Math 380, Spring 2018, Assignment 8

Mathematical proofs, like diamonds, are hard as well as clear, and will be touched with nothing but strict reasoning.

- John Locke, Second Reply to the Bishop of Worcester

Read:

1. Section 2.4.
2. Section 2.5.

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

1. $\left\langle LT(I)\right\rangle$ (the leading term ideal of an ideal $I$).
2. Grőbner basis (for an ideal $I$).

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

1. Dickson's Lemma.
2. Ascending chain condition for monomial ideals.
3. Hilbert basis theorem.
4. Theorem concerning the existence of Grőbner bases.
5. Ascending chain condition for arbitrary ideals.

Carefully describe the following algorithms:

1. Algorithm to decide whether $f\in\left\langle g_1,\dots,g_s\right\rangle$ when $g_1,\dots,g_s$ form a Grőbner basis for the ideal that they generate.

Solve the following problems:

1. Section 2.5, problems 1, 7, 8, 10, 17, and 18.

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

Solutions:

2.5.1: $\langle\text{LT}(g_1),\text{LT}(g_2),\text{LT}(g_3)\rangle=\langle xy^2,xy,x\rangle$ with lex order. I need to find a polynomial $g\in I$ such that $\text{LT}(g)\notin\langle xy^2,xy,x\rangle$. A monomial is contained in a monomial ideal iff it is divisible by one of the generators of the monomial ideal. Note that every generator of $\langle xy^2,xy,x\rangle$ contains a factor of $x$. Thus, any $g\in I$ with a leading term not containing $x$ will work for our purposes. Consider $g=g_2-yg_3=y^2z^4-z^2$ which has leading term $\text{LT}(g)=y^2z^4$. This is not divisible by any of the generators of $\langle xy^2,xy,x\rangle$, so it is not in this ideal.

2.5.7: A Groebner basis is a set of polynomials $g_1,\ldots,g_s\in I$ such that $\langle \text{LT}(I)\rangle=\langle\text{LT}(g_1),\ldots,\text{LT}(g_s)\rangle$. We want to know if the polynomials generating $I=\langle x^4y^2-z^5,x^3y^3-1,x^2y^4-2z\rangle$ with grlex order form a Groebner basis for $I$. Taking the leading terms of each generator, the question is whether $\langle \text{LT}(I)\rangle=\langle x^4y^2,x^3y^3,x^2y^4\rangle$. Consider \[ y^2(x^4y^2-z^5)-x^2(x^2y^4-2z)=-y^2z^5+2x^2z\in I \implies y^2z^5\in \langle \text{LT}(I)\rangle\;. \] But $y^2z^5\notin \langle x^4y^2,x^3y^3,x^2y^4\rangle$ because it is not divisible by any of the generators which implies that $\{x^4y^2-z^5,x^3y^3-1,x^2y^4-2z\}$ is not a Groebner basis for $I$.

2.5.8: We want to determine whether $\{x-z^2,y-z^3\}$ is a Groebner basis with lex order for the ideal $I$ they generate. We will show that $\langle\text{LT}(g_1),\text{LT}(g_2)\rangle=\langle x,y\rangle=\langle\text{LT}(I)\rangle$, and so the set is a Groebner basis. A general element of $I$ is of the form $a_1(x-z^2)+a_2(y-z^3)$ where $a_i\in\mathbb{R}[x,y,z]$. It is easy to see from this general form that the leading term of any element in $I$ has at least one power of $x$ or $y$. This will necessarily also be true for each term of any polynomial in $\langle \text{LT}(I)\rangle$. Since every term of any $q\in\langle\text{LT}(I)\rangle$ will be divisible by either $x$ or $y$ (or both), we have that $q\in\langle x,y\rangle$ which shows $\langle \text{LT}(I)\rangle\subseteq\langle x,y\rangle$. The reverse containment is trivial because $x=\text{LT}(g_1)\in\langle\text{LT}(I)\rangle$ and $y=\text{LT}(g_2)\in\langle\text{LT}(I)\rangle$.

2.5.10: Let $I=\langle g_1\rangle$ be a principal ideal. We want to show $\{g_1,\ldots,g_s\}$ is a Groebner basis for $I$. So we must show $\langle\text{LT}(I)\rangle=\langle\text{LT}(g_1),\ldots,\text{LT}(g_s)\rangle$. A general element of $I$ is the polynomial $f=hg_1$. The leading term is then $\text{LT}(f)=\text{LT}(hg_1)=\text{LT}(h)\text{LT}(g_1)$ where we used the result of previous homework problem showing $\text{LT}(ab)=\text{LT}(a)\text{LT}(b)$. Every term of any polynomial in the monomial ring $\langle \text{LT}(I)\rangle$ will be divisible by $\text{LT}(g_1)$ which shows $\text{LT}(I)\subseteq\langle\text{LT}(g1),\ldots,\text{LT}(g_s)\rangle$. For the other direction, $\text{LT}(g_1)$ is divisible by the generator of $\langle \text{LT}(I)\rangle$ formed when $h=1$ and $\text{LT}(g_i)$ is divisible by the generator formed when $h=g_i$.

2.5.17: (a) The decompositions $y+x^2-4=-(x^2-y)+2(x^2-2)$ and $x^2-2=(1/2)(x^2-y)+(1/2)(y+x^2-4)$ demonstrate the nontrivial polynomial inclusions necessary for the reverse containment argument.

(b) $\mathbb{V}(I)=\mathbb{V}(x^2-y,x^2-2)$. The second equation is clearly solved by $x=\pm\sqrt{2}$. When these values are substituted into the first, we see that $y=2$. Thus the variety is the points $\{(\pm\sqrt{2},2)\}$.

2.5.18: (a) Let $a\in\mathbb{V}(f,g)$. Then $f(a)=0$ and $f(a)=0=g_1(a)g_2(a)$. So either $g_1(a)=0$ or $g_2(a)=0$ which shows $a\in \mathbb{V}(f,g_1)\cup\mathbb{V}(f,g_2)$. OTOH, let $a\in\mathbb{V}(f,g_1)\cup\mathbb{V}(f,g_2)$. Then either $a\in\mathbb{V}(f,g_1)$ or $a\in\mathbb{V}(f,g_2)$. Say $a\in\mathbb{V}(f,g_1)$. Then $f(a)=0$ and $g_1(a)=0$. So $g(a)=g_1(a)g_2(a)=0$ which implies $a\in\mathbb{V}(f,g)$. It is similar for the other option.

(b) The decompositions $xz-x^4=(xz-y^2)+(y+x^2)(y-x^2)$ and $xz-y^2=(xz-x^4)-(y+x^2)(y-x^2)$ demonstrate the nontrivial polynomial inclusions necessary for the reverse containment argument.

(c) $\mathbb{V}(y-x^2,xz-x^4)=\mathbb{V}(y-x^2,x)\cup\mathbb{V}(y-x^2,z-x^3)$. The first of these varieties is the z-axis because the solution set of the system of equations $\{x=0,y=x^2\}$ is $\{0,0,z\}$ for any $z\in\mathbb{R}$. The second variety is the twisted cubic which we have seen before. So the original variety is the union of the z-axis and the twisted cubic.

Endorsing the solutions of 1, 7, 8, 10, 17, and 18 above. -Steven.Jackson (talk) 13:11, 14 May 2018 (EDT)