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How do i solve a 2 variable polynomial over 1 variable

asked 2020-05-26 18:23:29 +0100

JGC gravatar image

So i have a polynomial over 2 variables:

R.<X, Y> = GF(8009)[]
Phi = -X^2*Y^2 + X^3 + 1488*X^2*Y + 1488*X*Y^2 + Y^3 \
    - 162000*X^2 + 40773375*X*Y - 162000*Y^2 \
    + 8748000000*X + 8748000000*Y - 157464000000000

I want to know for what Y values the polynomial has a solution with X = 33. I've tried using the solve method:

Phi(33, Y).roots(Y)

But this yields the error

AttributeError: 'sage.rings.polynomial.multi_polynomial_libsingular.MPolynomial_libsingular' object has no attribute 'roots'

I also tried the following:

Phi(33).roots()

It yields the error

TypeError: number of arguments does not match number of variables in parent

So what is the correct way to do this?

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3 answers in a row, yeah !

tmonteil gravatar imagetmonteil ( 2020-05-26 18:36:43 +0100 )edit

3 Answers

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answered 2020-05-26 18:34:19 +0100

Sébastien gravatar image

updated 2020-05-26 18:37:45 +0100

The method univariate_polynomial of a multivariate polynomial object depending on a single variable returns a polynomial in the appropriate univariate polynomial ring:

sage: Phi(X=33).univariate_polynomial().parent()
Univariate Polynomial Ring in Y over Finite Field of size 8009

Then:

sage: Phi(X=33).univariate_polynomial().roots()
[(898, 1)]
sage: Phi(X=33).univariate_polynomial().roots(multiplicities=False)
[898]
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answered 2020-05-26 18:35:47 +0100

rburing gravatar image

The polynomial Phi can only be evaluated at two inputs (using the function call syntax). The method roots is only available for single variable polynomials. Your Phi(33,Y) is (in SageMath) still a multivariable polynomial (in which X does not appear). To convert it to a single variable polynomial in Y, use the univariate_polynomial method:

sage: f = Phi.subs(X=33).univariate_polynomial(); f
Y^3 + 6150*Y^2 + 5541*Y + 1175
sage: f.parent()
Univariate Polynomial Ring in Y over Finite Field of size 8009
sage: f.roots()
[(898, 1)]
sage: f.roots(multiplicities=False)
[898]

Alternatively:

sage: R.ideal([Phi, X-33]).variety()
[{Y: 898, X: 33}]
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answered 2020-05-26 18:34:26 +0100

tmonteil gravatar image

This is because Phi(33, Y) is still a polynomial in two variables:

sage: Phi(33, Y).parent()
Multivariate Polynomial Ring in X, Y over Finite Field of size 8009

So, you have to turn it into a one-variable polynomial first:

sage: S.<Y> = GF(8009)[]
sage: S(Phi(33, Y))
Y^3 + 6150*Y^2 + 5541*Y + 1175

sage: S(Phi(33, Y)).parent()
Univariate Polynomial Ring in Y over Finite Field of size 8009

sage: S(Phi(33, Y)).roots()
[(898, 1)]
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Asked: 2020-05-26 18:23:29 +0100

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Last updated: May 26 '20