Example of initial ideal

Question

I have a problem with understand some example, which I present below.

Let ideal $I = (x_1^2 + 3x_1x_2, 2x_1^2 + x_2^2)$. The initial monomial of both generators is $x_1^2$. However, twice the first generator minus the second generator, we obtain $6x_1x_2 +x_2^2$. Up to this point I understand.

But why it has initial monomial $x_1 x_2$? I though that it has initial monomial $6x_1 x_2$. Why we don't have $6$ constant?

And next I don't understand this part: No linear combination with constant coefficients of the generators has smaller initial monomial, so $in(I)_2 = (x_1^2, x_1x_2)$. Why $in(I)_2$ has above form?

Then, it is automatic that $in(I)_3$ contains $(x_1 ^3, x_1 ^2x_2, x_1x_2^2)$, but by direct computation we can find an element of $I$ which is equal to $x_2^3$, and thus has initial monomial $x_2^3$. Thus $in(I) = (x_1,x_2)^3$, and $in(I) = (x_1^2,x_1x_2,x_2^3)$.

And why $in(I)_3$ is not contain $(x_1 ^3, x_1 ^2x_2, x_1x_2^2, x_2^3)$? And why initial monomial is $x_2^3$?

Definition: If $f \in S^kV$ is a homogeneous polynomial, write $f = \sum_I a_Ix^I$. Let $I_m = max({I | a_I \neq O})$. Initial monomial of $f$ is $in(f) = x^{I_m}$. If $I \subset S$ is a homogeneous ideal, then the initial ideal of $I$ is the ideal $in(I)$ generated by ${in(f) | f \in I}$.

It comes from https://link.springer.com/content/pdf/10.1007/978-3-0346-0329-4.pdf Proposition 1.11

Answer 1

The example assumes $x_1 > x_2$ in the monomial ordering.

1. Your quoted definition of $\operatorname{in}(f) = x^{I_m}$ uses the notation $x^I = x_i^{i_1} \cdot \ldots x_n^{i_n}$ for a multi-index $I = (i_1,\ldots,x_n)$, so the result is a product of powers of $x_i$, without a constant factor in front. As noted in the comments, even if you would include the nonzero constant, it would not change the ideal $\operatorname{in}(I) = \langle \operatorname{in}(f) \mid f \in I\text{ and }f \neq 0 \rangle$.

2. To determine $\operatorname{in}(I)_2$, the degree-$2$ part of $\operatorname{in}(I)$, note that the two generators $f_1 = x_1^2 + 3x_1x_2$ and $f_2 = 2x_1^2 + x_2^2$ of $I = \langle f_1, f_2 \rangle$ are homogeneous of degree $2$, and by definition each element of $I$ is of the form $f = a_1\cdot f_1 + a_2 \cdot f_2$ where $a_1, a_2$ are polynomials. To obtain $\operatorname{in}(f)$ of degree $2$, the degree-$2$ monomial must come from one of the two parts, so $a_1$ or $a_2$ must have a nonzero constant term, and a priori the possibilities for $\operatorname{in}(f)$ are the monomials we see in $f_1,f_2$, namely $\{x_1^2, x_1x_2, x_2^2 \}$. The book explains how to obtain $x_1^2 = \operatorname{in}(f_1)$ and $x_1x_2 = \operatorname{in}(2f_1 - f_2)$. The reason $x_2^2$ cannot be obtained as $\operatorname{in}(f)$ is that a nonzero constant coefficient $c_2$ in $a_2$ (so $c_2$ is the coefficient of the monomial $x_2^2$ in $f$) implies also getting the term $2c_2 x_1^2$ in $f$, and here $x_1^2$ is bigger than $x_2^2$ in the monomial ordering, so to get rid of that term one would have to have $-2c_2$ as the constant term in $a_1$, but this would introduce another term $-6c_2x_1x_2$, where $x_1x_2$ would also be bigger than $x_2^2$ in the monomial ordering, so finally $x_2^2$ can never be the largest monomial in $f$.

3. To determine $\operatorname{in}(I)_3$, note first the property $\operatorname{in}(fg) = \operatorname{in}(f)\cdot\operatorname{in}(g)$ on the preceding page of the linked book. It follows that one can obtain elements of $\operatorname{in}(I)_3$ by multiplying elements of $\operatorname{in}(I)_2 = \{x_1^2, x_1x_2\}$ by monomials of degree $1$. (To be explicit: for example, if $\operatorname{in}(f_1)=x_1^2$, then $\operatorname{in}(x_1 f_1) = x_1^3$.) In this way you obtain $\{x_1^3,x_1^2x_2,x_1x_2^2\} \subset \operatorname{in}(I)_3$. By direct computation you obtain that using linear algebra $$(ax_1 + bx_2)f_1 + (cx_1 + dx_2)f_2 = x_2^3,$$ can be solved for constants $(a,b,c,d)$, so $x_2^3 \in \operatorname{in}(I)_3$. Thus $\operatorname{in}(I)_3 = \langle x_1^3,x_1^2x_2,x_1x_2^2,x_2^3 \rangle$.

4. Because all monomials of degree $3$ belong to $\operatorname{in}(I)_3$, all monomials of degree $d \geqslant 3$ belong to $\operatorname{in}(I)$, again due to the property $\operatorname{in}(fg) = \operatorname{in}(f)\cdot\operatorname{in}(g)$. To obtain our final list of generators for $\operatorname{in}(I)$, we take all the ones found in degrees $2$ and $3$, and we remove the redundant ones: $$\operatorname{in}(I) = \langle x_1^2, x_1x_2, x_2^3 \rangle.$$ Here $x_1^3$ would be redundant because it is a multiple of $x_1^2$, and $x_1^2x_2$ and $x_1x_2^2$ would be redundant because they are both multiples of $x_1x_2$.

More generally (and less ad-hoc) one would first compute a Gröbner basis $G = (g_1,\ldots,g_r)$ of $I$ which makes it easy to compute $\operatorname{in}(I) = \langle \operatorname{in}(g_1), \ldots, \operatorname{in}(g_2) \rangle$.