Example of a finite-dimensional algebra of infinite representation type suppose that $K$ is algebraically closed field. I'm trying to understand why the algebra $A=K[x, y]/(x^2, y^2)$ is of infinite representation type, that is, it has infinite non-isomorphic indecomposable modules:
For each $n\geq 1$ and $\lambda \in K$, let $M=K^{2n}$ and consider the action of $A$ on $M$ with the following representation : $x\mapsto \begin{bmatrix}0&I_n\\0&0\end{bmatrix}$, $y\mapsto \begin{bmatrix}0&J_n(\lambda)\\0&0\end{bmatrix}$, where $J_n(\lambda)$ denotes the Jordan block of size $n$ with $\lambda$'s are diagonal. Thus $M$ forms an $A$-module.  Question: I can't see why the modules $M$ are indecomposable. Indeed, for $n=2$, I can show the indecomposability of $M$ by using some ad hoc arguments, but I fail to generalize it for all $n$. Any help would be so much appreciated.
 A: Replacing $y$ by $y-\lambda x$, we may assume $\lambda=0$.  Let us write the standard $K$-basis of $M$ as $\{e_1,\dots,e_n,f_1,\dots,f_n\}$, so we have $xe_i=ye_i=0$, $xf_i=e_i$, $yf_i=e_{i-1}$ for $i>1$, and $yf_1=0$.
Now for any $A$-module $N$, let $F(N)$ be the set of elements $u\in N$ such that there exists a homomorphism $\varphi:M\to N$ such that $\varphi(e_1)=u$.  Explicitly $F(N)$ is the set of $u\in N$ such that there exist $u_1,\dots,u_n,v_1,\dots v_n\in N$ with $u_1=u$, $xu_i=yu_i=0$, $xv_i=u_i$, $yv_i=u_{i-1}$ for $i>1$, and $yv_1=0$.  It is easy to see that $F$ is an additive functor from the category of $A$-modules to itself, and in particular we have $F(P\oplus Q)=F(P)\oplus F(Q)$ for any modules $P$ and $Q$.
Now suppose we have a decomposition $M=P\oplus Q$.  Note that $F(M)$ is just the span of $e_1$: clearly $F(M)$ contains $e_1$, by taking $\varphi$ to be the identity.  Conversely, if $u_i$ and $v_i$ are elements of $M$ as above, then since $yv_1=0$, $v_1=af_1+\sum b_ie_i$ for some $a,b_i\in K$.  We then have $u_1=xv_1=ae_1$, so $u_1$ is in the span of $e_1$.
Thus $F(P)\oplus F(Q)$ is the span of $e_1$, and so WLOG $F(P)$ is the span of $e_1$ and $F(Q)=0$.  We will now show that $P=M$ and $Q=0$.  Since $e_1\in F(P)$, there exist $u_i,v_i\in P$ as above with $u_1=e_1$.  Since $yv_2=u_1$, we have $v_2=f_2+af_1+\sum b_ie_i$ for some $a,b_i\in K$ and so $u_2=xv_2=e_2+ae_1$.  In particular, $e_2=u_2-au_1\in P$.  Continuing similarly, we find that $u_3=e_3+ae_2+a'e_1$ for some $a'\in K$, and so $e_3\in P$ as well.  In the same way we can inductively show that $e_i\in P$ for all $i$.
It follows that for any $q\in Q$, $xq\in P$ (since $xq$ is in the span of the $e_i$).  This means $xq=0$.  But $P$ contains the entire kernel of $x$ (namely, the span of the $e_i$), so this means $q=0$.  Thus $Q=0$, and $P=M$, as desired.
(If you prefer, you can formulate this whole argument directly in terms of the projection $\varphi:M\to P$ without referring to the functor $F$.  The argument of the third paragraph shows $\varphi(e_1)=ae_1$ for some $a\in K$.  Since $\varphi^2=\varphi$, either $a=0$ or $a=1$.  Swapping $P$ and $Q$ if necessary, we may assume $a=1$.  You can then run the argument of the fourth paragraph, with $u_i=\varphi(e_i)$ and $v_i=\varphi(f_i)$.) 
