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Factoring bivariate polynomials w.r.t. a single variable

asked 2011-06-09 02:00:21 -0500

Harry gravatar image

updated 2015-01-14 09:27:17 -0500

FrédéricC gravatar image

the function factor() works in this fashion:

sage: x, y = PolynomialRing(GF(17), 2, ['x','y']).gens()

sage: f =  9*y^6 - 9*x^2*y^5 - 18*x^3*y^4 - 9*x^5*y^4 + 9*x^6*y^2 + 9*x^7*y^3 + 18*x^8*y^2 - 9*x^11

sage: f.factor()

(-9) * (x^5 - y^2) * (x^6 - 2*x^3*y^2 - x^2*y^3 + y^4)

Is there a possibility to factorize a bivariate polynomial in x,y with respect to a single variable only (e.g. y) and get the answer in the form of (y - f(x)) as factors?

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Comments

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Surely there is no essential difference between the factorization in $F[x,y]$ which you see and the one in $F(x)[y]$ which you want, by Gauss's Lemma? The fact that none of the displayed factors is linear in $y$ just means that $f$ has no roots in $F(x)$ when you consider it as a polynomial in $F(x)[y]$. To get roots (and hence linear factors) you would need to enlarge $F(x)$ to its algebraic closure.

John Cremona gravatar imageJohn Cremona ( 2011-08-19 03:24:11 -0500 )edit

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answered 2015-01-12 09:47:00 -0500

MvG gravatar image

Try this:

sage: g = f.polynomial(y)
sage: g.change_ring(g.base_ring().fraction_field()).factor()
(9) * (y^2 + 16*x^5) * (y^4 + 16*x^2*y^3 + 15*x^3*y^2 + x^6)

I'm not sure whether that's any better than your original

sage: f.factor()
(-8) * (-x^5 + y^2) * (x^6 - 2*x^3*y^2 - x^2*y^3 + y^4)

since as John already indicated, there is no fundamental difference.

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answered 2011-06-21 16:57:50 -0500

radokirov gravatar image

This should do it

R = PolynomialRing(GF(17), 'x')
x = R.gens()[0]
y = PolynomialRing(R, 'y').gens()[0]
f =  9*y^6 - 9*x^2*y^5 - 18*x^3*y^4 - 9*x^5*y^4 + 9*x^6*y^2 + 9*x^7*y^3 + 18*x^8*y^2 - 9*x^11
f.factor()

Unfortunately, it seems that the functionality to factor over a generic polynomial ring is not implemented.

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Asked: 2011-06-09 02:00:21 -0500

Seen: 207 times

Last updated: Jan 12 '15