Enumerating points in 0-dimensional ideals over Qbar

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asked 6 years ago

Mathieu Dutour Sikiric
21 ●1 ●2 ●4

updated 6 years ago

tmonteil
27483 ●32 ●205 ●519 http://wiki.sagemath.o...

I would like to find the points of a 0-dimensional ideal over Qbar. That is I do not want just the rational points,

The problem is that I found multiple problem while doing that.

1) While the code

R.<t1,t2,t3,e1,e2,e3> = PolynomialRing(QQbar,6, order="degrevlex(3),lex(3)")

is legal, the code

R.<t1,t2,t3> = PolynomialRing(QQbar,6, order="degrevlex(3)")

is not for reasons that esapes me.

2) The code

R.<t1,t2,t3,e1,e2,e3> = PolynomialRing(QQbar,6, order="degrevlex(3),lex(3)")
tvars = [t1,t2,t3]
eltsyms = [R((SymmetricFunctions(QQbar).elementary())[i].expand(3,alphabet=tvars)) for i in range(4)]

is not legal, it is if we replace QQbar by QQ. Why?

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answered 6 years ago

tmonteil
27483 ●32 ●205 ●519 http://wiki.sagemath.o...

updated 6 years ago

There is clearly an issue. Actully, your first example istelf has an issue:

sage: R.<t1,t2,t3,e1,e2,e3> = PolynomialRing(QQbar,6, order="degrevlex(3),lex(3)")
sage: t1
<repr(<sage.rings.polynomial.multi_polynomial_element.MPolynomial_polydict at 0x7f6cf1fb5500>) failed: TypeError: Argument 'self' has incorrect type (expected sage.rings.polynomial.polydict.ETuple, got tuple)>

As for the second example, on the left side of the equality you put 3 indeterminates, while on the right side, you require 6 indeterminates.

I have no idea where it comes from, it requires some more exploration. Thanks for reporting.

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Asked: 6 years ago

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Last updated: Sep 23 '18

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