Problem 1.25 of Etingof's Introduction to Representation theory asks us to verify that if $A := \mathbb{C}[x,y]/I_2$ where $I_2$ is the ideal generated by homogeneous polynomials of degree at least 2, and $V := A^\ast$, the set of linear maps $A \to \mathbb{C}$, made into an $A$-module via $a f ( a') := f ( a a')$, then $V$ is indecomposable as an $A$-module but not cyclic.

My solution for indecomposability is very hands-on. Is there a better way to do this question?

My attempt:

$V$ not cyclic: If $V = Af$ for some $f$, then for any $a = \lambda + \mu x + \nu y \in A$ one needs to solve for $\lambda, \mu, \nu$ the equation system $$\begin{pmatrix} af(1) \\ af(x) \\ af(y) \end{pmatrix} = \begin{pmatrix} f(1) & f(x) & f(y) \\ f(x) & 0 & 0 \\ f(y) & 0 & 0 \end{pmatrix} \begin{pmatrix} \lambda \\ \mu \\ \nu \end{pmatrix}$$ for arbitrary values of $af(1), af(x), af(y)$. This is not possible as the determinant of the right hand matrix is 0.

$V$ indecomposable: Suppose $V = U + W$. Under the basis $\{1,x,y\}$ of $A$, let $\delta_1, \delta_x, \delta_y \in V$ have matrices $(1\;0\;0)$, $(0\;1\;0)$, $(0\;0\;1)$ resp. Write $\delta_x = u_x - w_x$ some $u_x \in U$ and $w_x \in W$, with matrices $(\lambda\;(\mu+1)\;\nu)$ and $(\lambda\;\mu\;\nu)$ resp.

My original solution then goes on like this: There are four cases, each of which shows $U+W$ is not direct:

  1. If $\mu, \mu + 1 \neq 0$, have $(\mu + 1)^{-1} (xu_x) = \mu^{-1} (xw_x) = \delta_1 \neq 0$.
  2. If $\nu \neq 0$, have $\nu^{-1} (y u_x) = \nu^{-1} (yw_x) = \delta_1 \neq 0$.
  3. If either $\mu, \mu+1 =0$, and also $\nu = 0$ and $\lambda \neq 0$, wlog $\delta_1 \in U$ and $(\lambda\;1\;0) \in W$. Then $x (\lambda\; 1\;0) = \delta_1$.
  4. If either $\mu, \mu + 1 = 0$ and also $\nu = \lambda = 0$, then wlog $U \ni \delta_x$, hence also $U \ni \delta_1 = x\delta_x$. Now perform the same analysis on $\delta_y = u_y - w_y$. If we land in cases 1-3 we are done. Otherwise either $U$ or $W$ contains $\delta_y$, corresponding to $U = V$ or $\delta_1 \in U\cap W$.

Quicker method by jflipp: If $\dim{U} = 1$, then $U$ is a common eigenspace of $1,x,y$, ie $U = \langle \delta_1 \rangle$. We land in cases 3+4 above, since $-\delta_1 = xw_x \in W$. Argue similarly if $\dim{W} = 1$. For the other combinations of dimensions, $U+W$ is obviously not direct.


Here's an argument that might not be significantly simpler than yours, but at least it's a little bit different.

Let's denote the identity element of $A$ by $e$ to distinguish it from the scalar value $1$ (Euler's number won't occur in my argument). Then, $A$ is a $\mathbb C$-vectorspace with basis $e,\ x,\ y$. Multiplication is given by $x^2 = xy = yx = y^2 = 0$ ($e$ is the identity, as I just said). $V$ is a $\mathbb C$-vectorspace with basis $\delta_e,\ \delta_x,\ \delta_y$ where $\delta_e(e) = 1,\ \delta_e(x) = \delta_e(y) = 0$ and similarly for $\delta_x$ and $\delta_y$. The actions of $e,\ x,\ y$ on $\delta_e,\ \delta_x, \delta_y$ are easy to compute, we have for example $(x\delta_x)(e) = 1,\ (x\delta_x)(x) = 0,\ (x\delta_x)(y) = 0$, and so $x\delta_x = \delta_e$. If we perform similar computations for all other combinations of elements and write the general element $v = \epsilon\delta_e + \xi\delta_x + \eta\delta_y \in V$ as a column vector $\begin{pmatrix} \epsilon \\ \xi \\ \eta \end{pmatrix}$, then the actions of $e,\ x,\ y$ on $V$ are described by left multiplication with the following matrices. $$ e \leftrightarrow \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$ $$ x \leftrightarrow \begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}$$ $$ y \leftrightarrow \begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$$ In the following, we identify $e,\ x,\ y$ with their respective matrices.

