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.
-- William
[1] https://en.wikipedia.org/wiki/Schanuel%27s_conjecture
>
> John Cremona
>
>>
>> Regards,
>>
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--
William (http://wstein.org)
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