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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.

Any thoughts?

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  • $\begingroup$ How do you go from $\pi(1)^2=\pi(1)$ to $\pi(1)=1+I$? You seem to think that $1+I$ needs to be in one of $A_1$ or $A_2$, but the decomposition $A/I=A_1\oplus A_2$ doesn't imply that, rather it implies $1+I=e_1+e_2$ where $e_1\in A_1$ and $e_2\in A_2$ are a pair of orthogonal indempotents. $\endgroup$ – anon Dec 18 '17 at 5:33
  • $\begingroup$ If they are idempotents, then when $A_1$ is viewed as a subspace of $A / I$, shouldn't the only two possible idempotents be $0$ and $1$? $\endgroup$ – incertia Dec 18 '17 at 5:48
  • $\begingroup$ So you have to show $A/I$ doesn't have any other idempotents. $\endgroup$ – anon Dec 18 '17 at 5:51
  • $\begingroup$ This is trivial because the degree goes up if it is non-constant and if $k^2 = k$ then we can multiply by the inverse to get $k = 1$. $\endgroup$ – incertia Dec 18 '17 at 5:53
  • $\begingroup$ The degree is defined for elements of $A$, but what about elements of $A/I$? One element in $A/I$ can be represented by different polynomials in $A$ with different degrees. And what's this you're saying about "the inverse" of a nonconstant polynomial? $\endgroup$ – anon Dec 18 '17 at 5:54
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I believe I have fixed the idempotency issue. The proof is now as follows.

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 $A_1, A_2 \subseteq A / I$. If $\pi_i(1) = 0$, then $\pi_i(P) = \pi_i(1 \cdot P) = 0 \cdot \pi_i(P) = 0$ and $A_i = 0$. Also compute $\pi_i(1) \cdot \pi_i(1) = \pi_i(1 \cdot 1) = \pi_i(1)$, so $\pi_i(1)$ is some idempotent in $A_i$. We claim the only idempotents of $A_i$ are $0, 1$. We also claim that if $1 \in A_i$, then $A_i = A / I$.

Let $\iota_1, \iota_2$ be the inclusions of $A_1, A_2$ into $A_1 \oplus A_2$. Write $1 = e_1 + e_2 = \iota_1 (e_1') + \iota_2 (e_2')$, where $e_1, e_2$ are idempotents in $A_1 \oplus 0, 0 \oplus A_2$, implying that $e_1 e_2 = 0$. If $e_1$ is constant, then $e_2 = 0$ and vice versa. Thus suppose that $e_1, e_2$ both have non-zero constant term otherwise $e_1^N = e_2^N = 0$, contradicting idempotency. Since they both have non-zero constant term, we can pick a non-zero $\lambda \in K$ such that $e_1' + \lambda e_2'$ has a constant term of zero. Compute $$ 0 = (e_1 + \lambda e_2)^N = e_1^N + \lambda^N e_2^N = e_1 + \lambda^N e_2. $$ Similarly, we can arrive at $0 = e_1 + \lambda^{2N} e_2$ to get $\lambda^N = \lambda^{2N}$, or $\lambda^N = 1$. This implies $e_1 + e_2 = 0$, a contradiction.

Assume WLOG that $1 \in A_1$ (and thus implying $A_2 = 0$). 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 $A_1 = A / I$. This finishes the proof that $A / I$ is indecomposable.

Special care needs to be taken for $K = \mathbb{F}_2$, because we really only have $2 e_1 e_2 = 0$, which is always true in $\mathbb{F}_2$. If $e_1 = 1$, then $e_2 = 0$ by subtraction and vice versa. Then we can suppose they are non-constant and follow the same logic to deduce that they have non-zero constant term.

One can also note that if $A / I$ is decomposed into a direct sum, one of the summands must be $0$ so the other must be $A / I$.

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