Comment #1 on issue 2329 by [email protected]: Should Poly use an algebraic
domain by default?
http://code.google.com/p/sympy/issues/detail?id=2329
There are reasons for this: algebraic number domains are very slow
especially for the case when there are many extensions and algorithms for
automatic discovery of extensions aren't very smart.
btw. What you show is not a syntax for specifying extensions, this is:
In [1]: Poly(t**2 + I, t, extension=I)
Out[1]: Poly(t**2 + I, t, domain='QQ<I>')
In [2]: Poly(t**2 + I, t, extension=True)
Out[2]: Poly(t**2 + I, t, domain='QQ<I>')
In [3]: Poly(t**2 + I, t, domain='QQ<I>')
Out[3]: Poly(t**2 + I, t, domain='QQ<I>')
In [4]: _.get_domain().is_Algebraic
Out[4]: True
In [5]: Poly(t**2 + I, t, domain='ZZ[I]')
Out[5]: Poly(t**2 + I, t, domain='ZZ[I]')
In [6]: _.get_domain().is_Algebraic
Out[6]: False
ZZ[I] is a polynomial in I, not an algebraic number field (ZZ<I> will give
you an error).
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