Why is the A6 preprojective algebra of wild representation type? As mentioned in the title, I would like to know a proof of the "well known" fact that the A6 preprojective algebra is of wild representation type.
Ideally, I would like to see an explicit two-parameter family of indecomposable modules.
If you happen to prefer type D to type A and want to give an answer there, that would be appreciated too.
 A: Inspired by the reference Julian gave in his comment, here's an explicit example of a two-parameter family of indecomposable representations.
First, I'll describe a one-parameter family of indecomposable representations for the preprojective algebra of type $A_5$.
Let $k$ be the coefficient field, and $\alpha\in k$. Then let $M(\alpha)$ be the representation
$$\begin{array}{ccccccccc}
&\begin{pmatrix}0\\1\end{pmatrix}&&\begin{pmatrix}1&0\\0&0\end{pmatrix}
&&\begin{pmatrix}0&0\\1&\alpha\end{pmatrix}
&&\begin{pmatrix}1&0\end{pmatrix}\\
k&\rightleftarrows&k^2&\rightleftarrows&k^2&\rightleftarrows&k^2&
\rightleftarrows&k\\
&\begin{pmatrix}1&0\end{pmatrix}&&\begin{pmatrix}0&0\\1&1\end{pmatrix}
&&\begin{pmatrix}0&0\\1&0\end{pmatrix}
&&\begin{pmatrix}0\\\alpha\end{pmatrix}\\
\end{array}$$
It's a straightforward calculation to check that $M(\alpha)$ is indecomposable and that if $\alpha\neq\beta$ then any homomorphism $M(\alpha)\to M(\beta)$ is zero at the right hand vertex, so that a direct sum decomposition of $M(\alpha)\oplus M(\beta)$ into two non-zero summands is uniquely determined at the right hand vertex.
Now extend $M(\alpha)\oplus M(\beta)$ to a representation $N\left(\{\alpha,\beta\}\right)$ of the preprojective algebra of type $A_6$ by adding a vertex on the right hand end:
$$\begin{array}{ccccccccccc}
&&&&&&&&&\begin{pmatrix}0&0\end{pmatrix}\\
k\oplus k&\rightleftarrows&k^2\oplus k^2&\rightleftarrows&k^2\oplus k^2&\rightleftarrows&k^2\oplus k^2&
\rightleftarrows&k\oplus k&\rightleftarrows&k\\
&&&&&&&&&\begin{pmatrix}1\\1\end{pmatrix}\\
\end{array}$$
Then it follows easily from the properties stated for the representations $M(\alpha)$ that $N\left(\{\alpha,\beta\}\right)$ is indecomposable and $N\left(\{\alpha,\beta\}\right)\not\cong N\left(\{\alpha',\beta'\}\right)$ unless $\{\alpha,\beta\}=\{\alpha',\beta'\}$.
