Lecture. Algebraic extensions [l3]

Apr 6, 2026

Combining with Theorem 1207 yields the following proposition:

Proposition 1. finiteness implies algebraicity [1404]

$K / F$ is finite $\implies$ $K / F$ is algebraic.

Proposition 2. minimal polynomial for $\alpha$ over $F$ [1405]

Let $K / F$ be a field extension and $\alpha \in K$ be an element algebraic over $F$. There exists a unique monic irreducible polynomial $m_{\alpha, F}(x) \in F[x]$ such that $m_{\alpha, F}(\alpha) = 0$.

We call the polynomial $m_{\alpha, F}(x)$ the minimal polynomial for $\alpha$ over $F$, and call $\deg m_{\alpha, F}(x)$ the degree of $\alpha$ over $F$.

Proof. sketch [1405A]

Existence: let $p(x)$ be the very polynomial in the proof of Proposition 1403. Let $a$ be the leading coefficient of $p(x)$. Then $m_{\alpha, F}(x) = a^{-1} p(x)$. It is obviously monic, and the irreducibility is inherited from $p(x)$.

Uniqueness: use the same notation as in the proof of Proposition 1403. Then $\ker \varphi = \left( p(x) \right) = \left( m_{\alpha, F}(x) \right)$. Suppose there is another monic irreducible polynomial $m_{\alpha, F}'(x) \in F[x]$ such that $m_{\alpha, F}'(\alpha) = 0$. Then $m_{\alpha, F}'(x) \in \left( m_{\alpha, F}(x) \right)$, hence $m_{\alpha, F}'(x) = m_{\alpha, F}(x) q(x)$ for some $q(x) \in F[x]$. Since $m_{\alpha, F}'(x), m_{\alpha, F}(x)$ are monic, $q(x)$ should also be monic. If $\deg q > 0$, then it contradicts irreducibility of $m_{\alpha, F}'(x)$. So $q(x) = 1$, which implies $m_{\alpha, F}' = m_{\alpha, F}$.

Proposition 3. $f(\alpha)=0$ if and only if $m_{\alpha, F} \mid f$ [1406]

Let $K / F$ be a field extension and $\alpha \in K$. Suppose that $f(x) \in F[x]$. Then $f(\alpha) = 0$ if and only if $m_{\alpha, F}(x) \mid f(x)$ in $F[x]$.

In this case, $f(x)$ is monic and irreducible if and only if $f(x) = m_{\alpha, F}(x)$.

Proof. sketch [1406A]

$\implies$: Just use the same argument in the proof of Proposition 1403. We have $f(x) \in \ker \varphi = \left( m_{\alpha, F}(x) \right)$. So $m_{\alpha, F}(x) \mid f(x)$ in $F[x]$.

$\impliedby$: Similarly, $m_{\alpha, F}(x) \mid f(x) \implies f(x) \in \left( m_{\alpha, F}(x) \right) = \ker\varphi$ and therefore $f(\alpha) = 0$.

Proposition 4. field isomorphisms regarding algebraicity [1407]

Let $K / F$ be a field extension and $\alpha \in K$.

If $\alpha$ is algebraic over $K$, then the map$$F[x] / \left( m_{\alpha, F} \right) \to F(\alpha) (\subseteq K), \qquad \overline{a(x)} \mapsto a(\alpha)$$is a well-defined field isomorphism, and $[F(\alpha) : F] = \deg m_{\alpha, F}(x)$.
If $\alpha$ is transcendental over $F$, then the map$$F(x) \to F(\alpha) (\subseteq K), \qquad \frac{a(x)}{b(x)} \mapsto a(\alpha)b(\alpha)^{-1}$$is a well-defined field isomorphism, and $[F(\alpha) : F] = \infty$.

Proof. sketch [1407A]

By the proof of Proposition 1405, the kernel of the following ring homomorphism $$\varphi : F[x] \to K, \qquad a(x) \mapsto a(\alpha)$$ is $\left( m_{\alpha, F}(x) \right)$. By the First Isomorphism Theorem, $$F[x] / \left( m_{\alpha, F}(x) \right) \to \mathrm{Im}\varphi, \qquad \overline{a(x)} \mapsto a(\alpha)$$ is a well-defined isomorphism. Moreover, in the proof of Proposition 1403, we have shown $\mathrm{Im}\varphi = F(\alpha)$. And by Theorem 1206, $[F(\alpha) : F] = \deg m_{\alpha, F}(x)$.
Suppose $\frac{a(x)}{b(x)} = \frac{c(x)}{d(x)}$, then $$a(x)d(x) = b(x)c(x) \implies a(\alpha)d(\alpha) = b(\alpha)c(\alpha) \implies a(\alpha)b(\alpha)^{-1} = c(\alpha)d(\alpha)^{-1}.$$ So the map is well-defined. Also, it is clearly a field homomorphism. By definition of $F(\alpha)$, it is surjective. Denote the map by $\phi$. If $\phi$ is not injective, then there exists $\frac{a(x)}{b(x)}\in F(x)$ such that $a(\alpha) b(\alpha)^{-1} = 0 \implies a(\alpha) = 0$. Then $\alpha$ is algebraic, contradicting $\alpha$ is transcendental. So $\phi$ is an isomorphism.

Now, if $[F(\alpha) : F] < \infty$, then by Proposition 1403, $\alpha$ is algebraic over $F$, a contradiction. So $[F(\alpha) : F] = \infty$.