I can't see a way around these calculations. You have to get familiar with your objects.

Now, let's assume that $V$ is a cyclic $A$-module, i.e. there is a $v = \begin{pmatrix}a\\b\\c\end{pmatrix} \in V$ with $a,\ b,\ c \in \mathbb C$ such that $V = Av$. The general form of an element of $Av$ is $$ (\epsilon e + \xi x + \eta y) v = \begin{pmatrix}\epsilon a + \xi b + \eta c \\ \epsilon b \\ \epsilon c \end{pmatrix}$$ with $\epsilon,\ \xi,\ \eta \in \mathbb C$. Now consider the $\mathbb C$-linear map $$\pi:V\rightarrow\mathbb C^2,\ \begin{pmatrix}\epsilon\\\xi\\\eta\end{pmatrix} \mapsto \begin{pmatrix}\xi\\\eta\end{pmatrix},$$ i.e. $\pi$ is the projection on the second and third coordinates. On the one hand, we have $\pi(V) = \mathbb C^2$ and so $\dim\pi(V) = 2$. On the other hand, we see from the general form of an element of $Av$ above that $\dim\pi(Av) \leq 1$. This is a contradiction to $V=Av$. So we must have $V\neq Av$, and $V$ is in fact non-cyclic.

Now, let's assume that $V$ is decomposable, i.e. there's a decomposition $V = P \oplus Q$ with $P \neq 0 \neq Q,\ AP \subseteq P,\ AQ \subseteq Q$. From $\dim V = 3$, we get that either $P$ or $Q$ has dimension $1$. Let's assume $\dim P = 1$. This means that $P$ is contained in a common eigenspace of $x$ and $y$. By inspecting the matrices, we find that both have sole eigenvalue $0$ and $ker\ x = \begin{pmatrix}\mathbb C\\0\\ \mathbb C \end{pmatrix},\ ker\ y = \begin{pmatrix}\mathbb C\\ \mathbb C \\0\end{pmatrix}$. So the only common eigenspace is $ker\ x \cap ker\ y = \mathbb C\begin{pmatrix}1\\0\\0\end{pmatrix}$. So we must have $P = \mathbb C\begin{pmatrix}1\\0\\0\end{pmatrix}$. Now consider again the projection $\pi$ from above. From $\pi(V) = \mathbb C^2$ and $\pi(P) = 0$, we conclude $\pi(Q) = \mathbb C^2$. This implies that $Q \cap \pi^{-1}(\{\begin{pmatrix}1\\0\end{pmatrix}\})\neq\emptyset$. In other words, there is an element $w = \begin{pmatrix}\zeta\\1\\0\end{pmatrix} \in Q$ for some $\zeta \in \mathbb C$. But then we get $0 \neq xw = \begin{pmatrix}1\\0\\0\end{pmatrix} \in P$ by matrix multiplication, and also $xw \in Q$ since $w \in Q$ and $AQ \subseteq Q$. That's a contradiction. So $V$ must be an indecomposable $A$-module.
The point here is that for any $u \in V \setminus P$, we can find a $z \in A$ such that $0 \neq zu \in P$. So $P$ can't have an $A$-invariant complement in $V$.

  • $\begingroup$ Thanks for your feedback. I fixed ker $x$ and ker $y$. But I still think that the last two lines of my paragraph are valid for any $u\in V\setminus P$, because for any such $u$ we have $\pi(u) \neq 0$. This implies $0 \neq xu \in P$ or $0 \neq yu \in P$. $\endgroup$
    – jflipp
    Nov 18 '14 at 13:22
  • $\begingroup$ I see what you mean now, and you're right. My apologies. $\endgroup$
    – user71815
    Nov 18 '14 at 14:18

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