On Wikipedia, I read that
A field extension $L/K$ is algebraic if and only if its transcendence degree is $0$.
$[\Rightarrow]$ Suppose $L/K$ is an algebraic field extension, i.e., for every $l \in L$, there exists a non-zero polynomial $g(X)$ in $K[X]$ such that $g(l) = 0$. The transcendence degree of $L/K$ is the largest cardinality of an algebraically independent subset of $L$ over $K$. Since every $l \in L$ satisfies some non-trivial polynomial equation with coefficients in $K$, there is no algebraically independent subset of $L$ over $K$. By definition, the transcendence degree of $L/K$ is $0$.
$[\Leftarrow]$ Suppose $L/K$ has zero transcendence degree, i.e., the largest cardinality of an algebraically independent subset of $L$ over $K$ is zero. In other words, every non-empty subset of $L$ is not algebraically independent over $K$. In particular, consider $\{l\}$ for every $l \in L$. There exists non-zero $h(X) \in K[X]$ with $h(l) = 0$ by algebraic independence, showing that every element of $L$ is algebraic over $K$.
Could someone verify the proof? Thanks!
Best Answer
What you mean is : "consider ${l}$ for any $l \in L$".
Aside from that detail, I can't spot a flaw in your proof, as long as you use the definition of the transcendence degree that was given you in your lessons
I personally dislike this definition because it is not immediate that such a "largest cardinal" exists. Instead, I suggest this solid definition:
In particular, if the transcendence degree of $L/K$ is $0$, then indeed any nonempty subset of $L$ is algebraic over $K$.