Describing the ideals for which $\operatorname{dim}_F(F[x,y)]/I) = 4$ 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?
 A: 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 $U(\lambda)$.
