Given a field $F$. I want to describe the ideals $I$ such that $\operatorname{dim}_F(F[x,y]/I) = 4$ (with Groebner Basis). I have an understanding of Groebner Basis but I lack an intuition of what the expression $\operatorname{dim}_F(F[x,y]/I) = 4$ describes. What is a $F[x,y]/I$ in a $F$-vector space? Any hints?

  • 1
    $\begingroup$ I cannot see what you mean by «what is a $F[x,y]/I$ in a $F$-vector space?»... $\endgroup$ – Mariano Suárez-Álvarez Aug 27 '12 at 16:12
  • 3
    $\begingroup$ I do not know what intuition you expect, but what you want to do is to list all ideals $I$ in the ring $F[x,y]$ such that the quotient $F[x,y]/I$, which is an $F$-vector space, has dimension $4$. Can you at least make a list of some examples of such ideals? $\endgroup$ – Mariano Suárez-Álvarez Aug 27 '12 at 16:14

You want to learn about the theory of the Hilbert scheme of points in the plane. A good starting point is Chapter 18.1 in Combinatorial Commutative Algebra, by Miller and Sturmfels. What I write here is a slightly more user friendly, and less detailed, version of that. I assume that you are happy with Groebner bases, since you mention them in the question.

If $I$ is any ideal of $k[x,y]$, with $\dim k[x,y]/I=n$, then Groebner basis methods construct a monomial ideal $I_0$ so that the $n$ monomials not in $I_0$ (called the standard monomials), are a basis for the quotient $k[x,y]/I$. The number of monomial ideals $k[x,y]$ with $\dim k[x,y]/I_0=n$ is $p(n)$, the partition function. This is easy to see: The partition $(\lambda_1, \lambda_2, \ldots, \lambda_r)$ corresponds to the ideal $\langle x^{\lambda_1}, x^{\lambda_2} y, \ldots, x^{\lambda_r} y^{r-1}, y^r \rangle$. For any partition $\lambda$, $S(\lambda)$ be the corresponding set of standard monomials. For example, when $n=4$, we have $$\begin{matrix} I_0 & S(\lambda) & \lambda \\ \langle x^4, y \rangle & x^3, x^2, x, 1 & 4 \\ \langle x^3, xy, y^2 \rangle & y, x^2, x, 1 & 3+1 \\ \langle x^2, y^2 \rangle & xy, y, x, 1 & 2+2 \\ \langle x^2, xy, y^3 \rangle & y^2, y, x, 1 & 2+1+1 \\ \langle x, y^4 \rangle & y^3, y^2, y, 1 & 1+1+1+1 \\ \end{matrix}$$

Let $U(\lambda)$ be the set of ideals $I$ such that $S(\lambda)$ is a basis for $k[x,y]/I$. By the Groebner basis argument, every ideal which you wish to understand is in at least one of the $U(\lambda)$; most ideals are in all of them. So the final answer to your question comes from taking the $5$ sets $U(\lambda)$ above and figuring out how to glue them together. When you make this gluing, the resulting object is called the Hilbert scheme.

Note that $I$ being in $U(\lambda)$ does not mean that $I_0$ is the initial ideal of $I$; one can also study the question of describing the set of ideals with given initial ideal (for a specified term order) but this is a less fundamental problem.

The easiest sets to describe are $U(4)$ and $U(1,1,1,1)$. Every ideal in $U(4)$ is of the form $$\langle x^4-ax^3-bx^2-cx-d,\ y-ex^3-fx^2-gx-h \rangle$$ where the parameters $a$ through $h$ may be chosen freely. If you are comfortable with this language, $U(4) \cong \mathbb{A}^8$. To get $U(1,1,1,1)$, just switch the roles of $x$ and $y$.

Miller and Sturmfels compute (Example 18.5) that every ideal in $U(2,1,1)$ is of the form $$\langle x^2 - a y^2 - bx - py -q,\ xy - c y^2 - dy - ex - r,\ y^3 - f y^2 - gy - hx - s\rangle$$ where $$\begin{array}{rcl} p &=& fc^2 + ec^2 - fa+ae-bc+2cd \\ q &=& fec^2-c^3h-fae+gc^2+ae^2+ach-bec+2ecd-ga-bd+d^2 \\ r &=& -e^2c - c^2h+ah - ed \\ s &=& -f^2e+e^3+2ech-ge-bh+dh \\ \end{array}$$ and $a$ through $h$ are free parameters.

Ignoring the details, note that every generator of the ideal expresses an element of $I_0(2,1,1)$ in the basis $S(2,1,1)$. It appears that this requires $12$ parameters, but there are relations between them which means that there are only $8$ parameters. So, again, describing an ideal in $U(2,1,1)$ involves $8$ free parameters, but in a very nontrivial way.

Miller and Sturmfels also compute $U(2,2)$ (Example 18.6). Again, the generators of the ideal express elements of $I_0(2,2)$ in the basis $S(2,2)$. One winds up with $16$ parameters, and this time one can only eliminate $7$ of them. To describe a general ideal in $U(2,2)$, one must give $9$ parameters $(a,b,c,d,e,f,g,h,w)$ obeying an equation of the form $$w(1-ac) = \mbox{a polynomial in $a$ through $h$}.$$ Note that $U(2,2)$ is rational, meaning that $w$ can be expressed as a rational function of the other variables, but you have to watch out because that rational function might sometimes give $0/0$.

In general, $U(\lambda)$ is a smooth rational variety of dimension $2n$. (This is not obvious, and is very special to the case of polynomials in two variables.) There are algorithms to compute $U(\lambda)$, but the answers get complicated very fast. I think that there is a lot of work to do be done getting better descriptions of the $U(\lambda)$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.