Math 380, Spring 2018, Assignment 9

The moving power of mathematical invention is not reasoning but the imagination.

- Augustus de Morgan

Read:

1. Section 2.6.
2. Section 2.7.

Carefully define the following terms, then give one example and one non-example of each:

1. $\overline{f}^{(g_1,\dots,g_s)}$ (the normal form of $f$ modulo the ordered set $(g_1,\dots,g_s)$).
2. Least common multiple (of two monomials).
3. $S(f,g)$ (the syzygy polynomial determined by $f$ and $g$).

Carefully state the following theorems (you do not need to prove them):

1. Bound on the multidegree of $S(f,g)$.
2. Cancellation lemma (this is Lemma 2.6.5 in the text).
3. Buchberger's $S$-pair criterion.

Carefully describe the following algorithms:

1. Buchberger's algorithm (to compute a Grőbner basis for a given ideal $\left\langle f_1,\dots,f_s\right\rangle$).

Solve the following problems:

1. Section 2.6, problems 2, 5, 6, and 9.
2. Section 2.7, problem 2(a), using lex order only.
3. (Parameter elimination again) Using lex order in which $t>x>y$, compute a Grőbner basis for the ideal $\left\langle t-x, ty-1\right\rangle$. What does this tell you about the image of the parametrization $x=t,\quad y=1/t$?

--------------------End of assignment--------------------

Questions:

I really don't know what to do for the last part of question $3$. What does a parameterization derived from a set of polynomials which form a Groebner basis have over a parameterization derived otherwise?

Buchberger's algorithm leads to the Grőbner basis $(t-x, ty-1, -xy+1)$, which tells you that any point in the image of the parametrization satisfies the equation $xy=1$. In fact this equation turns out to characterize the image exactly. -Steven.Jackson (talk) 13:58, 14 May 2018 (EDT)

Solutions:

2.6.2: (a) Performing the division algorithm with divisors $g_1=x+z$ and $g_2=y-z$, we get $q_1=y$, $q_2=-z$, and $r=-z^2$.

(b) Performing the division algorithm with the reversed divisors, $g_1=y-z$ and $g_2=x+z$, we get $q_1=x$, $q_2=z$, and $r=-z^2$. The quotients are different, but the remainders are the same. That the remainders are the same is a general property when we have a Groebner basis.

2.6.5: We compute the syzygy polynomial with lex order for each pair.

(a) \[ S(4x^2z-7y^2,xyz^2+3xz^4)=-3x^2z^4-\frac{7}{4}y^3z \] (b) \[ S(x^4y-z^2,3xz^2-y)=\frac{1}{3}x^3y^2-z^4 \] (c) Assuming $i$ is the imaginary unit and not a variable. \[ S(x^7y^2z+2ixyz,2x^7y^2z+4)=2ixyz-2 \] (d) Assuming $3^z$ is a typo in the second polynomial because otherwise it wouldn't be a polynomial. \[ S(xy+z^3,z^2-3z)=3xyz+z^5 \]

2.6.6: Yes. Take the pair from part (d) of the previous problem. With lex order $S(f,g)=3xyz+z^5$. With grlex order, \[ S(f,g)=\frac{z^3}{z^3}(z^3+xy)-\frac{z^3}{z^2}(z^2-3z)=xy+3z^2\;. \]

2.6.9: Apply the S-pair criterion, that is, show that $\overline{S(-x^2+y,-x^3+z)}^{\{-x^2+y,-x^3+z\}}\neq 0$. First we compute $S(-x^2+y,-x^3+z)=-xy+z$. Then, using the division algorithm, we see that $r=-xy+z\neq 0$

2.7.2a: We use Buchberger's algorithm. It is tedious to write it all out, but it converges after two additions to the basis. The final Groebner basis is $\{x^2y-1,xy^2-x,x^2-y,y^2-1\}$. With this, the normal form of all 6 syzygy polynomials is zero.

Endorsing all solutions above. -Steven.Jackson (talk) 13:58, 14 May 2018 (EDT)

3: $\{t-x,ty-1\}$ is already a Groebner basis for the ideal it generates. We check this with the S-pair criterion. The syzygy polynomial is $S(t-x,ty-1)=ty-xy$ and the remainder upon dividing this by $\{t-x,ty-1\}$ is $0$. I don't know what this tells us about the parameterization, maybe that it doesn't miss any points in the variety?

I think the $S$-pair should be $-xy+1$. -Steven.Jackson (talk) 13:58, 14 May 2018 (EDT)