Let $A = k[X_1,.., X_n]/I$ where $k$ is a field and $I$ is some ideal.
Suppose $A$ is a field and $A/k$ is not an algebraic extension. Let $x_i = X_i + I$. I want to show that we can find (relabeling the variables if necessary) there exist $\{ x_1,.., x_r \}$ that is algebraically independent over $k$ and each $x_{r+1},.., x_n$ is algebraic over $k(x_1, .., x_n)$?
If $k[X_1,.., X_n]/I$ is a field non algebraic extension of $k$, how to show there exists an algebraically independent subset of $\{X_1+I,..,X_n+I\}$
abstract-algebracommutative-algebrafield-theory
Best Answer
"Zariski's lemma" states that, for any field $k$, if a field is finitely generated as a $k$-algebra then it is a finite field extension of $k$. Proof of this is below.
However, this means that the circumstances of your problem cannot arise; $A$ is finitely generated as a $k$-algebra (by the $x_i$), and hence – since it is a field – must be a finite and thus algebraic extension of $k$.
Lemma 1: If $E$ and $F=E(x)$ are fields, and $F$ is finitely generated as an $E$-algebra, then $F$ is a finite field extension of $E$.
Proof: Say $F=E[f_1,\ldots,f_n]$, and suppose for contradiction that $F$ is not finite over $E$. Then $x$ must be transcendental over $E$, and so $F$ is the field of rational functions in $x$ over $E$. This means that each $f_i=p_i/q_i$ for some $p_i,q_i\in E[x]$. Consider the element $r:=\frac{1}{x\prod_{i=1}^nq_i+1}\in F$; there must be some polynomial $g\in E[t_1,\ldots,t_n]$ such that $r=g(f_1,\ldots,f_n)$. We can clear denominators in $g(f_1,\ldots,f_n)$ by multiplying through with a suitable power of $\prod_{i=1}^nq_i$, giving $s:=(\prod_{i=1}^nq_i)^kg(f_1,\ldots,f_n)\in E[x]$ for some $k\in\mathbb{N}$, and hence $\frac{1}{x\prod_{i=1}^nq_i+1}=\frac{1}{(\prod_{i=1}^nq_i)^k}s$. But this means $(\prod_{i=1}^nq_i)^k=(x\prod_{i=1}^nq_i+1)s$, and this is a contradiction, since $x\prod_{i=1}^nq_i+1$ and $\prod_{i=1}^nq_i$ (and hence $x\prod_{i=1}^nq_i+1$ and $(\prod_{i=1}^nq_i)^k$) are coprime in $E[x]$.
Lemma 2: Let $A\subseteq B\subseteq C$ rings with $A$ Noetherian. If $C$ is finitely generated as an $A$-algebra and $C$ is finitely generated as a $B$-module, then $B$ is finitely generated as an $A$-algebra.
Proof: See eg here.
Theorem (Zariski's lemma): Suppose $F:E$ is a field extension and that $F$ is finitely generated as an $E$-algebra. Then $F$ is finite over $E$.
Proof: Say $F=E[x_1,\ldots,x_n]$; we prove by induction on $n$. The case $n=0$ is clear. Now suppose the result holds for $n-1$, and let $E'=E(x_n)\subseteq F$. By the tower law, $[F:E]=[F:E'][E':E]$, and so it suffices to show (i) that $F$ is a finite extension of $E'$ and (ii) that $E'$ is a finite extension of $E$. For (i), note that $F=E'[x_1,\ldots,x_{n-1}]$, and so – since $E'$ is a field – by the induction hypothesis we have that $F$ is a finite field extension of $E'$. For (ii), note that (i) allows us to apply lemma 2 and show that $E'$ is finitely generated as an $E$-algebra. By lemma 1, this means that $E'=E(x_n)$ is in fact a finite extension of $E$, so (ii) holds and we are done.