I want to find all ideals $I$ in $\mathbf{C}[x,y]$ with $\sqrt{I}=(x,y)$ and $\dim_{\mathbf{C}}\mathbf{C}[x,y]/I=2$.
I have no clue how to about it, I mean I can write down some examples, $I=(x,y^2)$ say, but in the general I failed. What I know about $I$ is that it contains some power of $x$ and some power of $y$, and that it is f.g., and in a sense the powers should not be that high so as to assure the dimension to be 2. Could someone help me?
Remark: I have seen a related question for the dimension $4$ case, but I have no idea what a Gröbner basis is, I want a hands on approach.
Let $I$ be such an ideal. Then $\{1,x,y\}$ is a $\mathbb{C}$-linearly dependent spanning set of $R := \mathbb{C}[x,y]/I$ (since $1 \not \in I$), so $ax + by + c \in I \subseteq (x,y)$ for some $a,b,c \in \mathbb{C}$ not all $0 \implies c = 0$, so there is a nonzero linear form $ax + by \in I$. By a linear change of coordinates $x' := ax + by, y' := cx + dy$ with $ad - bc \ne 0$ we may assume $x \in I$.
Thus $\{1, y\}$ is a $\mathbb{C}$-basis for $R$. Thus $y^2 - ry - s \in I \subseteq (x,y)$ for some $r, s \in \mathbb{C} \implies s = 0$. For $n := \min\{ t \mid y^t \in I\} \ge 2$, multiplying by $y^{n-2}$ gives $y^n \equiv ry^{n-1} \pmod I \implies r = 0$, so in fact $y^2 \in I$. Then $(x,y^2) \subseteq I$, so there is a surjection of $2$-dim $\mathbb{C}$-vector spaces $\mathbb{C}[x,y]/(x,y^2) \twoheadrightarrow R$, which is an isomorphism, so $I = (x,y^2)$.
Thus, the $(x,y)$-primary ideals $I \subseteq \mathbb{C}[x,y]$ of degree $2$ are associated to a matrix in $GL_2(\mathbb{C}$) via $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \leftrightarrow (ax + by, (cx + dy)^2)$; i.e. $I = (f_1, f_2^2)$, where $f_1, f_2$ are independent linear forms vanishing at $0$.
Edit: As Georges points out, there is some redundancy here, as $(f_1, f_2^2) = (r_1f_1, r_2(f_2 - r_3f_1)^2)$ for $r_i \ne 0$. Thus $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ gives the same ideal as $\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$ (if $a \ne 0$) or $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ (if $a = 0$), and these matrices, parametrized by $\mathbb{P}^1$ (the first row), give rise to distinct ideals.