Suppose $a$ and $b$ are reals such that $a^b=b^a$. If $a$ is algebraic, is $b$ algebraic too?

6$\begingroup$ No, by GelfondSchneider. Take $a=3$ and suitable $b$ for instance. $\endgroup$– WojowuMay 19, 2019 at 16:49

2$\begingroup$ Picking up on Wojuwo's comment: it's easier for me to contemplate $a^{1/a} = b^{1/b}$. By GelfondSchneider, if $b$ is algebraic and irrational, then $b^{1/b}$ will be transcendental. So in the case $a = 3$, we would need a rational $b \neq 3$ to satisfy $b^{1/b} = 3^{1/3}$, and then it's just a matter of the fundamental theorem of arithmetic to rule out this possibility. $\endgroup$– Todd Trimble ♦May 19, 2019 at 18:15

$\begingroup$ @ToddTrimble I think it's actually (slightly) easier to keep the problem as stated. If $3^b=b^3$ and $b\neq 3$, it's clear that $b$ is not an integer, and hence that $b$ is irrational (as else $b^3$ is rational and $3^b$ isn't) and hence transcendental (GelfondSchneider). (also, it's Wojowu, not Wojuwo :) ) $\endgroup$– WojowuMay 19, 2019 at 19:31

$\begingroup$ @Wojowu : thank you for your comment. Maybe you can post it as an answer so that I can accept it. $\endgroup$– Sylvain JULIENMay 19, 2019 at 19:51

1$\begingroup$ If you say so, Wojowu. I mean, thanks for the additional explanation, but I said easier for me, and that might still be true even after your addition. Chacun a son gout, or however it goes. $\endgroup$– Todd Trimble ♦May 19, 2019 at 20:08
1 Answer
The answer is no. For instance, let $a=3$ and $b\neq 3$ be the real number satisfying $3^b=b^3$. Clearly $b$ is not an integer. It follows that $b$ is irrational  indeed, if $b$ was a noninteger rational, $3^b$ would be irrational, while $b^3$ would be rational. Finally, $b$ is transcendental, since otherwise $b$ would be algebraic irrational, $b^3$ would be algebraic and $3^b$ would be transcendental by GelfondSchneider.

2$\begingroup$ please do not answer offtopic questions. The question was not yet recognized as offtopic when you posted the answer apparently, so the best course of action may be to delete your answer. $\endgroup$– user140761May 20, 2019 at 6:07

$\begingroup$ oh, actually, you were among people who put it on hold, so my last comment is irrelevant probably. $\endgroup$– user140761May 20, 2019 at 6:09

2$\begingroup$ GelfondSchneider isn't so common background, so I'd say even if it makes sense to close the question here as nonresearch level, it's reasonable to have posted an answer (actually this maybe rather should have been migrated to MathSE). $\endgroup$– YCorMay 20, 2019 at 22:35