Theorem 5. isomorphism induced in the middle [1408]

Let $K_{1} / F_{1}$ and $K_{2} / F_{2}$ be field extensions, $\varphi : F_{1} \to F_{2}$ be a field isomorphism and let $p_{1}(x) = a_{n}x^{n} + \cdots + a_{1}x + a_{0} \in F_{1}[x]$ be an irreducible polynomial, and let $p_{2}(x) = \varphi(a_{n})x^{n} + \cdots + \varphi(a_{1})x + \varphi(a_{0}) \in F_{2}[x]$. If $\alpha \in K_{1}$ is a root of $p_{1}(x)$ and $\beta \in K_{2}$ is a root of $p_{2}(x)$, then $$ \Psi: F_{1}(\alpha) \to F_{2}(\beta), \quad c_{m}\alpha^{m} + \cdots + c_{1}\alpha + c_{0} \mapsto \varphi(c_{m})\beta^{m} + \cdots + \varphi(c_{1})\beta + \varphi(c_{0}) $$ for $c_{i} \in F_{1}$, is well-defined and is a field isomorphism.

Exegesis 5.1. diagram illustrations and a special case [1408A]

$F_{1} = F_{2} = F, K_{1} = K_{2} = K$ implies that if $\alpha, \beta \in K$ are both roots of an irreducible polynomial $p(x) \in F[x]$, then $$ F(\alpha) \cong F(\beta), \quad a(\alpha) \mapsto a(\beta), \quad a(x) \in F[x] $$ is well-defined and is an isomorphism.

Example 5.1.1. An example for $F=\mathbb{Q}$ and $K=\mathbb{C}$ [1408AC]

$$ \mathbb{Q}\left( \sqrt[3]{ 2 } \right) \cong \mathbb{Q}\left( \sqrt[3]{ 2 } e^{ 2 \pi i / 3 } \right). $$

Proof. sketch [1408B]

WLOG, assume $p_{1}(x)$ and $p_{2}(x)$ are monic. Then by Proposition 1407, $$ \begin{align*} &\varphi_{1} : F_{1}[x] / \left( p_{1}(x) \right) \to F_{1}(\alpha), \qquad a(x) \mapsto a(\alpha), \\ &\varphi_{2} : F_{2}[x] / \left( p_{2}(x) \right) \to F_{2}(\beta), \qquad a(x) \mapsto a(\beta) \end{align*} $$ are field isomorphisms. Also, define $$ \sigma : F_{1}[x] / \left( p_{1}(x) \right) \to F_{2}[x] / \left( p_{2}(x) \right) , \quad c_{m}x^{m} + \cdots + c_{0} \mapsto \varphi(c_{m}) x^{m} + \cdots + \varphi(c_{0}), $$ which is well-defined and is an isomorphism induced by the isomorphism $\varphi$.

Then $\varphi_{2} \circ \sigma \circ \varphi_{1} : F_{1}(\alpha) \to F_{2}(\beta)$ is a field isomorphism and we have $\Psi = \varphi_{2} \circ \sigma \circ \varphi_{1}^{-1}$.

Proposition 6. necessary and sufficient conditions for $K / F$ to be finite or algebraic [14010]

Let $K / F$ be a field extension.

$K / F$ is finite if and only if there exist $\alpha_{1}, \dots, \alpha_{n} \in K$ algebraic over $F$ such that $K = F(\alpha_{1}, \dots, \alpha_{n})$.
$K / F$ is algebraic if and only if there exists $S \subseteq K$ such that every $\alpha \in S$ is algebraic over $F$ and $K = F(S)$.

A finite extension requires a finite number of algebraic generators ($\alpha_{1}, \dots, \alpha_{n}$).

An algebraic extension requires a set $S$ of algebraic generators, but $S$ can be infinite.

Not all algebraic extensions are finite. For example, the algebraic closure of $\mathbb{Q}$, denoted as $\overline{\mathbb{Q}}$, is an algebraic extension over $\mathbb{Q}$, but $[\overline{\mathbb{Q}} : \mathbb{Q}] = \infty$.

Proof. sketch [14010A]

$\implies$: Suppose $[K : F] = n < \infty$. Let $\alpha_{1}, \dots, \alpha_{n}$ be a basis of $K$ as an $F$-vector space. Then $K = F(\alpha_{1}, \dots, \alpha_{n})$. Moreover, $F(\alpha_{i}) \subseteq K$ implies $[F(\alpha_{i}) : F] \leq [K : F] < \infty$, then by Proposition 1403, $\alpha_{j}$ is algebraic over $F$.

$\impliedby$: By Theorem 1207, $$[K : F] = [F(\alpha_{1}, \dots, \alpha_{n}) : F] \leq [F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] [F(\alpha_{1}, \dots, \alpha_{n-1}) : F].$$ By HW2 Prob 1, $[F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] \leq [F(\alpha_{n}) : F]$. So $$[K : F] \leq [F(\alpha_{n}) : F][F(\alpha_{1}, \dots, \alpha_{n-1}) : F]$$ Therefore, by induction we have $[K : F] \leq [F(\alpha_{n}) : F][F(\alpha_{n-1}) : F] \cdots [F(\alpha_{1}) : F]$, in which every $[F(\alpha_{i}) : F] < \infty$.
$\implies$: Take $S = K$.

$\impliedby$: Given $\beta \in K = F(S)$, $\beta \in F(\alpha_{1}, \dots, \alpha_{n})$ for some $\alpha_{i} \in S$. The $\alpha_{i}$’s are algebraic, so $F(\alpha_{1}, \dots, \alpha_{n}) / F$ is finite. Hence $[F(\beta) : F]$ is finite since $F(\beta) \subseteq F(\alpha_{1}, \dots, \alpha_{n})$. Then by Proposition 1403, $\beta$ is algebraic over $F$. Therefore $K / F$ is algebraic.