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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!

stoic-santiago
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    This just seems like a logical shenanigan. If there is a nonempty independent subset, then clearly the extension is not algebraic over $K$. Conversely, If the extension is not algebraic over $K$, then there has to be a nonempty transc basis. – AlvinL Aug 09 '22 at 06:22

1 Answers1

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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

"The transcendence degree of $L/K$ is the largest cardinality of an algebraically independent subset of $L$ over $K$"

I personally dislike this definition because it is not immediate that such a "largest cardinal" exists. Instead, I suggest this solid definition:

Definition and property

The transcendence degree of $L/K$ is the cardinal $|X|$ of a maximal subset $X$ of $L$ that is algebraicaly independent over $K$. The quantity $|X|$ does not depend on the choice of a maximal subset $X$ of $L$.

In particular, if the transcendence degree of $L/K$ is $0$, then indeed any nonempty subset of $L$ is algebraic over $K$.

Olivier Roche
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