Assume that $R_i$ is a commutative $k$-algebra ($k$ is a field of characteristic zero) having finite Krull dimension $n_i$, $1 \leq i \leq 2$, and $k \subset R_1 \subseteq R_2$.
Further assume that $n_1=n_2 \geq 1$.
I wonder if it follows that $R_2$ is algebraic over $R_1$, or there exists a counterexample?
Where by an algebraic commutative rings extension $S \subseteq T$, I mean that every element $t \in T$ is algebraic over $S$, namely, there exist $s_m,\ldots,s_1,s_0 \in S$, $m \geq 1$, such that $s_mt^m+\cdots+s_1t+s_0=0$.
My motivation is the result about fields: If $k \subseteq R_1 \subseteq R_2$ are three fields with $R_i$ having finite transcendence degree $n_i$, $n_1=n_2$, implies that $R_2$ is algebraic over $R_1$.
Remarks: (1) Perhaps the $n_1=n_2=1$ case has a positive answer, while each of the $n_1=n_2 \geq 2$ cases has a negative answer? (2) See also this question.
Thank you very much!