Earlier today I gave an exam on the theory of Gröbner bases and I could not solve the last bonus question. Here it is:
Find the universal Gröbner basis for the ideal $I = \left<x - y^nz \,\, | \,\, n \in \mathbb{N}\right> \subset \mathbb{Q}[x, y, z]$.
Since the concept of a universal Gröbner basis is not very well known, here is the definition:
Definition. The universal Gröbner basis of an ideal I is the union of all reduced Gröbner bases $G_<$ of the ideal $I$ as $<$ runs over all term orders and is denoted by $\mathscr{U}_I$.
I thought about this problem together with some friends for a while now but we could not find a clever idea to begin with. I know that the solutions to such exercices can be very long. I'm not looking for an entire solution, only the appropriate method/steps to solve it, i.e. how to correctly think about this problem.
In general, finding universal Gröbner bases is a hard problem. As far as I know, we only have algorithms for specific classes of ideals, e.g. (certain families of) toric ideals. However, the ideal $I = \langle x-z, yz-z \rangle$ in question here is simple enough that we can compute its universal Gröbner basis $\mathscr{U}_I$ by hand without too much effort. In particular, no clever idea is used: we take a naive approach, and it happens to work.
First, let's compute some reduced Gröbner bases of $I$ to get an idea what $\mathscr{U}_I$ looks like. Taking $<_1$ to be the lexicographic order with $x > y > z$ and $<_2$ to be the lexicographic order with $z > x > y$, we get \begin{equation*} G_{<_1} = \{x-z,yz-z\} \quad\text{and}\quad G_{<_2} = \{z-x,xy-x\}. \end{equation*} In other words, $G = \{x-z,z-x,xy-x,yz-z\} \subseteq \mathscr{U}_I$. Trying other term orders, we always get one of the two reduced bases above, which suggests that the inclusion $G \subseteq \mathscr{U}_I$ is perhaps an equality after all.
The goal now is to prove that $G$ is a Gröbner basis of $I$ for any term order $<$. Indeed, if this is true, then the reduced Gröbner basis of $I$ with respect to $<$ is either $G_{<_1}$ or $G_{<_2}$ (depending on whether $x > z$ or $x < z$), from which it follows that $G = \mathscr{U}_I$. Naturally, our proof will rely on Buchberger's criterion:
Theorem (Buchberger's criterion). Let $I$ be an ideal of $k[x_1,\dots,x_n]$. Then a basis $G = \{g_1,\dots,g_s\}$ of $I$ is a Gröbner basis if and only if $S(g_i,g_j) \rightarrow_G 0$ for all $i \neq j$.
Here, the notation $f \rightarrow_G 0$ means that $f$ has a standard representation with respect to $G$, i.e. we can write \begin{equation*} f = h_1g_1 + \dots + h_sg_s, \end{equation*} where $\deg(f) \geq \deg(h_ig_i)$ whenever $h_ig_i \neq 0$.
Let's get to work. From now on, we will denote by $g_1,g_2,g_3,g_4$ the polynomials $x-z,z-x,xy-y,yz-z$. We want to show that $S(g_i,g_j) \rightarrow_G 0$ for all $1 \leq i < j \leq 4$. Notice that \begin{equation*} S(g_1,g_2) = S(g_3,g_4) = 0,\quad S(g_1,g_3) = S(g_2,g_3) \quad\text{and}\quad S(g_1,g_4) = S(g_2,g_4), \end{equation*} so we only need to consider $S(g_1,g_3)$ and $S(g_1,g_4)$. There are two cases to consider, depending on whether $x > z$ or $x < z$. In each case, we compute the resulting $S$-polynomials and find standard representations for them. All of this is summarized in the following table:
\begin{array}{|c|c|c|c|c|} \hline \text{Properties of } < & S(g_1,g_3) & S(g_1,g_3) \rightarrow_G 0 & S(g_1,g_4) & S(g_1,g_4) \rightarrow_G 0 \\ \hline x > z & x-yz & (x-z)-(yz-z) & xz-yz^2 & z(x-z)-z(yz-z) \\ \hline x < z & xz-x^2y & x(z-x)-x(xy-x) & z-xy & (z-x)-(xy-x) \\ \hline \end{array}
For example, if $x > z$ then $S(g_1,g_3) = g_1-g_4$ is a standard representation since \begin{equation*} \deg(x-yz) = \max\{\deg(x),\deg(yz)\} = \max\{\deg(x-z),\deg(yz-z)\}. \end{equation*} Therefore, we conclude from Buchberger's criterion that $G$ is a Gröbner basis of $I$ for the term order $<$, as desired.
Remark. In the definition of $I$, it was assumed that $0 \in \mathbb{N}$. If $I$ was meant to be $\langle x-yz, y^2z-yz \rangle$ instead, then the same approach works and yields \begin{equation*} \{x-yz,yz-x,xy-x,x^2-xz,xz-x^2,y^2z-yz\} \end{equation*} as universal Gröbner basis for $I$.