This is exercise 7 from Ideals, Varieties, and Algorithms by Cox, Little, and Shea.
If we use grlex order with $x > y> z$, is $\{x^4 y^2 – z^5, x^3 y^3 – 1, x^2 y^4 – 2z\}$ a grobner basis for the ideal generated by these polynomials?
Definition: Given a non-zero ideal $I \subseteq k[x_1, \cdots, k_n]$ and a monomial ordering on $k[x_1, \cdots, x_n]$, a set $G = \{g_1, \cdots, g_t\} \subseteq I$ is called a Grobner basis (or standard basis) for $I$ if $\langle LT(g_1), \cdots, LT(g_t)\rangle = \langle LT(I) \rangle$.
I am specifically confused on how to write $\langle LT(I) \rangle$.
By glex order, I write $\langle LT(g_1), \cdots, LT(g_t)\rangle = \langle LT(I) \rangle = \langle x^4 y^2, x^3 y^3, x^2y^4\rangle$. I must check that this is a subset of $\langle LT(I) \rangle$ and vice-versa.
But how do I write $\langle LT(I) \rangle$?
Best Answer
Just for clarity, the exercise is in section 2.5 of the book.
The inclusion $\langle LT(g_1),\ldots,LT(g_t)\rangle \subset \langle LT(I) \rangle$ always holds, as $\{g_1,\ldots,g_t\} \subset I$ and $\langle LT(I) \rangle$ is the ideal generated by elements of $LT(I) = \{LT(f) \mid f \in I, f \neq 0\}$. You have correctly identified the leading terms in the left-hand side. The question is whether there exists nonzero $f \in I$ such that $LT(f) \not\in \langle LT(g_1),\ldots,LT(g_t) \rangle$.
That is, can you make a linear combination (with polynomial coefficients) of $g_1,\ldots,g_t$ such that the result has its leading term not in $\langle LT(g_1),\ldots,LT(g_t) \rangle$? A natural idea is to make a linear combination such that the "would-be" leading term (looking at the leading terms of the summands) would cancel, so that you get a "possibly new" leading term. In this particular case we have e.g. $LT(g_1) = x^4y^2$ and $LT(g_2) = x^3y^3$, so we can multiply $g_1$ by $y$ and $g_2$ by $x$ to get the same leading terms $LT(yg_1) = x^4y^3$ and $LT(xg_2) = x^4y^3$, and now we consider the linear combination $f=yg_1 - xg_2 \in I$. In fact $f=yg_1 - xg_2 = y(x^4y^2 - z^5) - x(x^3y^3-1) = -yz^5 + x$, so its leading term is $LT(f) = -yz^5$, and this is not contained in the ideal $\langle x^4y^2, x^3y^3,x^2y^4 \rangle$. Recall that checking membership of a monomial in a monomial ideal is easy; see Lemma 2 of section 2.4 in the book. So we have found a nonzero $f \in I$ such that $LT(f) \not\in \langle LT(g_1),LT(g_2),LT(g_3) \rangle$, and hence $\{g_1,g_2,g_3\}$ is not a Groebner basis.
This heuristic idea leads to the definition of the S-polynomial later in the book (the next paragraph 2.6), and from there to Buchberger's criterion and Buchberger's algorithm.