On 22 October 2016 at 14:44, William Stein <[email protected]> wrote: > On Sat, Oct 22, 2016 at 6:11 AM, John Cremona <[email protected]> wrote: >> On 22 October 2016 at 09:37, Ralf Stephan <[email protected]> wrote: >>> sage: 2*(QQbar(1)) >>> 2 >>> sage: 2^(QQbar(1)) >>> ... >>> TypeError: no canonical coercion from Algebraic Field to Rational Field >>> >>> Why does the one work, the other not? Is it a bug? >> >> I don't see that as a bug. Any product of an integer and an element >> of QQbar is defined, and is again an element of QQbar, but not any >> integer raised to a QQbar exponent. I think it is a rather hard >> question to determine for which algebraic numbers a is 2^a algebraic! > > This made me curious whether or not 2^a is ever algebraic when a in > QQbar is not rational. It seems to me that Schanuel's conjecture > [1] (really an unproved conjecture of Lang) implies that if a is not > rational, then 2^a is not algebraic. > > Proof: The case n=2 of Schanuel's conjecture asserts that if z1, z2 > are complex numbers that are linearly independent over QQ, then the > field > > K = QQ(z1, z2, exp(z1), exp(z2)) > > has transcendence degree at least 2. Set > > z1 = a*log(2) and z2 = log(2), > > which are linearly independent over QQ, since a is not in QQ. Then > > K = QQ(z1, z2, exp(z1), exp(z2)) > = QQ(a*log(2), log(2), exp(a*log(2)), exp(log(2))) > = QQ(a*log(2), log(2), 2^a) > > The subfield QQ(a*log(2), log(2)) has transcendence degree 1 over QQ, > since a is algebraic over QQ and log(2) is transcendental. The > conjecture implies that K has transcendence degree at least 2, so 2^a > can't be algebraic over QQ(log(2)), hence 2^a is not an algebraic > number.
That looks good to me! So Ralf, do you want to catch the case where the exponent in QQbar is actually rational and then do what QQbar already knows how to do (raise to a rational power), returning an element of QQbar; and raise an Error of some kind in the other case? Or return some other type? John > > -- William > > [1] https://en.wikipedia.org/wiki/Schanuel%27s_conjecture > > >> >> John Cremona >> >>> >>> Regards, >>> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "sage-support" group. >>> To unsubscribe from this group and stop receiving emails from it, send an >>> email to [email protected]. >>> To post to this group, send email to [email protected]. >>> Visit this group at https://groups.google.com/group/sage-support. >>> For more options, visit https://groups.google.com/d/optout. >> >> -- >> You received this message because you are subscribed to the Google Groups >> "sage-support" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to [email protected]. >> To post to this group, send email to [email protected]. >> Visit this group at https://groups.google.com/group/sage-support. >> For more options, visit https://groups.google.com/d/optout. > > > > -- > William (http://wstein.org) > > -- > You received this message because you are subscribed to the Google Groups > "sage-support" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To post to this group, send email to [email protected]. > Visit this group at https://groups.google.com/group/sage-support. > For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/sage-support. For more options, visit https://groups.google.com/d/optout.
