Math 380, Spring 2018, Assignment 14

I must study politics and war that my sons may have liberty to study mathematics and philosophy.

- John Adams, letter to Abigail Adams, May 12, 1780

Solve the following problems:

1. Working in $\mathbb{R}^2$, consider the variety $V=\mathbb{V}(x^2y^2-1)$.

(a) Make a sketch of $V$. What do you think its "dimension" ought to be?

(b) Compute a Grőbner basis for the ideal $I=\left\langle x^2y^2-1\right\rangle$.

(c) Make a diagram of $\left\langle\mathrm{LT}(I)\right\rangle$, showing the monomials of this ideal as integer points in the plane.

(d) For each non-negative integer $d$, let $H(d)$ denote the number of monomials of total degree $d$ or less that lie outside $\left\langle\mathrm{LT}(I)\right\rangle$. Compute $H(d)$ for $0\leq d\leq 10$. (The function $H(d)$ is called the Hilbert function associated with $I$.)

(e) Make a graph of $H(d)$ as a function of the integer $d$. Show that, although there is no line passing through all the points of this graph, there is a line that passes through all its points for all sufficiently large $d$. That is, show that $H(d)$ is eventually linear.

2. Repeat the problem above, replacing $V$ by the variety $\mathbb{V}(x^2-1, y^2-1)$. Show in the end that in this case the Hilbert function is eventually constant.

3. (Optional) Devise an algorithm that loops through the leading terms of a Grőbner basis of an ideal, and decides whether its Hilbert function is eventually constant. In other words, devise an algorithm that decides whether a given variety is finite.

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Solutions:

1: (a) The sketch is the plot of the functions $y=1/x$ and $y=-1/x$, so there are lines in all four quadrants of the x-y plane. Intuitively this is a $1$-dimensional variety.

(b) There is one polynomial generator so it is already a Groebner basis: $G=\{x^2y^2-1\}$.

(c) By the definition of a GB, $\langle\text{LT}(I)\rangle=\langle x^2y^2\rangle$. Plotting this monomial ideal in the normal way, there is one corner at the grid position $(2,2)$.

(d) $H(0)=1$, $H(1)=3$, $H(2)=6$, $H(3)=10$, and then $H(4)$ through $H(10)$ result from adding $4$ to the previous number: $14,18,22,26,30,34$, and $38$.

(e) The asymptotic "adding $4$" implies the slope becomes constant, that is, that the $H$ function is asymptotically linear. From class, we associated this with defining what we mean by a $1$-dimensional variety.

2: (a) The variety is the four points $\{\pm 1, \pm 1\}$. A reasonable definition should designate this a $0$-dimensional ideal.

(b) Once again, we already have a Groebner basis which can be quickly verified by calculating the normal form of the only syzygy polynomial and finding it to be zero. $G=\{x^2-1,y^2-1\}$.

(c) By the definition of a GB, $\langle\text{LT}(I)\rangle=\langle x^2,y^2\rangle$. Plotting this monomial ideal in the normal way, there are two corners to the "boat": one at the grid position $(0,2)$ and the other at $(2,0)$.

(d) $H(0)=1$, $H(1)=3$, $H(2)=4$, $H(3)=4=H(4)=H(5)=\cdots$.

(e) From $d=3$ on, $H(d)=4$, that is, it is "asymptotically constant." This means the variety is $0$-dimensional.