I would like to determine all the ideals of $\mathbb{C}[x,y]/(xy)$. I thought of using the quotient maps
$$p_x\colon\mathbb{C}[x,y]/(xy)\to\mathbb{C}[x,y]/(y),$$ $$p_y\colon\mathbb{C}[x,y]/(xy)\to\mathbb{C}[x,y]/(x),$$
to form the (injective) ring map
$$f\colon\mathbb{C}[x,y]/(xy)\to\mathbb{C}[x,y]/(y)\times\mathbb{C}[x,y]/(x),$$
defined by $f=p_x\times p_y$. This seems to give information about prime (and radical) ideals, because the geometry on the right-hand side is quite simple, but I'm not sure if it helps in determining the other ideals.
Difficulty: Since $\mathbb{C}[x,y]/(y)$ and $\mathbb{C}[x,y]/(x)$ are unital, all ideals on the right-hand side are of the form $I\times J$, where $I$ and $J$ are ideals of $\mathbb{C}[x,y]/(y)$ and $\mathbb{C}[x,y]/(x)$ respectively. However, this does not mean that ideals of the subring $\mathrm{im} f$ are of this form. For example, the ideal of $\mathrm{im} f$ generated by $f(x+y)$ is not a product (otherwise it would contain $(p_x(x),0)$, for example).
Question: Is there a way of using $f$ to determine all the ideals of $\mathbb{C}[x,y]/(xy)$? Or perhaps another way works?
The starting point is to classify all ideals that correspond to subschemes supported at $(0,0)$. In an algebraic language, this is the same as classifying ideals that are contained in $(x,y)$ and contain some power of $(x,y)$ (Now I am working in the ring $\mathbb C[x,y]/(xy)$).
Proof: Let $k, l$ be minimal such that $x^k, y^l \in I$, an assume that $I \neq (x^k, y^l)$. Then there must be an element $f$ of the form $a_1 x + \ldots + a_{k-1}x^{x-1} + b_1y + \ldots + b_{l-1}y^{l-1}$. We can assume wlog that one of the $a_i$ is not $0$. Let $u = \min\{i \mid a_i \neq 0\} $. Then $u=k-1$. Why? well, look at what happens when we compute $x f$: $$ xf = a_1x^2 + \ldots + a_{k-2}x^{k-1} + x^k\cdot \text{something} = a_ux^{u+1} + \ldots + a_{k-2}x^{k-1} + x^k\cdot \text{something} $$ Therefore, $x^{k-1-u}f= b_u x^{k-1} + x^k \cdot \text{something}$, and so $x^{k-1}$ would be in the ideal, a contradiction with the assumptions on $k,l$. BY the mirror argument, we see that $b_1, \ldots , b_{l-2}$ are all $0$, and so our element must be of the form $f = a x^{k-1} + by^{l-1}$. Note that $x^k, y^l \in (f)$. If there was another further element $g$, which would again be of the form $g= cx^{k-1} + dy^{l-1}$, $(c,d)$ would have to be a multiple of $(a,b)$, since otherwise, one could find $x^{k-1}$ or $y^{l-1}$ in the ideal by solving a linear system. This proves the claim.
Knowing this, one can classify all ideals corresponding to subschemes of dimension $0$.
Now, ideals corresponding to irreducible subschemes of dimension $1$. If $Y$ is such a subscheme, in particular $Y$ is an irreducible subscheme of dimension $1$ in $\mathbb C^2$, and a well known theorem says that all these subschemes are principal, i.e., correspond to a principal ideal $(f)$. Since $Y= V(f) \subset V(xy) = V(x) \cup V(y)$, we must have that $V(f) = V(x)$ or $V(y)$. Assuming the first, this implies that $\sqrt{(f)} = (x)$ in $\mathbb C[x,y]$, which forces $(f)$ to be a power of $x$ or a power of $y$.
So, the final classification of all ideals in $\mathbb C[x,y]$ is as follows:
Type 1 ideals correspond to subschemes whose irreducible components are (as sets) $V(y)$ and a finite set of points, which necessarily lie in $V(x) \setminus \{(0,0)\}$. If the points are $(0, b_1), \ldots , (0,b_q)$, each of them can come with some multiplicity $m_1, \ldots , m_q$, and there is also a multiplicity for $V(y)$, say it is $m$. The associated ideal is $$ I = \left(y^m\prod_{i=1}^q(y-b_i)^{m_i}\right) $$
Type 2 ideals are the same as type 1 with the roles of $x,y$ interchanged: $$ I = \left(x^n\prod_{i=1}^p(x-a_i)^{n_i}\right) $$
Type 3 ideals are those which are $0$-dimensional. By the above argument, these are in bijection with the set of points $a_1, \ldots a_p, b_1, \ldots , b_q$ in $\mathbb C ^*$ with multiplicities $n_1, \ldots , n_p,m_1\ldots m_q$, as well as a choice of an ideal $J = (x^n, y^m)$ (Type 3A) or $J=(x^n+ay^m)$ (Type 3B). The corresponding ideal is $$ I = \left(\prod_{i=1}^p(x-a_i)^{n_i} \prod_{j=1}^q(y-b_j)^{m_j}\right)\cdot (x^n , y^m) = \left(x^n\prod_{i=1}^p(x-a_i)^{n_i}, y^m \prod_{j=1}^q(y-b_j)^{m_j}\right), $$ or $$ I = \left(\prod_{i=1}^p(x-a_i)^{n_i} \prod_{j=1}^q(y-b_j)^{m_j}\right)\cdot (x^n +a y^m) = \left(x^n\prod_{i=1}^p(x-a_i)^{n_i}+a y^m \prod_{j=1}^q(y-b_j)^{m_j}\right) $$
Therefore, all the ideals are of the form $(bf(x)+ag(y))$ or $(f(x), g(y))$ for some unique (up to scalar) polynomials $f,g \in \mathbb C[x]$ and $a,b \in \mathbb C$
Type 1, Type 2 and Type 3B ideals can be read off from the map that you suggested, and you could say that Type 3A too, but I think that the proof that these are all the ideals does not follow from just looking at the ring map $\mathbb C[x,y]/(xy) \to \mathbb C[y] \times \mathbb C[x]$ (someone correct me if I am wrong)