---- > [!theorem] Theorem. ([[Zariski's Lemma]]) > Let $k \subset L$ be a [[field]] [[field extension|extension]]. > If $L$ is of [[subalgebra generated by a subset|finite-type]][^1] as a $k$-[[algebra]], then $L$ is a [[finite algebra|finite]][^2] $k$-[[algebra]], i.e., $L$ is a finite-[[dimension|dimensional]] $k$-[[vector space]]. ^theorem > [!proof]+ Proof. ([[Zariski's Lemma]]) > By [[Noether's normalization theorem]], we have a [[finite algebra|finite extension]] $k[x_{1},\dots,x_{n}] \subset L$ for some $k$-[[algebraically independent]] $x_{1},\dots,x_{n} \in L$. [[integral algebra|Recall]] that finite-type + integral $\iff$ finite, hence this is in fact an [[integral algebra|integral extension]]. Since $k[x_{1},\dots,x_{n}]$ and $L$ are each [[integral domain|integral domains]] and $L$ is a [[field]], by [[integral extensions, units, and fields]], it follows that $k[x_{1},\dots,x_{n}]$ is a [[field]]. But this forces $n=0$: a [[polynomial 4|polynomial algebra]] in more than $0$ variables is never a field, for the variables are not [[unit|units]]. ---- #### [^1]: That is, $L$ is [[subalgebra generated by a subset|finitely generated]] as a $k$-[[algebra]]. [^2]: That is, $L$ is [[submodule generated by a subset|finitely generated]] as a $k$-[[module]]. Thus, while it is always true that finite $\implies$ finite-type, for [[field extension|field extensions]] the converse holds: finite-type $\implies$ finite. ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```