ANT week 1.1

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Definition. Let $K/F$$K/F$ be a field extension, we say $\alpha \in K$$\alpha \in K$ is transcendental if the evaluation map $\phi :F[x]\to K$$\varphi: F[x] \to K$ defined by $\phi (f(x))=f(\alpha )$$\varphi(f(x)) = f(\alpha)$ is injective. Otherwise we say $\alpha$$\alpha$ is algebraic over $F$$F$.

Definition. Let $\alpha$$\alpha$ be an algebraic element over $F$$F$. Then the kernel of $\phi$$\varphi$ is non-zero and it is a principal ideal since $F[x]$$F[x]$ is a PID. We define the monic generator of $\ker (\phi )$$\ker(\varphi)$ to be the minimal polynomial of $\alpha$$\alpha$, denoted as $m_{\alpha }(x)$$m_\alpha(x)$.

Proposition. $F(\alpha )=F[\alpha ]\cong F[x]/(m_{\alpha }(x))$$F(\alpha) =F[\alpha]\cong F[x]/(m_\alpha(x))$.

Proof. By the first isomorphism theorem, we have

$$F[x]/\ker (\phi )\cong \text{Im}(\phi )=F[\alpha ].$$

As a subring of $K$$K$, $F[\alpha ]$$F[\alpha]$ must be an integral domain, hence $\ker (\phi )=(m_{\alpha }(x))$$\ker(\varphi) = (m_\alpha(x))$ is a prime ideal hence a maximal ideal since $F[x]$$F[x]$ is a PID. Hence $F[\alpha ]$$F[\alpha]$ is a field and equal to $F(\alpha )$$F(\alpha)$.

Corollary. The minimal ploynomial $m_{\alpha }(x)$$m_\alpha(x)$ is irreducible. Otherwise if $m_{\alpha }(x)=g(x)h(x)$$m_\alpha(x) = g(x)h(x)$ with $\mathrm{deg}g,\mathrm{deg}h\ge 1$$\deg g, \deg h \geq 1$, this means $F[x]/(m_{\alpha }(x))$$F[x]/(m_\alpha(x))$ has zero divisors, contradicting that it's a field.

Proposition. Let $K/F$$K/F$ be a finite extension, then it is an algebraic extension.

Proof. Let $\alpha \in K$$\alpha \in K$ be any element, assume it is not algebraic, then we have an injective map

$$\phi :F[x]\to K,f(x)\mapsto f(\alpha ).$$

Hence we could view $F[x]$$F[x]$ as the subspace of $K$$K$ via the injection, but ${\dim }_{F}K$$\dim_FK$ is finite, contradiction. $◻$$\square$