If $\alpha \in E$ is algebraic over a $F$ and $E = F(\alpha)$, then $E$ is an algebraic extension.

Question

This seems like it should be straightforward but I can't figure out how to show this. Any help would be appreciated.

Answer 1

If $\alpha\in E$ is algebraic over $F$, then $[F(\alpha):F]=\deg(\mu_\alpha)<+\infty$, where $\mu_{\alpha}$ is the minimal polynomial of $\alpha$ over $F$. Hence, if $E=F(\alpha)$, then $E/F$ is finite dimensional, whence algebraic.

If you want to justify the equality $[F(\alpha):F]=\deg(\mu_\alpha)$, notice that $F[\alpha]\cong F[X]/(\mu_\alpha)$. Indeed, use first isomorphism theorem for rings applied to the evaluation map $f\mapsto f(a)$. Furthermore, $F[\alpha]=F(\alpha)$, since $\mu_{\alpha}$ being irreducible over $F$ implies that $F[\alpha]$ is a field.

If you are not convinced that $\mu_\alpha$ is irreducible over $F$, proceed by contradiction. Otherwise, there exists $(P,Q)\in F[X]^2$ with $\deg(P),\deg(Q)\geqslant 1$ such that $\mu_{\alpha}=PQ$. In particular, $P(\alpha)Q(\alpha)=0$ and $F$ being a field, without loss of generality one can assume that $P(\alpha)=0$. Therefore, by definition, $P\in\{f\in F[X]\textrm{ s.t. }f(\alpha)=0\}=(\mu_{\alpha})$, which is a contradiction, since $Q$ has degree at least $1$.

If you are wondering why a finite dimensional extension $E/F$ is algebraic, let $x\in E$ and notice that $\{1,x,\cdots,x^{[E:F]}\}$ contains more elements than the dimension of $E/F$ and therefore is $F$-linked. This means that there exists a nonzero $P\in F[X]$ such that $P(x)=0$.

Answer 2

Since $\alpha$ is algebraic, there is some non zero $P\in F[x]$ such that $P(\alpha)=0$. Then $\{1,\alpha,\ldots, \alpha^n\}$ (where $n$ is the degree of $P$) is a linearly dependent finite set in the $F$-vector space $E$.

The actual dimension of $E$ is the degree of the minimal polynomial of $\alpha$, which must divide $P$.

Answer 3

Since $F(\alpha)$ is a finite extension of $F$ (see C. Falcon and ajotatxe's answers), it suffices to show that finite extensions are algebraic.

Suppose $[E:F] = n < \infty $. Let $x \in E$. Since $E$ is a vector space over $F$ of dimension $n$, any collection of $n+1$ elements of $E$ are linearly dependent over $F$. Therefore the set $\{1,x,\dots,x^n\}$ is linearly dependent over $F$. That is, there exist $a_0,a_1,\dots a_n \in F$ (at least one of the $a_i \neq 0$) such that $$a_0 + a_1x + \dots a_nx^n = 0. $$

Thus $x$ is algebraic over $F$. Since $x$ was chosen arbitrarily in $E$, $E$ is an algebraic extension of $F$.