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Solving polynomial equations with Groebner basis in $\mathbb{R}$

asked 2016-05-11 12:44:15 +0200

KittyL gravatar image

I am trying to solve the following system of polynomial equations: $$x^2+y^2+z^2=4\ x^2+2y^2=5\ xz=1$$

I used the following command to get a Groebner basis

P.<x,y,z> = PolynomialRing(QQ,order='lex')
PList = [x^2+y^2+z^2-4, x^2+2*y^2-5, x*z-1]
I = ideal(PList)
B = I.groebner_basis(); B
[x + 2*z^3 - 3*z, y^2 - z^2 - 1, z^4 - 3/2*z^2 + 1/2]

Now I want to use B.subs() to plug the solutions for $z$ to solve for $x,y$. It worked well for $z=\pm 1$.

[x - 1, y^2 - 2, 0]

but not for the $1/\sqrt{2}$ since this is in polynomial ring $\mathbb{Q}[x,y,z]$. If I change the original QQ to RR, the numbers become all decimals.

[x - 1.41421356237310, y^2 - 1.50000000000000, 0]

How can I still get exact solutions when using RR? I know I can compute this example by hands, but I'd like to know how to use Sage to solve this kind of problems.

Thank you for any help!

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answered 2016-05-11 16:51:27 +0200

tmonteil gravatar image

You can change the ring over which your polynomial ring is defined, to be the field of algebraic numbers (note the QQbar instead of QQ in the first line):

sage: P.<x,y,z> = PolynomialRing(QQbar, order='lex')
sage: PList = [x^2+y^2+z^2-4, x^2+2*y^2-5, x*z-1]
sage: I = ideal(PList)
sage: B = I.groebner_basis(); B
verbose 0 (3369:, groebner_basis) Warning: falling back to very slow toy implementation.
[x + 2*z^3 + (-3)*z, y^2 - z^2 - 1, z^4 + (-3/2)*z^2 + 1/2]


sage: B.subs(z=1)
[x - 1, y^2 - 2, 0]

sage: B.subs(z=1/sqrt(2))
[x - 1.414213562373095?, y^2 - 3/2, 0]
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Thank you for your reply! That looks much better! Is there a way to get $\sqrt{2}$ instead of $x-1.414....$? Sorry I don't have enough points to upvote.

KittyL gravatar imageKittyL ( 2016-05-11 17:48:19 +0200 )edit

Note that 1.414213562373095? is only a screen representation, but it is a genuine algebraic number, not a floating-point approximation. You can see this because of the question mark at the end of the number. You can check it as follows:

sage: c = B.subs(z=1/sqrt(2))[0].constant_coefficient()
sage: c
sage: c.parent()
Algebraic Field
sage: c^2 == 2
sage: c.minpoly()
x^2 - 2

Not every algebraic number has a representation with radicals such as "sqrt(2)", but for the algebraic numbers that do, there are plans to implement such a representation, but it is not done yet.

tmonteil gravatar imagetmonteil ( 2016-05-11 18:29:51 +0200 )edit

Thank you! That is very helpful!

KittyL gravatar imageKittyL ( 2016-05-11 20:39:08 +0200 )edit

answered 2016-05-11 14:55:18 +0200

vdelecroix gravatar image

You should not use RR which stands for floating point approximations of real numbers.

The simples to get exact solutions is to use QQbar (= algebraic closure of QQ)

sage: sqrt2 = QQbar(2) ^ (1/2)
sage: sqrt2
sage: sqrt2.n(100)
sage: sqrt2.minpoly()
x^2 - 2

But mixing it with groebner bases I got

sage: B.subs(z=QQbar(2).sqrt())
Traceback (most recent call last):
ValueError: Cannot coerce irrational Algebraic Real 1.414213562373095? to Rational
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Thank you for your reply! I can actually use $z=QQbar(2)^{1/2}$. But I wish they would give the answer $\sqrt{2}$.

KittyL gravatar imageKittyL ( 2016-05-11 17:50:21 +0200 )edit

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Asked: 2016-05-11 12:44:15 +0200

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Last updated: May 11 '16