I'm studying Gröbner bases and am using the following definitions and results:
- Let $I$ be an ideal in $R = K[x_1,…,x_n]$ for some field $K$ and some $n>0$ and fix a monomial ordering on $R$. A finite set of
polynomials $\{g_1, . . . , g_s\} \in I$ is a Gröbner basis of $I$ (with respect to the given monomial
order) if the initial ideal of I is generated by the initial terms of the polynomials $g_i$. - Let $G$ be a Gröbner basis of an ideal $I\subset R$. For any polynomial $p\in R$, there exists a unique polynomial $\overline{p}\in R$ such that $p-\overline{p}\in I$ and only standard monomials (i.e. monomials that don't belong to the initial ideal of $I$) in $\overline{p}$. We call $\overline{p}$ the normal form of $p$.
- Given two polynomials $g,g'$ with initial terms $m$ and $m'$ respectively. We define their $S$-polynomial as
$$S(g,g') = \frac{\text{lcm}(m,m')}{m}g – \frac{\text{lcm}(m,m')}{m'}g'.$$ - Buchberger's criterion: A set of polynomials $G$ is a Gröbner basis for $I$ is and only if every $S$ polynomial formed by pairs $g,g'$ in $G$ has normal form zero w.r.t. $G$.
Now I'm struggling with the following question:
Find an ideal in two variables whose Gröbner basis with respect to the
graded lexicographic order consists of 3 polynomials with exactly 13 common solutions
(counting them with multiplicity).
I am able to find 3 polynomials that have exactly $13$ common solutions, namely
$$ f = (x-\alpha_1)\cdot…\cdot(x-\alpha_{13}),\ g = (y-\alpha_1)\cdot…\cdot (y-\alpha_{13}),\ h = x-y.$$
Here $\alpha_i$ just denote some real numbers.
These however don't form a Gröbner basis, because I think that $S(f,g)$ has a non-zero normal form.
Does anyone have an idea how to find correct polynomials?
Best Answer
The simplest systems of polynomial equations are the ones that can be solved using back-substitution. Here is one way to contruct such a system with exactly $m \geq 1$ common solutions: pick your favorite polynomial \begin{equation*} f(y) = (y-\alpha_1)\cdot\dots\cdot(y-\alpha_{m-1}) \end{equation*} with $\alpha_1,\dots,\alpha_{m-1} \neq 0$ all distinct and consider the system $V(G)$, where $G = \{x,yf(y)\}$. It follows from Buchberger's criterion that $G$ is a Gröbner basis since $S(x,yf(y))$ is divisible by $x$. Therefore, $I = \langle x,yf(y) \rangle$ is an ideal in $K[x,y]$ with Gröbner basis $G$ such that $V(G) = \{(0,0),(0,\alpha_1),\dots,(0,\alpha_{m-1})\}$ has exactly $m$ solutions. The only problem is that $G$ has $2$ elements instead of $3$, but this can be easily remedied: just add any other polynomial in $I$ to $G$, e.g. consider $G' = \{x,yf(y),xy\}$. Then $G'$ is a Gröbner basis for $I$ consisting of $3$ polynomials with exactly $m$ common solutions, as desired.
There is still something unsatisfactory with the solution presented above: the Gröbner basis $G'$ is not reduced. In other words, we would like to answer the question:
One idea to achieve this is to enlarge $G$ by adding a polynomial that is not in $I$ but that vanishes at all the $m$ common solutions of $G$. Actually, it helps to first modify $G$ to $\{x^2,y^2f(y)\}$. We may then consider the basis $G' = \{x^2,xy,y^2f(y)\}$ and the ideal $I = \langle x^2,xy,y^2f(y) \rangle$ it generates. Since $S(x^2,xy) = 0$, $S(x^2,y^2f(y))$ is divisible by $x^2$ and $S(xy,y^2f(y))$ is divisible by $xy$, it follows from Buchberger's criterion that $G'$ is a Gröbner basis of $I$. Moreover, note that $G'$ is reduced and consists of $3$ polynomials with exactly $m$ common solutions.
Finally, the same argument can be used to prove that the set \begin{equation*} \{x^{n-1},x^{n-2}y,\dots,xy^{n-2},y^{n-1}f(y)\} \end{equation*} is a reduced Gröbner basis for the ideal it generates which consists of $n \geq 2$ polynomials with exactly $m \geq 1$ common solutions.