# Proving every algebraically independent subset of $F$ is contained in a transcendence basis.

I am studying transcendence bases and need a help on this problem.

Every algebraically independent subset of $$F$$ is contained in a transcendence basis.

Here are the relevant definitions:

Definition: Let $$F$$ be an extension field of $$K$$ and $$S$$ a subset of $$F$$. $$S$$ is algebraically dependent over $$K$$ if for some positive integer $$n$$ there exists a nonzero polynomial $$f\in K[x_1, \cdots, x_n]$$ such that $$f(s_1,\cdots, s_n)=0$$ for some distinct $$s_1, \cdots, s_n\in S$$. $$S$$ is algebraically independent over $$K$$ if $$S$$ is not algebraically dependent over $$K.$$

Definition: Let $$F$$ be an extension field of $$K.$$ A transcendence base of $$F$$ over $$K$$ is a subset $$S$$ of $$F$$ which is algebraically independent over $$K$$ and is maximal in the set of all algebraically independent subsets of $$F$$.

• Do you know how to prove that every linearly independent set in a vector space can be extended to a basis for the vector space? Can you try to modify such a proof to fit the question you ask here? – Gerry Myerson Mar 26 at 2:00
• @GerryMyerson Thanks very much for the idea! I solved the problem. – rgb12 Mar 26 at 2:26
• Good! You can write it up, and post it as an answer. (But first maybe check the Related questions to see whether something like it has already been posted here.) – Gerry Myerson Mar 26 at 2:30
• Are you still here, aloe? – Gerry Myerson Mar 27 at 11:35
• I'm voting to close this question as off-topic because OP has lost interest. – Gerry Myerson Mar 28 at 11:49