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This is an elementary question (coming from an undergraduate student) about algebraic numbers, to which I don't have a complete answer.

Let $a$ and $b$ be algebraic numbers, with respective degrees $m$ and $n$. Suppose $m$ and $n$ are coprime. Does the degree of $a+b$ always equal $mn$?

I know that the answer is "yes" in the following particular cases (I can provide details if needed) :

1) The maximum of $m$ and $n$ is a prime number.

2) $(m,n)=(3,4)$.

3) At least one of the fields $\mathbf{Q}(a)$ and $\mathbf{Q}(b)$ is a Galois extension of $\mathbf{Q}$.

4) There exists a prime $p$ which is inert in both fields $\mathbf{Q}(a)$ and $\mathbf{Q}(b)$ (if $a$ and $b$ are algebraic integers, this amounts to say that the minimal polynomials of $a$ and $b$ are still irreducible when reduced modulo $p$).

I can also give the following reformulation of the problem : let $P$ and $Q$ be the respective minimal polynomials of $a$ and $b$, and consider the resultant polynomial $R(X) = \operatorname{Res}_Y (P(Y),Q(X-Y))$, which has degree $mn$. Is it true that $R$ has distinct roots? If so, it should be possible to prove this by reducing modulo some prime, but which one?

Despite the partial results, I am at a loss about the general case and would greatly appreciate any help!

[EDIT : The question is now completely answered (see below, thanks to Keith Conrad for providing the reference). Note that in Isaacs' article there are in fact two proofs of the result, one of which is only sketched but uses group representation theory.]

François Brunault
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    A former colleague and (current) friend of mine was thinking about this problem a couple of years ago. I believe he was able to prove it in some special cases and was open enough to the possibility of a counterexample to do some computer experimentation. More than that I don't remember, but the moral is: this is a much harder question than one might think! I will try to contact him and see if he is willing to come here and weigh in on the matter. – Pete L. Clark Jun 02 '10 at 16:34
  • This would be easy Galois theory if I could prove the following combinatorial result: Let $G$ be a finite group and $A$ and $B$ sets with transitive $G$ actions, of relatively prime orders. (So the $G$ -action on $A \times B$ is necessarily transitive.) Let $\sim$ be a $G$-invariant equivalence relation on $A \times B$ such that $(a,b) \sim (a,b')$ implies $b=b'$ and $(a,b) \sim (a',b)$ implies $a=a'$. Then $\sim$ is the trivial equivalence relation. Does anyone have a counter-example to the combinatorial claim? – David E Speyer Jun 02 '10 at 16:58
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    Prompted by Francois's example below, G is S_3. A={1,2,3} and B={+,-}, with action by the sign representation of G. Equivalence classes {1+,2-}, {2+,3-} and {3+,1-}. Somehow, we need to use the fact that our binary operation is addition, not an arbitrary cancellable relation. – David E Speyer Jun 02 '10 at 17:55

  • Is it true that R has distinct roots? This question is irrelevant. The relevant question is: "is it true that R is irreducible?"

     – potap Sep 26 '13 at 10:55
  • @potap The roots of $R$ are the numbers $a'+b'$ where $a'$ resp. $b'$ runs over the conjugates of $a$ resp. $b$. Saying that $a+b$ has degree $mn$ is equivalent to say that these $mn$ numbers are pairwise distinct. – François Brunault Feb 05 '19 at 20:28

4 Answers4

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The following answer was communicated to me by Keith Conrad:

See:

M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25 (1970), 638--641.

http://alpha.math.uga.edu/~pete/Isaacs70.pdf

Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are relatively prime, then $K(a,b)$ = $K(a+b)$, which answers the students question in the affirmative. His proof shows the same conclusion holds under the weaker assumption that

$[K(a,b):K] = [K(a):K][K(b):K]$.

since Isaacs uses the relative primality assumption on the degrees only to get that degree formula above, which can occur even in cases where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.

