Gröbner bases: Polynomial equations. Solution $x$ to $G \cap k[x_1, .., x_i]$ imply solution to $G \cap k[x_1, .., x_i, x_{i+1}]$, $x$ plugged in.

Question

I'm have been studying Gröbner bases for a while now and seen a few examples in my textbook / exercises.

Let $\mathcal k$ be a field and $\mathcal k[x_1,..,x_n]$ a polynomial ring.

I wish to solve a system of equations $f_1, .., f_k \in \mathcal k[x_1,..,x_n]$ where $G$ a Gröbner basis for $I = <f_1, ..,f_n>$ (ideal generated by $f_1,..,f_n$)

$f_1(x_1,..,x_n) = 0, .., f_k(x_1, .. ,x_n) = 0$

In the examples / exercises all solutions to $G \cap \mathcal k[x_1,.., x_i]$ also imply that these solutions work for solving polynomial equations $G \cap \mathcal k[x_1,.., x_i, x_{i+1}]$ all the way up to $ i = n$.

I haven't experienced that a solution $x$ for $G \cap \mathcal k[x_1,.., x_i]$ make the system of the equations $G \cap \mathcal k[x_1,.., x_i, x_{i+1}]$ unsolveable, when $x$ plugged in.

So does a solution $x$ for $G \cap \mathcal k[x_1,.., x_i]$ imply that there exist $x_{i+1} \in \mathcal k$ such that $G \cap \mathcal k[x_1,.., x_i, x_{i+1}]$ is solved for $x, x_{i+1}$ ?

Please give a counter example if possible, thanks.

Answer 1

Consider $k[x_1,x_2]$, $f_1=x_1$, $f_2=x_2^2+1$. We have that $G:=\{f_1,f_2\}$ is already a Groebner basis of the ideal $\left<f_1,f_2\right>$.

We have that $G\cap k[x_1]=\{f_1\}$ and we get as solution of $f_1=0$, $x_1=0$.

But $G\cap k[x_1,x_2]=\{f_1,f_2\}=\{x_1,x_2^2+1\}$. If $k$ doesn't contain square roots of $-1$ then the equation $x_2+1=0$ is not going to have solutions.

Answer 2

For $\mathbb C$, the Extension Theorem tells us when we can extend a partial solution to a complete one.

Theorem 1 (The Extension Theorem)
Let $I=<f_1,...,f_s>\subset \mathbb C[x_1,...,x_n]$. Then for each $1\le i \le s$, write $f_i=g(x_2,...,x_n)x_1^{N_i} + $ terms in which $x_1$ has degree $< N_i$ where $N_i\ge 0$ and $g_i \in \mathbb[x_2,...,x_n]$ is nonzero. Let $I_1$ be the first elimination ideal of $I$ and suppose we have a partial solution $(a_2,...,a_n) \in V(I_1)$. If $(a_2,..,a_n) \not\in V(g_1,...,g_s)$, then $\exists a_1 \in \mathbb C s.t. (a_1,...,a_n) \in V(I)$

That is, if any of our partial solutions cause the leading coefficients of the ideal we extend to to simultaneously vanish, then we can't extend it.

The accepted answer does have a solution over the complex field, but not the reals. A better example is

$$ xy = 1, \\ xz = 1 $$

The Gröbner basis is $\{y-z, xz-1\}$, so the first elimination ideal is $y-z$ from which we get the partial solution $\forall a\in \mathbb R, y=z=a$. However, since $y$ and $z$ are the leading coefficients of our original system, the partial solution $(0,0)$ causes them to vanish simultaneously, so we cannot extend this partial solution.