Algebraic Extensions

Lemma: Let K be a field and $(F_i)_{i \in I}$ be a collection of subfields of K. Then $\bigcap_{i \in I}F_i$ is a subfield of K (proof is left as an exercise to the reader).

Generated field: let K be a field, $A \subseteq K$. The field generated by A is the smallest subfield of K that contains A, which if it exists, is exactly $\bigcap_{F \leq K, A \subseteq F}F$).

Notation: if $K \vert F$ is a field extension and $\alpha_1, \alpha_2, ..., \alpha_n \in K$, then $F(\alpha_1, \alpha_2, ..., \alpha_n)$ is the field generated by F and $\alpha$. If $K = F(\alpha), \alpha \in K$ then $K \vert F$ is simple and $\alpha$ is a primite element.

Proposition: let $F \vert K$ is a field extension, $\alpha \in K$, and $P \in F[x]$ irreducible with $P(\alpha) = 0$ then $F[x]/(p) \cong F(\alpha) \leq K$, $x \mod p \mapsto \alpha$.

Proof: let $\phi : F[x] \rightarrow F(\alpha)$ be a ring homomorphism defined by $P \mapsto P(\alpha)$. ker$(\phi): P \in ker(\phi) \rightarrow (P) \in ker(\phi)$. since P is irreducible and F[x] is a PID, (P) is maximal, hence $ker(\phi) = (P)$ or $F(\alpha)$. Note that $ker(\phi) \neq F(\alpha)$ because $\phi(1x^0) \neq 0$ so $ker(\phi) = (P)$. By the first isomorphism theorem $F[x]/(p) \cong Im(\phi) \leq F(\alpha)$, where $Im(\phi) = F(\alpha)$.

Corollary: Let F, K, P and $\alpha$ be defined as above and deg(P) = n, then $F(\alpha) = \{\sum_{i=0}^{n-1}a_i\alpha^i : a_i \in F \}$ and $[F(\alpha): F] = n$.

let $K \vert F$ be a field extension. $\alpha in K$ is algebraic if there is a $P \in F[x]$ such that $P(\alpha) = 0.$ If $\alpha \in K$ is not algebraic it is transendental. $K \vert F$ is algebraic if every $\alpha \in K$ is algebraic over F.

A polynomial is said to me monic if the leading coefficient is 1.

Proposition: Let $K \vert F$ be a field extesion, $\alpha \in K$ algebraic. Then there exists $M_{\alpha, F} \in F[x]$ irreducible monic such that $M_{\alpha, F}(\alpha) = 0.$ Also, $\{P \in F[x]: P(\alpha) = 0\} = (M_{\alpha, F})$.

Proof: $I(\alpha) = \{P \in F[x] : P(\alpha) = 0 \}$, $I(\alpha) = ker(ev_{\alpha} \vert F[x])$ so $I(\alpha) \subseteq F[x]$ is an ideal. Since $F[x]$ is a PID, $I(\alpha) = (Q)$ for some $Q \in F[x]$. $Q = \sum_{i=0}^n a_iX^i$ and $M_{\alpha, F} = a^{-1}_nQ$ so $I(\alpha) = (M_{\alpha, F})$ because Q and $M_{\alpha, F}$ are associated. So $M_{\alpha, F}(K) = 0$. By the first isomorphism theorem $F[x]/I(\alpha) \cong Im(ev_{\alpha} \vert F[x]) \leq K$. Since $Im(ev_{\alpha} \vert F[x])$ is integral, $I(\alpha)$ is prime, therefore Q is irreducible and $M_{\alpha, F}$ is irreducible.

Corollary: let $L \vert K \vert F$ be field extensions and $\alpha \in L$ be algebraic over F, then $M_{\alpha, K} \vert M_{\alpha, F}$ in $K[x]$.

Proof: $\alpha$ is a root of $M_{\alpha, F}$ so $M_{\alpha, F} \in (M_{\alpha, k}) \leq K[x]$.

$M_{\alpha, F}$ is a minimal polynomial of $\alpha$ over F.

Remark: $F[x]/(M_{\alpha, F}) \cong F(\alpha)$ so $[F(\alpha): F] = deg M_{\alpha, F} = deg(\alpha/F)$

Examples: $i/\mathbb{R}$ has minimal polynomial $x^2+1$ (deg 2), $\sqrt{2}/\mathbb{Q}$ has minimal polynomial $x^2-2$ (deg 2), $\sqrt(2)/\mathbb{R}$ has minimal polynomial $x - \sqrt(2)$ (deg 1), j/Q has minimal polynomial $x^2+x+1$ (deg 2)

Proposition: $\alpha$ is algebraic over F if and only if $[F(\alpha): F]$ is finite.

Proof: the forward direction is proved above: $[F(\alpha):F] = deg(M_{\alpha, F})$ which is finite. Conversely, let $n = [F(\alpha):F]$. $1, \alpha, \alpha^2, ..., \alpha^n$ are not linearly independent over F, therefore there exists $\lambda_i \in F$ not all 0 such that $\sum \lambda_i \alpha^i = 0$, hence $P = \sum \lambda_i \alpha^i$ so $P(\alpha) = 0$.

Corollary: let $K \vert F$ be a field extension and $\alpha \in K$ algebraic. Then $F(\alpha) \vert F$ is algebraic.

Proof: let $\beta \in F(\alpha)$, then $F \leq F(\beta) \leq F(\alpha)$ so $[F(\alpha):F] \geq [F(\beta):F] \leq \infty$ so $\beta$ is algebraic over F.