Show that a sequence of polynomials cannot be a Gröbner basis wrt. any term ordering

Question

Consider the ideal $I=\left\langle x^{2}+y^{2}, x^{3}+y^{3}\right\rangle \subseteq \mathbb{Q}[X, Y]$ and the basis $G=(x^2+y^2,x^3+y^3)$ with lexicographic ordering $x\geq y$. Consider two $S$-polynomials given by:

\begin{equation} \begin{aligned} &S_{1}=S\left(x^{2}+y^{2}, x^{3}+y^{3}\right)=\frac{x^{3}}{x^{2}}\left(x^{2}+y^{2}\right)-\frac{x^{3}}{x^{3}}\left(x^{3}+y^{3}\right) \\ &=x^{3}+x y^{2}-x^{3}-y^{3}=x y^{2}-y^{3}=x\left(x^{2}+y^{2}\right)-1 \cdot\left(x^{3}+y^{3}\right) \in I \end{aligned} \end{equation}

and \begin{equation} \begin{aligned} &S_{2}=S\left(x^{2}+y^{2}, x y^{2}-y^{3}\right)=\frac{x^{2} y^{2}}{x^{2}}\left(x^{2}+y^{2}\right)-\frac{x^{2} y^{2}}{x y^{2}}\left(x y^{2}-y^{3}\right) \\ &=x^{2} y^{2}+y^{4}-x^{2} y^{2}-x y^{3}=-x y^{3}+y^{4}=-x y\left(x^{2}+y^{2}\right)+y\left(x^{3}+y^{3}\right) \in I \end{aligned} \end{equation}

1. Use this to show that $y^4\in I$
2. Show that $G$ cannot be a Gröbner basis for I with any term ordering (It may be useful that I have already shown $F= \left(y^{4}, x y^{2}-y^{3}, x^{2}+y^{2}\right)$ is a reduced Gröbner basis for I) wrt. $x\geq y$).

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1. I wanted to express $y^4$ in terms of $S_1,S_2$ somehow and use the properties of an ideal. But I just can't seem to find an appropriate expression. I got this far $y^{4}=S_{2}+x y^{3}=S_{1}+S_{2}+x y^{3}-x y^{2}+y^{3}$

2. I'm really confused about this problem. Since I have expressed $S_1$ as a "linear" combination of the elements in G (so no residual), haven't I shown that $S_1$ reduces to zero wrt. G? But I have a theorem that states:

Buchberger theorem: A sequence $F=\left(f_{1}, \ldots, f_{m}\right)$ of polynomials is a Gröbner basis if and only if $S\left(f_{i}, f_{j}\right) \rightarrow_{F} 0$ for $1 \leq i<j \leq m$.

2. So I would argue that G is a Gröbner basis from this theorem. However, this is of course a contradiction to F being a reduced Gröbner basis (also a minimal). What am I misunderstanding? And how do I approach this problem, since there can be an infinite number of term orderings?

Answer 1

1. You found $S_1 = xy^2-y^3$ correctly, but you lost a sign in your calculation of $S_2$: it should be $S_2=xy^3 + y^4$.

   Then we clearly have $y^4 = \frac{1}{2}(S_2 - yS_1) \in I$.

   Note also that $S(S_2,S_1) = S_2 - yS_1 = 2y^4$.

2. You have expressed the $S_i$ as linear combinations with polynomial coefficients of the elements in $G$. This does not prove that the $S_i$ reduce to zero with respect to $G$, it only proves that the $S_i$ belong to $I$. The expression $S_i = q_{i,1} \cdot g_1 + q_{i,2} \cdot g_2$ with $q_{i,1},q_{i,2} \in \mathbb{Q}[x,y]$ has the same shape as you would get from a multivariate polynomial division of $S_i$ by $G$ with zero remainder, but you have not actually done the multivariate polynomial division of $S_i$ by $G$. The actual multivariate polynomial division of $S_i$ by $G$ proceeds by an algorithm, and in general the result (the quotients and remainder) depend on the monomials in $S_i$, the chosen monomial ordering, and the choices made in the algorithm (e.g. depending on the ordering of $G$).

From your calculation of a Gröbner basis $F$ of $I$ it follows that $\mathbb{Q}[x,y]/I$ is a $6$-dimensional $\mathbb{Q}$-vector space. Indeed, each nonzero equivalence class in $\mathbb{Q}[x,y]/I$ has a representative (namely the remainder after multivariate division by $F$ of an arbitrary representative) such that none of its monomials belong to $LM_{\text{lex}}(I) = \{ LM_{\text{lex}}(f) \mid f \in I, f \neq 0 \}$, where $LM_{\text{lex}}(f)$ denotes the leading monomial of $f \neq 0$ with respect to $\text{lex}$. Since $F=\{ f_1 = y^4, f_2 = xy^2 - y^3, f_3 = x^2+y^2 \}$ is a $\text{lex}$ Gröbner basis, we have $LM_{\text{lex}}(I) = \langle LM_{\text{lex}}(f_1), LM_{\text{lex}}(f_2), LM_{\text{lex}}(f_3) \rangle = \langle y^4, xy^2, x^2 \rangle$, and it follows that $\mathbb{Q}[x,y]/I$ has a basis consisting of the six monomials $1, x, y, xy, y^2, y^3$ which are not in $LM_{\text{lex}}(I)$.

Suppose $G$ is a Gröbner basis with respect to a monomial ordering $>$. Then $\mathbb{Q}[x,y]/I$ has a basis (as a $\mathbb{Q}$-vector space) consisting of monomials which are not in $LM_{>}(I) = \langle LM_{>}(g_1), LM_{>}(g_2) \rangle$. There are a priori four possibilities for $\langle LM_{>}(g_1), LM_{>}(g_2) \rangle$, namely $\langle x^2, x^3 \rangle$, $\langle x^2, y^3 \rangle$, $\langle y^2, x^3 \rangle$, and $\langle y^2, y^3 \rangle$. Clearly $\langle x^2, x^3 \rangle$ and $\langle y^2, y^3 \rangle$ are not possible, because it would mean that the quotient $\mathbb{Q}[x,y]/I$ would be an infinite-dimensional $\mathbb{Q}$-vector space. Suppose $LM_{>}(g_1) = x^2$ and $LM_{>}(g_2)=y^3$, i.e. $x^2 > y^2$ and $y^3 > x^3$. Then since the monomial ordering respects multiplication, we would have $y^4 > x^3y > xy^3 > x^4 > x^2y^2 > y^4$, which is a contradiction. The remaining case $\langle y^2, x^3 \rangle$ is handled by swapping $x$ and $y$. So $G$ is not a Gröbner basis with respect to any monomial ordering.