Computing the length of this module I am trying to find the length of $k[x,y]/(x^4,y^5)$ as a $k[x,y]$-module, where $k$ is a field. Therefore I am trying to find a composition series of $k[x,y]/(x^4,y^5)$ which is a strict chain $0=M_0\subset M_1\subset ...\subset M_n=k[x,y]/(x^4,y^5)$ of submodules of $k[x,y]/(x^4,y^5)$ such that $M_i$ is maximal in $M_{i+1}$ (with respect to inclusion). By correspondance this is equivalent to finding a strict chain $(x^4,y^5)=N_0\subset N_1\subset ...\subset N_n=k[x,y]$ such that $N_i$ is maximal in $N_{i+1}$. I tried to start with $(x^4,y^5)\subset (x^4,y^4)\subset (x^4,y^3)$ but I think the maximality property is not verified. Can anyone help me find such a chain?
 A: Let $R$ denote the ring $k[x,y]/(x^4,y^5)$. We know that the submodules of $R$ are precisely those $k[x,y]$ submodules annihilated by $(x^4,y^5)$, so it suffices to find a composition series of $M=R$ as an $R$ module.  By the correspondence of submodules, the same chain of submodules will be a composition series for both $R$ and $k[x,y]$.
Now, $R$ is local and therefore has a unique simple module isomorphic to $R/M\cong k$.  It’s also clearly 20 dimensional over $k$. Therefore, a composition series must have $20$ steps, each one stripping one dimension off at a time yielding a copy of $k$.
A: I have no idea whether there is a faster way to compute something like this (I am also very much a novice of commutative algebra), but we can make our lives a little easier by finding a composition series/chain of maximal length by finding all the intermediate ideals in the chain $(x^4, y^5) \subseteq (x, y)^4 \subseteq (x, y)^3 \subseteq (x, y)^2 \subseteq (x, y) \subseteq k[x, y]$. For brevity, let $A = k[x, y]$.
The first step is the most tedious. As a $k$-vector space, $(x, y)^4/(x^4, y^5)$ is generated minimally by the elements $$x^3y, x^3y^2, x^3y^3, x^3y^4, x^2y^2, x^2y^3, x^2y^4, xy^3, xy^4, y^4$$ i.e. it has dimension $10$ as a $k$-vector space. Then the chain of $A$-submodules $$0 \subseteq (x^3y^4) \subseteq (x^2y^4) \subseteq (xy^4) \subseteq (y^4) \subseteq (x^3y^3, y^4) \subseteq (x^2y^3, y^4) \subseteq (xy^3, y^4) \subseteq (x^3y^2, xy^3, y^4) \subseteq (x^2y^2, xy^3, y^4) \subseteq (x^3y, x^2y^2, xy^3, y^4)$$ is a chain of length $10$, since the inclusions are all strict. But any $A$-submodule is also a $k$-vector space, so this is necessarily a maximal chain. [Note: this chain was found by successively adding generators of $(x, y)^4$ carefully such that each addition is minimal and irredundant.]
Next, $(x, y)^3/(x, y)^4$ is generated as a $k$-vector space by the elements $$x^3, x^2y, xy^2, y^3,$$ and it is easy to see that $0 \subseteq (x^3) \subseteq (x^3, x^2y) \subseteq (x^3, x^2y, xy^2) \subseteq (x^3, x^2y, xy^2, y^3)$ is a maximal chain, so $(x, y)^3/(x, y)^4$ is of length $4$ as an $A$-module. In the same way, we see that $(x, y)^2/(x, y)^3$ is of length $3$ and $(x, y)/(x, y)^2$ is of length $2$.
Thus, $$\ell(A/(x^4, y^5)) = \ell(A/(x, y)) + \ell((x, y)/(x, y)^2) + \ell((x, y)^2/(x, y)^3) + \ell((x, y)^3/(x, y)^4) + \ell((x, y)^4/(x^4, y^5)) = 1 + 2 + 3 + 4 + 10 = 20.$$
Hope this helps, and that I haven't made a mistake.
