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Consider the equations, $$x^{2}+y^{2}+z^{2}-xt-t^{2}=0 \tag{1}$$ $$x^{2}+y^{2}+z^{2}+yt-2t^{2}=0 \tag{2}$$

Clearly, both equations are homogeneous. Solve for $t$ from the above equations. You will get $t=x+y$. Now, if you replace $t$ with $x+y$ in the above equations, you will get another second-degree equation, given by $$x^{2}-z^{2}+3xy=0 \tag{3}$$

As we can see, the third equation is also homogeneous. So why is homogeneity preserved even when we change variables?

dxiv
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Sasikuttan
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1 Answers1

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Given homogeneous polynomials equations $A(x,y,z,t)=0, B(x,y,z,t)=0$ it is possible to eliminate variable $z$ between them, resulting in an algebraic equation $F(x,y,t)=0$. Polynomial $F$ is generally the polynomial resultant $R = \operatorname{res}_z(A,B)$, which is a homogeneous polynomial because of the homogeneity properties of the resultant, specifically:

If $P$ and $Q$ are homogeneous multivariate polynomials of respective degrees $d$ and $e$, then their resultant in degrees $d$ and $e$ with respect to an indeterminate $x$, denoted $\operatorname{res}_{x}^{d,e}(P,Q)$ in § Notation, is homogeneous of degree $de$ in the other indeterminates.

Proving the property is fairly straightforward.

  • Dehomogenize the equations by defining $x'=\dfrac{x}{t}, y'=\dfrac{y}{t}, z'=\dfrac{z}{t}$, $A'(x',y',z') = A(x,y,z,1)$, $B'(x',y',z')=B(x,y,z,1)$.

  • Eliminate $z'$ between $A'(x',y',z')=0$ and $B'(x',y',z')=0$ by calculating the resultant $R'(x',y') = \operatorname{res}_{z'}\left(A'(x',y',z'),B'(x',y',z')\right)$ of degree $n \le \deg A' \cdot \deg B'$.

  • Homogenize back with $R(x,y,t) = t^n \,R' \left(\dfrac{x}{t}, \dfrac{y}{t}\right)\,$. By construction, $R$ is homogeneous.

If the resultant $R = T^n$ is a perfect $n^{th}$ power, and using the property that factors of homogeneous polynomial are homogeneous it follows that $T$ is also homogeneous.

OP's case is $P=A, Q=B,x \mapsto z, d=e=2$, and the resultant evaluates to:

$$ R(x,y,t) = \operatorname{res}_{z}^{2,2}\big(A(x,y,z,t),B(x,y,z,t\big) = t^2 (t - x - y)^2 $$

Looking for non-trivial solutions $t \ne 0$ leaves the second factor $t-x-y=0$, which is a homogeneous polynomial equation, as expected from the general statement.

dxiv
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