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A field extension $k\subseteq F$ is finitely generated if there exist $\alpha_a,\alpha_2,\dots,\alpha_n\in F$ such that $$F=k(\alpha_1)(\alpha_2)\dots(\alpha_n)$$ This is not the same as saying $F$ is a finite type $k$-Algebra. What if $\alpha_i$ is transcendental for some $i\in\{1,2,3\dots,n\}$?
Why is $F$ not a finite type $k$-Algebra? Isn't $F=k[\alpha_1,\alpha_2\dots,\alpha_n]$?
The element $\alpha^{-1}$ is contained in $k(\alpha)$ basically by definition. (Excluding the degenerate case $\alpha=0$.) As the notation $k(\alpha)$ means the smallest field containing $k$ and $\alpha$.
By contrast $k[\alpha]$ may well not contain $\alpha^{-1}$, informally as it is formed by the elements one gets plugging $\alpha$ into a polynomial with coefficients in $k$.
If $\alpha$ is algebraic the two happen to coincide but this is a result that needs to be proved. And the converse holds as well.
In case you actually want a proof that they are not equal for transcendental elements, rather than just the plausibility argument I gave above, then show that $\alpha^{-1}$ is actually not a polynomial expression in $\alpha$ by assuming the converse and deriving that $\alpha$ would be a root of a polynomial.
If $t$ is transcendental, $k(t)$ is finitely generated as a field extension of $k$, but it's not finitely generated as a $k$-algebra (if $k$ is too large). In fact if $k$ is uncountable, then the dimension of $k(t)$ as a vector space over $k$ is uncountable, whereas the dimension of a finitely generated $k$-algebra as a vector space over $K$ is at most countable (exercise). This observation can actually be used to prove special cases of the Nullstellensatz.
