The problem comes from Etingof et al and is worded as follows:
Let $A = K[x_1, x_2, \dots, x_n]$ and $I \unlhd A$ any ideal in $A$ containing all homogeneous polynomials of degree $\geq N$. Show that $A/I$ is an indecomposable representation of $A$.
My current proof is given below, but it feels kind of sketchy to me. I am pretty sure that looking at $P \cdot 1$ in both $A / I$ and $A_1$ is valid, but it feels weird.
Notice that $A / I$ is the algebra generated by all monomials of degree strictly less than $N$ such that when the product of two elements has degree $\geq N$, it is equivalent to zero.
Let $A / I \cong A_1 \oplus A_2$ be the decomposition with canonical projections $\pi_1, \pi_2$. $A_1, A_2$ are subrepresentations of $A / I$ so $A_1, A_2 \subseteq A / I$. Let $\pi$ be either projection. If $\pi(1) = 0$, then $\pi(P) = \pi(1 \cdot P) = 0 \cdot \pi(P) = 0$. Also compute $\pi(1) \cdot \pi(1) = \pi(1 \cdot 1) = \pi(1)$, so $\pi(1) = 1$. We claim that if $A_1$ or $A_2$ contains $1$, then it contains $A / I$.
Assume WLOG that $1 \in A_1$. Consider all elements $P \cdot 1$ where $P$ is a monomial of $A$ that generates $A / I$. Compute $P \cdot 1 = P$. This action of $P$ given by $\rho : A \to A / I$ ends up in $A / I$, so its value is precisely $P$. But $1 \in A_1$, and $A_1$ is a subrepresentation so it must also be in $A_1$, so $A_1 \cong A / I$.
The decomposition $A / I \cong A / I \oplus A / I$ is impossible so the only two decompositions are $A / I \cong A / I \oplus 0 \cong 0 \oplus A / I$, so $A / I$ is indecomposable.