Let $k$ be a field of characteristic zero, let $R$ be a (commutative) $k$-algebra with $1$, and $\{R_i\}_{i \in \mathbb{N}}$ (with $R_0:=R$) a descending chain of $k$-subalgebras of $R$, namely $R \supseteq R_1 \supseteq R_2 \supseteq \ldots$, with $R_i \neq k$ for all $i \in \mathbb{N}$.
Assume that $R$ has finite Krull dimension, $\dim(R)=n < \infty$.
Further assume that $\cap_{i \in \mathbb{N}} R_i=k$.
Does it follow that there exist indices $i_1,\ldots,i_n$ such that $\dim(R_{i_j})=j$, $1 \leq j \leq n$?
Of course, we can take $i_n=0$; indeed, $R_{i_n}=R_0=R$ is of Krull dimension $n$.
Example: $R=k[x,y]$, $R_1=k[y]$, $R_2=k[y^2]$, $R_3=k[y^4]$, $R_4=k[y^8]$, so $R_m=k[y^{2^{m-1}}]$, $m \geq 1$. $\dim(R)=2$, $\dim(R_m)=1$, $m \geq 1$. (How to prove that $\cap_{i \in \mathbb{N}} R_i=k$?). But this is just an example. Perhaps there exists a counterexample to my question?
Thank you very much!
Edit: After receiving a counterexample in the comments, I wish to restrict my question to the special case where $\dim(R)=2$ (or even to $R=k[x,y]$). In other words:
If $R=k[x,y]$, $R \supseteq R_1 \supseteq R_2 \supseteq \ldots$, with $\dim(R_i) \in \{1,2\}$ (this guarantees that $R_i \neq k$) and $\cap_{i \in \mathbb{N}}R_i =k$, does it follow that there exists $l \in \mathbb{N}$ such that $\dim(R_l)=1$? (If there exists such $l$, then for every $L > l$, we also have $\dim(R_L)=1$).
I think that a counterexample is $R_1=k[x^2,y^2]$, $R_2=k[x^4,y^4]$, $R_3=k[x^8,y^8]$, etc. Here, $\cap_{i \in \mathbb{N}}R_i=k$ and $\dim(R_i)=2$, for all $i \in \mathbb{N}$.
Therefore, I wish to further restrict my question to $R_i$, $i \geq 1$, of the following special forms:
$R_i=k[f_i,g_i]$, where the Jacobian of $f_i$ and $g_i$ is a non-zero scalar, namely, $Jac(f_i,g_i) \in k^{\times}$.
Notice that above we had $Jac(f_i,g_i)=Jac(x^{2^i},y^{2^i}) \in k[x,y]-k$, which is not a non-zero scalar; it is just a non-zero polynomial.