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Is it possible to generalize Degree of sum of algebraic numbers (especially Pete L. Clark's answer, based on Keith Conrad's answer) in the following way: Let $D$ be a (noetherian) UFD of zero characteristic which is not a field, and let $a$ and $b$ be integral over $D$. Denote the degree of the minimal polynomial of $a$, $b$, $a+b$ by $n$, $m$, $d$, respectively. Assume that $d=nm$. Is it true that $D[a+b]=D[a,b]$? What about other conditions (instead of $d=nm$) guaranteeing $D[a+b]=D[a,b]$?; for example (as noted in the above question), the maximum of $n$ and $m$ is a prime number.

Sorry if my question is somewhat trivial.

Community
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user237522
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2 Answers2

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When you don't work over a field your previous experience with field degrees can break down pretty badly because over a ring that is not a field (even over a ring as simple as $\mathbf Z$) a finite free module can be a submodule of another finite free module with the same rank and the two modules don't have to be the same. The simple property that $K \subset F \subset E$ and $[F:K] = [E:K]$ implies $E = F$ for finite extensions of fields is false when you replace fields with other integral domains. In terms of linear algebra, a subspace of an $n$-dimensional vector space with dimension $n$ must be the whole space, but this isn't generally true when you work with finite free modules over rings other than fields.

A counterexample to your question occurs in the simplest case $m = n = 2$ and $D = \mathbf Z$. Try $a = \sqrt{2}$ and $b = \sqrt{3}$. Although $\mathbf Q(\sqrt{2}+\sqrt{3}) = \mathbf Q(\sqrt{2},\sqrt{3})$, the ring $\mathbf Z[\sqrt{2}+\sqrt{3}]$ is not $\mathbf Z[\sqrt{2},\sqrt{3}]$. In fact, the first ring has index 4 in the second. More generally, if $a$ and $b$ are nonsquares in $\mathbf Z$ such that their ratio $a/b$ is also not a square, then $\mathbf Q(\sqrt{a}+\sqrt{b}) = \mathbf Q(\sqrt{a},\sqrt{b})$ but $\mathbf Z[\sqrt{a}+\sqrt{b}]$ is not $\mathbf Z[\sqrt{a},\sqrt{b}]$: the first ring has index $4|b-a|$ in the second.

I think it is hopeless to extend the property you read about over fields to other kinds of integral domains in any simple way.

KConrad
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  • Thank you very much for the nice counterexamples. I wonder if one can find conditions on $D$, $a$ and $b$ such that $D[a+b]=D[a,b]$? For example, if $a$ and $b$ are in the field of fractions of $D$ and $D[a+b] \subseteq D[a,b]$ is flat, then $D[a+b]=D[a,b]$ (since flat+integral=faithfully flat, and then applying Exercise 7.2 of Matsumura's book). Here it is enough to assume that $D$ is a domain, not necessarily a UFD. (Actually, I prefer not to assume that $D[a+b] \subseteq D[a,b]$ is flat). – user237522 May 14 '15 at 21:17
  • Do you have an actual application in mind (i.e., you have a situation where you really see such equality happening, and you seek a proof of it) or is this a random question you are asking? – KConrad May 14 '15 at 22:32
  • Truly, I have an actual application in mind. I am interested in the following situation: $D$ is a noetherian UFD. $a$ and $b$ are integral over $D$. $D \subseteq D[a+b]$ is flat. $D[a+b] \subseteq D[a,b]$ have the same field of fractions. We do not know if $D$ and $D[a+b]$ have the same field of fractions and we do not know if $D[a+b] \subseteq D[a,b]$ is flat. I wish to show that $D=D[a+b]$ or $D[a+b]=D[a,b]$. – user237522 May 15 '15 at 13:09
  • Why is my example not a counterexample to what you are asking? Take $D = \mathbf Z$, $a = \sqrt{2}$, and $b = \sqrt{3}$. The ring $\mathbf Z[\sqrt{2}+\sqrt{3}]$ is a flat $\mathbf Z$-module since it is free, and $\mathbf Z[\sqrt{2}+\sqrt{3}]$ lies strictly between $\mathbf Z$ and $\mathbf Z[\sqrt{2},\sqrt{3}]$. (When I was asking for an actual application I meant a concrete example -- specifying $D$, $a$, and $b$ -- where you wanted to prove such an equality. You still haven't provided one.) – KConrad May 15 '15 at 14:14
  • I did not claim that your example is not relevant; on the contrary, it's nice that it serves both as a counterexample to my original question and to the more specific case which I mentioned above. My concrete example involves polynomial rings; I will try to think about it myself for some more time before bringing it here (I prefer solving problems myself, asking questions only when I feel I have reached a point where I am not able to say more). – user237522 May 17 '15 at 15:39
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Take $D=k[x,y]$ ($k$ a field, $x$ and $y$ indeterminates), and $a=x^{1/n}$, $b=y^{1/n}$ for any $m,n\geq2$. Then $D[a,b]\cong k[u,v]/(u^n -x,v^m-y)$. Reducing modulo the ideal $J=(x,y)$ of $D$, we get the $D/J$-algebra $k[u,v]/(u^n ,v^m)$ which cannot be generated by one element, hence is not isomorphic to $D[a+b]\,/\,J\,D[a+b]$. Hence $D[a,b]\not\cong D[a+b]$ as $D$-algebras.

Laurent Moret-Bailly
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