Brian Bi
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## Section 2.5. Quotients

Problem 2.5.1 $$A/I$$ is isomorphic to the algebra generated by monomials with degree strictly less than $$N$$, where the product of monomials is zero if the sum of their degrees is at least $$N$$. Call this algebra $$A'$$. Suppose $$A'$$ is the direct sum of two subrepresentations, $$A' = A_1 \oplus A_2$$. At least one of the two must contain an element with nonzero constant term, therefore it must contain an element with constant term equal to unity. Without loss of generality, let this be $$a \in A_1$$. We claim that $$A_1 = A'$$.

If we set $$D = 0$$, this is equivalent to the claim that $$A_1$$ contains all monomials of degree $$d$$ where $$D \leq d \leq N-1$$. We prove this by induction. For the base case, consider that for any monomial $$P$$ of degree $$N-1$$, we have $$Pa = P$$, so $$P \in A_1$$; this establishes the claim for $$D = N-1$$. Now assume $$A_1$$ contains all monomials of degree at least $$D+1$$, and let $$P$$ be any monomial of degree $$D$$; then $$Pa = P + Q$$ where $$Q$$ consists only of terms of degree at least $$D+1$$. $$Pa = P + Q$$ is an element of $$A_1$$, and by the inductive hypothesis, so is $$Q$$, therefore, $$P \in A_1$$. This establishes the claim for $$D$$.

We conclude that $$A_1$$ indeed contains all monomials, so $$A_1 = A'$$ and $$A_2 = 0$$. Thus $$A'$$ is indecomposable, and so is $$A/I$$ which is isomorphic to it.

Problem 2.5.2
1. If there is a nonzero vector $$v \in V$$ which is not cyclic, then $$Av$$ is a nonzero proper subrepresentation of $$V$$ and $$V$$ is reducible.
2. If $$v \in V$$ is cyclic, then define $$I$$ to be the set of those elements of $$a$$ which annihilate $$v$$. This is clearly a left ideal of $$A$$. Define a homomorphism $$\varphi : V \to A/I$$ as follows: $$\varphi(w) = \pi(a)$$ where $$a$$ is any element of $$A$$ for which $$av = w$$. That this is well-defined follows from the fact that if $$a_1 v = a_2 v = w$$, then $$a_1 - a_2 \in I$$ so $$\pi(a_1) = \pi(a_2)$$. That it defines a homomorphism follows from the fact that for given $$w$$, if we have $$av = w$$, then $$bav = bw$$ for all $$b \in A$$, so that $$\varphi(bw) = \pi(ba) = b\pi(a) = b\varphi(w)$$. Finally, it is easy to see that $$\varphi$$ has a well-defined inverse, $$\varphi^{-1} : A/I \to W$$ given by $$\varphi^{-1}(\pi(a)) = av$$. So $$V$$ and $$A/I$$ are isomorphic.

For the other direction, if there exists an isomorphism $$\varphi : V \to A/I$$ for some left ideal $$I$$ of $$A$$, then let $$v = \varphi^{-1}(u)$$ where $$u$$ is the unit of $$A/I$$. Then if $$w \in V$$, choose $$a \in A$$ such that $$\pi(a) = \varphi(w)$$; we then have $$av = a\varphi^{-1}(u) = \varphi^{-1}(au) = \varphi^{-1}(\pi(a)) = w$$, so $$v$$ is cyclic.

3. For the example given, a basis for $$A$$ is given by $$\{\pi(1), \pi(x), \pi(y)\}$$, which we will simply refer to as $$\{1, x, y\}$$. A basis for $$V$$ is given by $$\{\d 1, \d x, \d y\}$$, which are the linear functionals defined to act as follows: \begin{align*} \d x(ax + by + c) &= a \\ \d y(ax + by + c) &= b \\ \d 1(ax + by + c) &= c \end{align*} Then, \begin{align*} \rho(x) (a\d x + b\d y + \d 1) &= a\d 1 \\ \rho(y) (a\d x + b\d y + \d 1) &= b\d 1 \\ \end{align*} from which we can immediately observe that for any nonzero $$v \in V$$, there is some $$a \in A$$ with $$av = \d 1$$, so $$V$$ cannot be the direct sum of nonzero subrepresentations. To show that $$V$$ is not cyclic, take $$v = a\d x + b\d y + \d 1$$. If $$a = 0$$, then $$\d x \notin Av$$. Otherwise, if $$b = 0$$, then $$\d y \notin Av$$. But if $$a, b \neq 0$$, then $$\d x, \d y \notin Av$$.