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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!

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    $\begingroup$ Maybe I misunderstood the question, but isn't any non-algebraic field extension a counterexample? $\endgroup$ – rschwieb Dec 4 '17 at 15:43
  • $\begingroup$ What do you mean by $R_2$ is algebraic over $R_1$, if they are not necessary fields? $\endgroup$ – mfox Dec 4 '17 at 15:50
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    $\begingroup$ I mean that every element $r \in R_2$ is algebraic over $R_1$, namely, that there exist $c_m,c_{m-1},\ldots,c_1,c_0 \in R_1$, $m \geq 1$, such that $c_mr^m+c_{m-1}r^{m-1}+\cdots+c_1r+c_0=0$. $\endgroup$ – user237522 Dec 4 '17 at 15:52
  • $\begingroup$ @user237522 you should add this definition in your question. This is not exactly integral extensions. (Which I thought it is) $\endgroup$ – Krish Dec 4 '17 at 15:57
  • $\begingroup$ Krish, I did not write that it is integral, I only wrote that it is algebraic. But I do not mind to add the exact definition of algebraic if this helps. $\endgroup$ – user237522 Dec 4 '17 at 15:59
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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 $P_j's$.

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  • $\begingroup$ A nice idea, thanks! (though I am not sure I understand all the details). What if we take the base field to be $\mathbb{C}$? $\endgroup$ – user237522 Dec 4 '17 at 18:40
  • $\begingroup$ I'm quite sure that $e^X$ is still not algebraic over $K[X]$. I'm confident that this is the case over any field $K$ of characteristic $0.$ $\endgroup$ – GreginGre Dec 4 '17 at 21:38
  • $\begingroup$ I have added a complete argument for $K=\mathbb{R}$ or $\mathbb{C},$ since the one I gave was not correct. $\endgroup$ – GreginGre Dec 4 '17 at 22:27
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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.

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  • $\begingroup$ MooS, I like your idea when the base field is finite/countable. Also thank you for the reference for the interesting MO answers. $\endgroup$ – user237522 Dec 5 '17 at 17:35

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