Maximal algebraic independent in commutative algebra over field In field extension, maximal algebraic independent elements in a set of generators (generate by means of fraction of generators) will also be maximal alebraic independent amoung all subsets. It is then natural to ask: is this still true in general case of commutative algebra over field? More concretely:
Field $F$ is contained in a commutative ring $R$, and $x_1,\dots,x_n,y$ are elements of $R$, such that:

*

*$R=F[x_1,\dots, x_n, y]=\{ \sum_{i_1\dots i_n,j} a_{i_1\dots i_n,j}x_1^{i_1}\dots x_n^{i_n}y^j|a_{i_1\dots i_n,j}\in F\}$

*$x_1,\dots,x_n$ are algebraic independent while $x_1,\dots,x_n,y$ are algebraic dependent

is it true that for any $r\in R$, $x_1,\dots,x_n,r$ are alebraic dependent?
Can anyone help me with this? Thank you so much.
 A: Yes.
Suppose $x_1,\ldots,x_n,r$ are algebraically independent over $F$. Then $S=F[x_1,\ldots,x_n,r]$ is a polynomial ring of dimension of $n+1$
(Note that if $R$ is a domain, this is immediate from the definition of transcendence degree of its fraction field over $F$.)

Claim 1: Let $z_1,\ldots,z_k\in R$, and let $R'=R/\operatorname{nil}(R)$, where $\operatorname{nil}(R)$ is the
nilradical of $R$. Then $z_1,\ldots,z_k$ are algebraically dependent
over $F$ if and only if their images in $R'$ are also.
Proof of claim: $(\Rightarrow)$ If $f(z_1,\ldots,z_k)=0$ in $R$, the same is true in $R'$.
$(\Leftarrow)$ Suppose $f(z_1,\ldots,z_k)=0$ in $R'$, i.e.
$f(z_1,\ldots,z_k)\in \operatorname{nil}(R)$. Then some power of
$f(z_1,\ldots,z_k)$ vanishes in $R$. QED

By Claim 1, we may assume that $R$ is reduced, i.e. its nilradical is 0. Therefore, replacing $R$ with its total ring of fractions, we may assume that $R$ is a finite direct product $K_1\times\cdots \times K_s$ of fields, where each field $K_i$ is a finitely-generated field extension of $F$. For each $i=1,\ldots,s$, let $\pi_i\colon R\to K_i$ be projection.

Claim 2: Let $z_1,\ldots,z_k\in R$. Then $z_1,\ldots,z_k$ are algebraically dependent over if and only if, for each $i$, the
elements $\pi_i(z_1),\ldots,\pi_i(z_k)$ are algebraically dependent
over $F$.
Proof of claim: $(\Rightarrow)$ This is immediate.
$(\Leftarrow)$ Suppose that for each $i$, the elements
$\pi_i(z_1),\ldots,\pi_i(z_k)$ are algebraically dependent over $F$.
Then for each $i$, there is a non-zero polynomial $f_i$ such that  $$
> f_i(\pi_i(z_1),\ldots,\pi_i(z_k)) = 0.$$ Then the $i$th component of
$f_i(z_1,\ldots,z_k)$ is zero. So, letting $g=f_1\cdots f_s$, we get
that $g(z_1,\ldots,z_k)=0$. Hence, $z_1,\ldots,z_k$ are algebraically
dependent. QED

By Claim 2, we know that:

*

*$\pi_i(x_1),\ldots,\pi_i(x_n)$ are algebraically independent over $F$ for each $i$; and


*$\pi_i(x_1),\ldots,\pi_i(x_n),\pi_i(y)$ are algebraically dependent over $F$ for each $i$.
Therefore, each $K_i$ has transcendence degree $n$ over $F$. Hence, for each $i$, we have $\pi_i(x_1),\ldots,\pi_i(x_n),\pi_i(r)$ are algebraically dependent over $F$. Thus, again by Claim 2, $x_1,\ldots,x_n,r$ are algebraically dependent over $F$.
