## Is there a straightforward way to solve unmixed, homogeneous systems of polynomials?

I came across this problem in my research. It might just be an easy algebraic geometry question, but I don't know much algebraic geometry.

Suppose we have a system of $k\leq n$ polynomials in $\mathbb{C}[x_1,\dots x_n]$, each of the form $c_1 x_1^a + \cdots + c_n x_n^a = 0$ for some fixed integer $a$. I can solve this system by substituting $y_i := x_i^a$, performing some linear algebra to solve for the $y_i$, and then substituting back to get the $x_i$ (taking care that each $y_i$ will produce $a$ different values of $x_i$).

The problem I'm curious about is what happens when we have a system like above where each polynomial is homogeneous and has no cross terms, but where now the degrees of the polynomials are not all the same. i.e. each equation is of the form $c_1 x_1^{a_j} + \cdots + c_n x_n^{a_j} = 0$ where the $a_j$ are possibly distinct integers, for $j=1,\dots, k$. Is there a straightforward way to solve such a system? And if so, is there a straightforward way to solve this over the algebraic torus $(\mathbb{C}\setminus 0)^n$?

Thanks for any help anyone can provide. I'm new at this overflow stuff.

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I'd look up the Gröbner basis: en.wikipedia.org/wiki/Gr%C3%B6bner_basis and for relatively painless a.g. background see the book Ideals, Varieties and Algorithms by Cox, Little and O'Shea. – jc Sep 21 2010 at 5:22
I doubt this is much simpler than solving a general system of polynomial equations. You might want to look at "Solving Systems of Polynomial Equations", by Sturmfels. – David Speyer Sep 21 2010 at 12:14