# An algebraic simple extension has finite degree

I know that finite extensions imply algebraic extensions (but not the converse) and that transcendental extensions are not necessarily finite extensions (even if they are simple extensions), so I am wondering if there is some mistake in my understanding that a simple algebraic extension is always a finite extension. Specifically, my claim is that:

If $\alpha$ is algebraic over F, then $[F(\alpha):F] < \aleph{}_0$ (finite degree)

My thoughts: Since $\alpha$ is algebraic, it has a unique minimal polynomial in $F$, whose existence implies that $[F(a):F]$ is finite since $F(a) \approx F[x]/\langle{}p(x)\rangle{}$.

Your thoughts are correct: since $\;\Bbb F(a)\cong\Bbb F[x]/\langle p(x)\rangle\;$ . Assume then that $\;\deg p=n\;$ , so we can write $\;p(x)=\sum\limits_{k=0}^n b_kx^k\;$ . Observe now the elements
$$\overline 1:=1+\langle p(x)\rangle,\,\overline x=x+\langle p(x)\rangle,\,\overline{x^2}:=x^2+\langle p(x)\rangle,\ldots,\overline{x^{n-1}}:=x^{n-1}+\langle p(x)\rangle \in\Bbb F(a)\;$$
The above $\;n\;$ elements must be linearly independent in $\;\Bbb F(a)\;$ , otherwise there exist scalars $\;c_0,c_1,...,c_{n-1}\in\Bbb F\;$, not all zero, s.t. $\;c_1\cdot\overline 1+c_1\overline x+\ldots+c_{n-1}\overline{x^{n-1}}=0\;$ , but this is possible iff
$$p(x)\,|\,\sum_{k=0}^{n-1}c_kx^k\iff \sum_{k=0}^{n-1}c_ka^k=0$$
which means $\;a\;$ is a root of a non-zero polynomial with degree $\;< n\;$ , which contradicts minimality of $\;\deg p\;$ ...
Finally, just observe that $\;\overline 1,\overline x,\ldots,\overline{x^n}\;$ are already linearly dependent, and you get your extension's degree, i.e. $\;\dim_{\Bbb F}\Bbb F(a)=n\;$