$$
If $r(x)\neq 0$, then
$$
r(a)=g(a)-f(a)q(a)=0-0=0,
$$
and at the same time
$$
\deg r(x)<\deg f(x).
$$
This contradicts the minimality of $f(x)$. Therefore necessarily
$$
f(x)\mid g(x).
$$
Theorem 2 (Dimension Formula). Let $E/F$ be a field extension, and let $K$ be an intermediate field of $E/F$, that is,
$$
F\subseteq K\subseteq E.
$$
Then
$$
[E:F]=[E:K][K:F].
$$
Proposition 3. Let
$$
F\subseteq E\subseteq K
$$
be fields, and let $a\in K$ be algebraic over $F$. Then
$$
[E(a):E]\le [F(a):F].
$$
Proof: Let $f(x)$ and $g(x)$ be the minimal polynomials of $a$ over $E$ and $F$, respectively. By Proposition 2, we have
$$
f(x)\mid g(x).
$$
Hence
$$
[E(a):E]=\deg f(x)\le \deg g(x)=[F(a):F].
$$
Proposition 4. Let
$$
F\subseteq E\subseteq K
$$
be fields, and let $a_1,a_2,\dots,a_n\in K$. Then
$$
[E(a_1,a_2,\dots,a_n):E]\le [F(a_1,a_2,\dots,a_n):F].
$$
Proof: We use induction on $n$.
When $n=1$, the conclusion follows from Proposition 3.
Assume that the conclusion holds for $n-1$. We now prove it for $n$. By the induction hypothesis and Proposition 3, we have
$$
\begin{aligned}
\left[E(a_1,a_2,\dots,a_n):E\right] &=[E(a_1,a_2,\dots,a_{n-1})(a_n):E(a_1,a_2,\dots,a_{n-1})]\cdot [E(a_1,a_2,\dots,a_{n-1}):E] \\\
&\le [F(a_1,a_2,\dots,a_{n-1})(a_n):F(a_1,a_2,\dots,a_{n-1})] \cdot [F(a_1,a_2,\dots,a_{n-1}):F] \\\
&=[F(a_1,a_2,\dots,a_n):F].
\end{aligned}
$$
Thus the conclusion also holds for $n$, and the induction is complete.
Proposition 5. Let $E/F$ be a finite extension, and let $a$ be algebraic over $F$. Then
$$
[E(a):F(a)]\le [E:F].
$$
Proof: By Proposition 3,
$$
[E(a):F(a)][E(a):E]\le [E(a):F(a)][F(a):F]=[E(a):F].
$$
On the other hand, by the dimension formula,
$$
[E(a):F]=[E(a):E][E:F].
$$
Therefore,
$$
[E(a):F(a)][E(a):E]\le [E(a):E][E:F].
$$
Cancelling $[E(a):E]$, we obtain
$$
[E(a):F(a)]\le [E:F].
$$
Theorem 3. Let $F$ be a field, and let $a_1,a_2,\dots,a_n$ be algebraic over $F$. Then
$$
[F(a_1,a_2,\dots,a_n):F]\le [F(a_1):F][F(a_2):F]\cdots [F(a_n):F].
$$
Proof: We use induction on $n$.
When $n=1$, the statement is obvious.
Assume that the conclusion holds for $n-1$. We now prove it for $n$. By Proposition 5 and the induction hypothesis,
$$
\begin{aligned}
\left[ F(a_1,a_2,\dots,a_n):F \right]
&=[F(a_1,a_2,\dots,a_{n-1})(a_n):F(a_n)][F(a_n):F] \\\
&\le [F(a_1,a_2,\dots,a_{n-1}):F][F(a_n):F] \\\
&\le [F(a_1):F][F(a_2):F]\cdots [F(a_n):F].
\end{aligned}
$$
Thus the conclusion holds for $n$, and the induction is complete.
Example. Compute the extension degree
$$
[\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}].
$$
Solution: Clearly, the minimal polynomials of $\sqrt[3]{2}$ and $\sqrt[4]{5}$ are $x^3-2$ and $x^4-5$, respectively. By Eisenstein’s criterion, both are irreducible. Hence
$$
[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3, \\\
[\mathbb{Q}(\sqrt[4]{5}):\mathbb{Q}]=4.
$$
Moreover,
$$
[\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}]
=
[\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}(\sqrt[3]{2})]
[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}],
$$
and
$$
[\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}]
=
[\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}(\sqrt[4]{5})]
[\mathbb{Q}(\sqrt[4]{5}):\mathbb{Q}].
$$
Therefore,
$$
3\mid [\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}]
$$
and
$$
4\mid [\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}].
$$
Since $(3,4)=1$, it follows that
$$
12\mid [\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}],
$$
and hence
$$
[\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}]\ge 12.
$$
On the other hand, by Theorem 3,
$$
[\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}]
\le
[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]
[\mathbb{Q}(\sqrt[4]{5}):\mathbb{Q}]
=12.
$$
Therefore,
$$
[\mathbb{Q}(\sqrt[3]{2},\sqrt[4]{5}):\mathbb{Q}]=12.
$$
Proposition 6. Let $E$ be an extension field of $F$, and let $a,b\in E$ be algebraic over $F$. Suppose that the degrees of their minimal polynomials over $F$ are $m$ and $n$, respectively, and that
$$
(m,n)=1.
$$
Then
$$
[F(a,b):F]=mn.
$$
Proof: On the one hand,
$$
[F(a,b):F]=[F(a,b):F(a)][F(a):F],
$$
and
$$
[F(a,b):F]=[F(a,b):F(b)][F(b):F].
$$
Thus
$$
m\mid [F(a,b):F], \qquad n\mid [F(a,b):F].
$$
Since $(m,n)=1$, it follows that
$$
mn\mid [F(a,b):F].
$$
Hence
$$
[F(a,b):F]\ge mn.
$$
On the other hand, by Theorem 3,
$$
[F(a,b):F]\le mn.
$$
Therefore,
$$
[F(a,b):F]=mn.
$$
The cover image of this article was taken in Irkutsk Oblast, Russia.
Degrees of Field Extensions