How do i computed the Groebner Basis for this ideal?

Question

I have the ideal $$\begin{split}I_{k} &= \langle\,\, x_{1}^{3}-1,\\ &\qquad x_{1}^{2}+x_{1}x_{2}+x_{2}^{2},\\ &\qquad x_{1}^{2}+x_{1}x_{3}+x_{3}^{2},\\ &\qquad x_{2}^{3}-1,\\ &\qquad x_{2}^{2}+x_{2}x_{3}+x_{3}^{2},\\ &\qquad x_{3}^{3}-1 \rangle \end{split}$$ with lex $x_{1} > x_{2} > x_{3}$.

I'm unsure how i compute the Groebner Basis for this, do i take the S polynomial for the 1st and 2nd EQ and divide the remainder by just equations 1 & 2 or do i divide the remainder by all the equations within the ideal?

How do i know which polynomials i am able to remove from the ideal? I've been trying to figure it out all day and all the examples i have seen only deal with 2 polynomials in the ideal?

Thanks for you help in advance.

Answer 1

In general, what you should do is to take the $S$-polynomial for pairs and compute the remainder modulo all of the other polynomials.

For example, for the first two polynomials, $x_1^3-1$ and $x_1^2+x_1x_2+x_2^2$, compute the $S$ polynomial by multiplying to make the lcm of the leading coefficients: $$ 1(x_1^3-1)-x_1(x_1^2+x_1x_2+x_2^2)=-x_1^2x_2-x_1x_2^2-1. $$ Then, compute the remainder of this polynomial under division by all the polynomials in the generating set. The leading term of this polynomial $-x_1^2x_2$ is not divisible by the leading term $x_1^3$ of the first polynomial, but it is divisible by the leading term of the second polynomial, $x_1^2$. In fact, we can compute $$ (-x_1^2x_2-x_1x_2^2-1)+x_2(x_1^2+x_1x_2+x_2^2)=x_2^3-1. $$ The leading term of this new polynomial is $x_2^3$ which is not divisible by the leading terms of the first three polynomials, but it is divisible by the fourth polynomial, and you get $$ x_2^3-1-(x_2^3-1)=0. $$ Therefore, the remainder of this $S$-polynomial under division is zero and there's nothing to add to the set. Now, you keep moving on through the pairs of polynomials.

Note that if you get that the leading term is not divisible by the leading terms of any of the generators, you put that leading term aside and try to reduce the remainder of the polynomial. This step is not strictly necessary, but can make your answers simpler.

You do not need to discard any polynomials, a superset of a Groebner basis is a Groebner basis, but you can discard polynomials which have zero remainder when divided by the other generators.

Answer 2

Let's label the polynomials: \begin{align*} p_1&=x_1^3-1\\ p_2&=x_1^2+x_1x_2+x_2^2\\ p_3&=x_1^2+x_1x_3+x_3^2\\ p_4&=x_2^3-1\\ p_5&=x_2^2+x_2x_3+x_3^2\\ p_6&=x_3^3-1. \end{align*} We'll keep this order throughout this answer. Therefore, the ideal of interest is $$ I=\langle p_1,p_2,p_3,p_4,p_5,p_6\rangle. $$

Since a lot of the leading terms divide each other, we can reduce this generating set. For instance, if we take the first polynomial and compute its remainder modulo the other polynomials, we see that the leading term of the second polynomial divides the leading term of the first polynomial, so we compute $$ p_1-x_1p_2=-x_1^2x_2-x_1x_2^2-1. $$ The leading term of the remainder is still divisible by the leading term of $p_2$, so we compute $$ (-x_1^2x_2-x_1x_2^2-1)+x_2p_2=x_2^3-1=p_4. $$ Therefore, $$ p_1=(x_1-x_2)p_2+p_4. $$ Therefore, $$ I=\langle p_2,p_3,p_4,p_5,p_6\rangle. $$ Similarly, $$ p_4=(x_2-x_3)p_5+p_6, $$ so it can be removed as well. This reduces your situation to $$ I=\langle p_2,p_3,p_5,p_6\rangle. $$ One more reduction that we can make is that $p_2$ and $p_3$ share the same leading term, so we can look at their difference, $$ p_2-p_3=x_1x_2-x_1x_3+x_2^2-x_3^2. $$ Therefore, if we let $q_3=x_1x_2-x_1x_3+x_2^2-x_3^2$, we see that $$ I=\langle p_2,q_3,p_5,p_6\rangle. $$ Also, observe that at every step, we have not reduced the ideal of initial terms, only increased it, so we are closer to a Groebner basis. This leads to only $6$ $S$-polynomials that might need to be calculated (at least with these polynomials, as we get more polynomials, more $S$-polynomials might be needed).

To see an example, let's work with $p_2$ and $q_3$. The leading term of $p_2$ is $x_1^2$ while the leading term of $q_3$ is $x_1x_2$. Therefore, their least common multiple is $x_1^2x_2$. Then, the $S$ polynomial of these two is $$ S(p_2,q_3)=\frac{x_1^2x_2}{x_1^2}p_2-\frac{x_1^2x_2}{x_1x_2}q_3=x_2p_2-x_1q_3. $$ Working this out, we get $$ S(p_2,q_3)=x_1^2x_3+x_1x_3^2+x_2^2. $$ Next, we compute the remainder of this polynomial by the other polynomials. We observe that the first term of the $S$-polynomial is divisible by the leading term of $p_2$, so we compute $$ (x_1^2x_3+x_1x_3^2+x_2^2)-x_3p_2=-x_1x_2x_3+x_1x_3^2-x_2^2x_3+x_2^2. $$ The first term of this polynomial is divisible by the leading term of $q_3$, so we get $$ (-x_1x_2x_3+x_1x_3^2-x_2^2x_3+x_2^2)+x_3q_3=x_2^2-x_3^3. $$ The leading term of what we have now is divisible by $p_5$, so we further reduce to get $$ (x_2^2-x_3^3)-p_5=-x_2x_3-x_3^2-x_3^3. $$ Now, none of the coefficients, except the last, are divisible by leading coefficients of terms that we have, so we further reduce by $$ (-x_2x_3-x_3^2-x_3^3)+p6=-x_2x_3-x_3^2-1. $$ Therefore, we let $q_7=-x_2x_3-x_3^2-1$ and our ideal is $$ I=\langle p_2,q_3,p_5,p_6,q_7\rangle. $$ This has increased the ideal generated by the leading terms, but there's nothing obvious to reduce since the leading terms don't divide one another. This has increased the number of $S$-polynomials to check, but, hopefully, things will reduce down the line. The next step is to look at $S(p_2,p_5)$.