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Suppose I have $F = k[a_1, ..., a_n]$ is a field. We know that $F$ is a finitely generated algebra over $k.$ However, if $F$ is an algebraic extension over the field $k$ does that mean the extension is finite? I think so. Here is my reasoning:

If $F$ is a field then $F$ can be rewritten as $F = k(a_1, ..., a_n)$ (is this true?). Thus, since each $a_i$ is algebraic over $k,$ $k(a_1)$ is a finite extension over $k.$ So is $k(a_1, a_2)$ and so on for a finite number of steps. Is my reasoning correct?

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Your arguments are correct.

But we should mention that there is even more. Even without knowing that $F$ is an algebraic extension, we get that the extension is finite. This is the famous Zariski's lemma leading to Hilbert's Nullstellensatz.

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  • $\begingroup$ Thank you! And yes I am trying to prove Zariski's lemma by showing that the extension is algebraic first. $\endgroup$
    – green frog
    Oct 30, 2017 at 5:28

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