Martin Sleziak
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Pete L. Clark
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    Note also that Isaacs finishes the proof of the characteristic 0 case early on; the hard work is the char p extensions. – David E Speyer Jun 02 '10 at 21:26
  • Great! This is exactly what I was looking for. Thanks to Keith Conrad for providing the reference. The extended result you mention is also very interesting. So, this completely answers the student's question. I think he too will appreciate your help! – François Brunault Jun 03 '10 at 05:46
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    This article is freely and legally available at http://www.ams.org/journals/proc/1970-025-03/S0002-9939-1970-0258803-3/home.html – Pierre-Yves Gaillard Feb 26 '12 at 11:09
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A counter-example to show that this result does not extend to characteristic $p$: Let $K= \mathbb{F}_p(s,t)$, the rank two transcendental extension of $\mathbb{F}_p$. Let $\alpha$ and $\beta$ be roots of $$\alpha^{p-1} - s=0$$ $$\beta^p - s \beta - t=0.$$

Then $\alpha$ and $\beta$ have degrees $p-1$ and $p$ over $K$. The element $\alpha + \beta$ obeys $$(\alpha+\beta)^p - s (\alpha+\beta) - t=0.$$

David E Speyer
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    Note that this is an example, not a counterexample, if $p=2$. – Gerry Myerson May 14 '12 at 00:40
  • It is worth noting that although $K(\alpha,\beta) \neq K(\alpha+\beta)$, $K(\alpha,\beta)/K$ is still simple extension: a proof can be found here ($\alpha$ and $\beta$ are both separable in this example). By the way, the minimal polynomial of $\alpha$ in $K(\alpha+\beta)$ is still $x^{p-1} - s$, and the minimal polynomial of $\beta$ in $K(\alpha+\beta)$ is $(\alpha+\beta-x)^{p-1} - s$. – Jianing Song Nov 06 '22 at 16:26
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For those who want to read Isaac's proof --- mentioned in Pete's answer --- in the language of Molière and Bourbaki, there is François Brunault's exposé.

Addendum (30/01/2011). Today I came across the following related article by Weintraub in the Monatshefte.

Glorfindel
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Chandan Singh Dalawat
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What is true for sure is that this statement (in fact, a more general one) holds over a finite field (see Brawley, J. V., Carlitz, L. Irreducibles and the composed product for polynomials over a finite field. MR0893074 (89g:11118)). In char=0, however, I don't have a good reference, even though I always thought it should be true as well (maybe Brawley and Carlitz have hints on it, but I have no access to their article: Ireland seems to fight recession by discharging online subscriptions to academic journals!)...

Vladimir Dotsenko
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  • Thank you for the reference. So, in the case the base field Q is replaced by a finite field, then not only deg(a+b)=mn but also deg(ab)=mn. Note that in the original setting deg(ab) can be less than mn : take for example the algebraic numbers a=2^{1/3} and b=j. – François Brunault Jun 02 '10 at 17:09
  • Here $j$ is a square root of $-1$? I don't think that works; I think $ab$ has degree $6$. What am I missing? – David E Speyer Jun 02 '10 at 17:33
  • Sorry for the notation, j is a cube root of 1. – François Brunault Jun 02 '10 at 17:36
  • Looking at the proof of their Theorem 2, the key fact they are using it that the Galois group of k(alpha, beta)/k is cyclic. I don't see any hints about how to get beyond this. – David E Speyer Jun 02 '10 at 17:48
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    Got it. Thanks. That should also reveal what is wrong with my approach ... – David E Speyer Jun 02 '10 at 17:49
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    @David : I think this gives a combinatorial counter-example to the claim above. Take $A$ (resp. $B$) to be the set of conjugates of $\sqrt[3]{2}$ (resp. of $j=e^{2i\pi/3}$) and $G=\operatorname{Gal}(\mathbf{Q}(A,B)/\mathbf{Q})$ with the usual action on $A$ and $B$. Then the relation $(x,y) \sim (x',y') \Leftrightarrow xy=x'y'$ works. – François Brunault Jun 02 '10 at 17:50