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convert expression to QuadraticField

asked 2013-05-16 04:52:39 -0600

updated 2013-05-16 04:55:39 -0600

I tried (in the Sage shell)

K = QuadraticField(-3)
a = K(sqrt(-3))

But that fails with a TypeError exception. What is the problem?

I know that I can get the generator by K.gen() but I want to be able to convert expressions to K.

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I don't know that it will be so easy to coerce from the symbolic ring to here in general, but more power to you if someone can... I do find the following behavior a little disturbing - try `sage: K.gen()`, which yields `a`, which no one ever asked for... this *is* documented in `QuadraticField?` but only as the default variable name, which of course isn't injected into the global namespace.

kcrisman gravatar imagekcrisman ( 2013-05-16 08:42:06 -0600 )edit

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answered 2013-05-16 22:48:47 -0600

Hi,

You should avoid usage of the symbolic ring as much as possible. It is useful to deal with functions, derivations and such but not to deal with arithmetic questions.

There is a class called AlgebraicConverter in sage.symbolic.expression_conversions which is implemented and called in such case. It works well for the field of algebraic numbers QQbar::

sage: QQbar(sqrt(-3))
1.732050807568878?*I
sage: QQbar(sin(pi/4))
0.7071067811865475?

But not for other number fields. Nevertheless it is possible to do

sage: K = QuadraticField(-3)
sage: K(3).sqrt()
a

The method sqrt above actually uses Pari in the background: if your number field contains such a root it returns it, otherwise the answer will be an element of the symbolic ring.

sage: K(3).sqrt().sqrt()
3^(1/4)
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vdelecroix gravatar imagevdelecroix ( 2013-05-16 22:51:46 -0600 )edit

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Asked: 2013-05-16 04:52:39 -0600

Seen: 42 times

Last updated: May 16 '13