= Solution
An element $\alpha$ is an <algebraic element> over $F$ if some nonzero <polynomial> in $F[X]$ vanishes at $\alpha$. A <field extension> is an <algebraic extension> if each of its elements is an <algebraic element> over $F$. If $[K:F]=d<\infty$, the $d+1$ <vectors> $1,\alpha,\ldots,\alpha^d$ are <linearly dependent>, producing such a <polynomial>. Thus every <finite field extension> is an <algebraic extension>.
The converse fails for the <field of algebraic numbers> $\overline{\mathbb Q}$. It is an <algebraic extension> of $\mathbb Q$ by definition, but it contains primitive $p$th <roots of unity> of degree $p-1$ for arbitrarily large primes $p$, by the permitted irreducibility of the <cyclotomic polynomial>. The <tower law> would make every such degree at most $[\overline{\mathbb Q}:\mathbb Q]$ if the latter were finite, a contradiction.
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