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  "body": "\n## Statement\n\nLet $L/K$ be a [[field extension|0WQ1A8SHTZ]].\nLet $S \\subseteq T \\subseteq L$ be subsets such that $S$ is [[algebraically independent|KV4PT6KW10]] over $K$ and $L$ is [[algebraic|09EYMKMSB9]] over $K(T)$.\nThen there is a [[transcendence basis|E45PA267S4]] $B$ of $L/K$ with $S \\subseteq B \\subseteq T$.\nIn particular, taking $S = \\emptyset$ and $T = L$, every field extension has a transcendence basis.\n\n## Remarks\n\nApply [[Zorn's lemma|04RXTQA5ZA]] to the algebraically independent sets $A$ with $S \\subseteq A \\subseteq T$, ordered by inclusion; a union of a chain is independent, since a polynomial relation involves only finitely many elements.\nLet $B$ be a maximal one.\nEach $t \\in T$ is algebraic over $K(B)$, since otherwise $B \\cup \\set{t}$ would be a larger independent set.\nSo $K(T)$ is algebraic over $K(B)$, and $L$ is algebraic over $K(B)$ by transitivity of algebraic extensions.\nFor a finitely generated extension $L = K(t_1, \\dots, t_n)$, Zorn's lemma is not needed, and one gets a transcendence basis among $t_1, \\dots, t_n$.\n\n## Explanations\n\n### L1 human\n\nEvery field extension can be split into a free part and an algebraic part.\nOne can choose the free variables among any given generators, and one can start from any free variables one already has.\n\n### L2 human\n\nThis is the analogue of the theorem that [[every vector space has a basis|00MS9VMQ9Z]], and the proof is the same Zorn's lemma argument.\nTogether with the exchange property, it gives that all transcendence bases have the same cardinality, which defines the [[transcendence degree|D7TR0YWR9H]].\nIt is used to write every extension as a [[purely transcendental extension|G5PVJMM37Z]] followed by an algebraic one.\n",
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