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Suppose that

\begin{align*} x&=t+t^{-1}+t^2s+t^{-2}s^{-1}+ts^{-1}+t^{-1}s-6\\ y&=t+t^{-2}+ts+s^{-1}-4\\ z&=t^{-1}+t^2+t^{-1}s^{-1}+s-4 \end{align*}

Find a polynomial $P(x, y, z)=0$ relating $x$, $y$ and $z$.

In general, if we are given \begin{align*} x_1&=P_1(t_1, \cdots, t_m)\\ \vdots&\\ x_n&=P_n(t_1, \cdots, t_m) \end{align*} where $P_1, \cdots, P_n$ are polynomials, is there a general way we can find polynomial relations of $x_1, \cdots, x_n$? I know that when $n=2$ and $m=1$, a polynomial relation is given by the resultant of $P_1(t_1)-x_1$ and $P_2(t_1)-x_2$.

No_way
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  • Well, you could try to homogenise the expressions for $x,y,$ and $z$ and compute their multivariate resultant... – A.P. Jul 17 '15 at 15:42
  • Could you please point out a reference on multivariate resultant? I don't know what that is. Thanks. – No_way Jul 17 '15 at 15:43
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    Have a look at the book by Cox, Little, and O' Shea titled Using Algebraic Geometry, chapter 3. (note: I just found out about this with a quick search, so I'm no expert, but it does sound like what you are looking for) – A.P. Jul 17 '15 at 15:50

2 Answers2

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As suggested in the comments, Cox, Little, and O'Shea have written about how to solve the implicitization problem, but in their first book Ideals, Varieties, and Algorithms. See my answer here for a reference to the result used below. (I think we could alternatively use multivariable resultants rather than Gröbner bases.)

Let $u = s^{-1}$ and $v = t^{-1}$. We form the ideal \begin{align*} I = (su - 1, tv - 1, x &- (t + v + t^2 s + v^2 u + t u + v s - 6),\\ y &- (t + v^2 + t s + u - 4), z - (v + t^2 + v u + s - 4)) \end{align*} in $\mathbb{Q}[s,t,u,v,x,y,z])$, and compute a Gröbner basis $G$ with respect to the lexicographical order. The last entry of $G$ is a Gröbner basis for the elimination ideal $I \cap \mathbb{Q}[x,y,z]$, and is the polynomial $P(x,y,z)$ you're looking for. Computing using SageMath, we find \begin{align*} P(x,y,z) = x^{5} + 2 x^{4} y + 2 x^{4} z + 50 x^{4} - x^{3} y^{2} z - 7 x^{3} y^{2} - x^{3} y z^{2} - 17 x^{3} y z - 25 x^{3} y - 7 x^{3} z^{2} - 25 x^{3} z + 625 x^{3} + x^{2} y^{4} - 2 x^{2} y^{3} z + x^{2} y^{3} - 49 x^{2} y^{2} z - 245 x^{2} y^{2} - 2 x^{2} y z^{3} - 49 x^{2} y z^{2} - 610 x^{2} y z - 1875 x^{2} y + x^{2} z^{4} + x^{2} z^{3} - 245 x^{2} z^{2} - 1875 x^{2} z + 2 x y^{5} - x y^{4} z + 48 x y^{4} + x y^{3} z^{3} + 19 x y^{3} z^{2} + 66 x y^{3} z + 455 x y^{3} + 19 x y^{2} z^{3} + 336 x y^{2} z^{2} + 1245 x y^{2} z + 1875 x y^{2} - x y z^{4} + 66 x y z^{3} + 1245 x y z^{2} + 3750 x y z + 2 x z^{5} + 48 x z^{4} + 455 x z^{3} + 1875 x z^{2} + y^{6} - y^{5} z^{2} - 10 y^{5} z + 14 y^{5} - 23 y^{4} z^{2} - 237 y^{4} z - 19 y^{4} + 8 y^{3} z^{3} - 73 y^{3} z^{2} - 1564 y^{3} z - 625 y^{3} - y^{2} z^{5} - 23 y^{2} z^{4} - 73 y^{2} z^{3} + 366 y^{2} z^{2} - 1875 y^{2} z - 10 y z^{5} - 237 y z^{4} - 1564 y z^{3} - 1875 y z^{2} + z^{6} + 14 z^{5} - 19 z^{4} - 625 z^{3} \, . \end{align*}

Viktor Vaughn
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  • Thanks so much for answering this question which has laid idle for many years! Just one more question: does the Gröbner basis algorithm or multivariate resultant give the polynomial relation of minimal degree? Sorry if this is answered in Cox-Little-O'Shea, but I do not have their book. – No_way Sep 28 '21 at 21:53
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    For Gröbner bases, the answer is yes. The polynomial $P$ you get from this method is a generator for the elimination ideal, so any other polynomial relation must be a multiple of $P$. I don't think this is necessarily true for resultants. The resultant will be an element of the elimination ideal, but might not generate the whole ideal. – Viktor Vaughn Sep 28 '21 at 23:38
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If the Jacobian

$\frac{d(x_1, \cdots, x_n)}{d(t_1, \cdots, t_m) }$

is non-zero there is no relationship like:

$ P(x, y, z)=0 $.

In fact, the system

\begin{align*} x_1&=P_1(t_1, \cdots, t_m)\\ \vdots&\\ x_n&=P_n(t_1, \cdots, t_m) \end{align*}

is invertible;

we can consider the $t_1, \cdots, t_m $ as dependent variables and the $x_1, \cdots, x_n$ as independent.

We can give $x_1, \cdots, x_n$ any values and this shows that there is no relation like:

$ P(x, y, z)=0 $.