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Consider these two polynomials:

$$f=x^2y+3xy-1$$ $$g=6x^2+y^2-4$$

I need to compute their resultant, denoted in my textbook as $h=Res(f,g,x)$.

Here's where I need help: setting up the Sylvester Matrix. I could be missing something obvious, but from the definitions I've read online, I kept getting a matrix like so:

$X =\begin{bmatrix} 1 & 0 & 0 & 6 & 0 & 0\\3 & 1 & 0 & 1 & 6 & 0\\-1 & 3 & 1 & -4 & 1 & 6\\0 & -1 & 3 & 0 & -4 & 1\\0 & 0 & -1 & 0 & 0 & -4\\\end{bmatrix}$

My understanding is that $X$ should be a square matrix- what am I missing?

Thanks in advance.

UPDATE:

Is this more like it?

$X =\begin{bmatrix} y & 0 & 6 \\-1 & y & 0 \\0 & -1 & y^2-4 \\\end{bmatrix}$

glS
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daOnlyBG
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    You are asked to compute the resultant, considering the polynomials as polynomials in the variable $x$. The coefficients aren't numbers, they are polynomials in $y$. Do it that way, you'll get a square matrix. – Gerry Myerson Dec 02 '14 at 11:40
  • @GerryMyerson thank you; do you mind looking at my update to check if that's correct? – daOnlyBG Dec 02 '14 at 11:54
  • It isn't, but Fredrik has set you straight. – Gerry Myerson Dec 02 '14 at 22:48

1 Answers1

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Write your polynomials as $$ yx^2+ (3xy)-1 $$ and $$ -6x^2+ (y^2-4) $$

As polynomials in $x$, these have degree two, so your matrix should be a $4 \times 4$ matrix. $$ \begin{pmatrix} y & 3y & -1 & 0 \\ 0 & y & 3y & -1 \\ -6 & 0 & y^2-4 & 0 \\ 0 & -6 & 0 & y^2-4 \end{pmatrix} $$

Taking the determinant gives $$ y^{6}-62 y^{4}-12 y^{3}+232 y^{2}+48 y+36 $$

This can also be computed in Macaulay2 by the commands:

i1 : R = QQ[x,y]

o1 = R

o1 : PolynomialRing

i2 : f1 = x^2*y+3*x*y-1

      2
o2 = x y + 3x*y - 1

o2 : R

i3 : f2 = -6*x^2+y^2-4

         2    2
o3 = - 6x  + y  - 4

o3 : R

i4 : resultant(f1,f2,x)

      6      4      3       2
o4 = y  - 62y  - 12y  + 232y  + 48y + 36

o4 : R
Fredrik Meyer
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