Algebraicity of $R_1 \subseteq R_2$, where $R_1$ and $R_2$ have same finite Krull dimension

Question

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!

Answer 1

Let $K$ be any field (of characteristic zero if you want). Take $R_1=K[X]$ and $R_2=K[[X]]$.

Then $R_1,R_2$ are $K$-algebras with Krull dimension $1$, and $K\subset R_1\subset R_2$.

However, I'm quite sure that the exponential formal series is not algebraic over $K[X]$ (at least, this is the case if you take $K=\mathbb{R}$, since the contrary would imply that $e$ is algebraic over $\mathbb{R}$, which can be seen by specializing a dependence relation at $X=1$, after dividing by a suitable power of $(X-1)$ if necessary)

Edit The arguments for $K=\mathbb{R}$ don't work, see my comment below. Here is a correct argument. Take an dependence relation $$P_0+P_1e^X+\cdots +P_n (e^X)^n=0,$$ with $P_i\in\mathbb{R}[X]$.

Multiply by $e^{-nX}$, substitute $X$ to a real parameter $x\in\mathbb{R}$, and let $x$ goes to $+\infty$. Then we get that $P_n(x)\to 0$ when $x\to +\infty$. This is impossible, unless $P_n=0.$ By induction, all the $P_i's$ are $0$, proving that $e^X$ is not algebraic over $\mathbb{R}[X]$.

This can be easily adapted to $K=\mathbb{C}$ by separating real and imaginary parts of the $P_j's$.

Answer 2

See this MO answers for more examples of transcendental power series and a proof for $e^X$ that works over any field of characteristic zero.

I would like to point out, that there is a trivial counting argument if $k$ is finite or countable (i.e. $\mathbb F_q$ or $\mathbb Q$):

If $k$ is countable, so is $k[x]$. If an integral domain is countable, there are clearly only finitely many algebraic elements over that ring. This is because there are only countable many polynomials and each of those has only finitely many roots.

But $k[[x]]$ is uncountable for any field $k$, since it is a $k$-vector space with uncountable basis.