Tensor product of modules over Kronecker algebra Let $\mathbf{k}$ be a field and $A=\begin{bmatrix}\mathbf{k}&0\\\mathbf{k}^2&\mathbf{k}\end{bmatrix}$ be the Kronecker algebra.
Let $M$ and $N$ be the left and right $A$-modules (respectively), given by

and

respectively. Then I want to find the tensor product $M\otimes_A N$.
To my understanding, we have $M=\mathbf{k}\oplus \mathbf{k}^2$ and $N=\mathbf{k}\oplus\mathbf{k}$, where the $A$-action is given by
$$\begin{bmatrix}\lambda&0\\(u_1,u_2)&\mu\end{bmatrix}\begin{pmatrix}m_1\\(m_2,m_3)\end{pmatrix}=\begin{pmatrix}\lambda m_1\\(u_1m_1+\mu m_2, u_2m_2+\mu m_3) \end{pmatrix}$$
and
$$\begin{pmatrix} n_1& n_2\end{pmatrix}\begin{bmatrix}\lambda&0\\(u_1,u_2)&\mu\end{bmatrix}=\begin{pmatrix}\lambda n_1+u_1n_2&\mu n_2\end{pmatrix}$$
But I don't know how to continue from here to find the tensor product.
 A: One trick is to use the following facts:

*

*$M\otimes_A N$ is easy to compute if $M$ is projective;

*the functor $-\otimes_A N$ is right exact.

This trick works more generally: assume that $A = kQ/I$, where $Q$ is a finite quiver and $I$ is an admissible ideal of the path algebra $kQ$.  Let $e_1, \ldots, e_n$ be the idempotents of $A$ corresponding to the vertices of $Q$.  Then the indecomposable projective right $A$-modules are $e_1A, \ldots, e_nA$.  Let $M_A$ and ${}_AN$ be finite-dimensional modules, viewed as quiver representations.
Assume for now that $M = e_iA$ is indecomposable and projective.  Then $M\otimes_A N = e_iA \otimes_A N \cong e_iN$ is the vector space at vertex $i$ in the representation $N$.
Assume next that $M = \bigoplus_{i=1}^n (e_iA)^{\oplus a_i}$ is an arbitrary projective $A$-module.  Then by the above $M\otimes_A N \cong \bigoplus_{i=1}^n (e_iN)^{\oplus a_i}$.
Assume finally that $M$ is arbitrary.  Let
$$
 P_1^M \to P_0^M \to M \to 0
$$
be a projective presentation of $M$.  Then applying the functor $-\otimes_A N$ yields an exact sequence
$$
 P_1^M\otimes_A N \xrightarrow{f} P_0^M\otimes_A N \to M\otimes_A N \to 0.
$$
The two first term are computed as above, and the map $f$ between them is easily computed.  It then suffices to take the cokernel of $f$ to obtain $M\otimes_A N$.
Let's apply this method to your specific example.  Let's take $Q = 1 \xleftarrow{\alpha, \beta} 2$ the Kronecker quiver (there are two arrows $\alpha$ and $\beta$ from $2$ to $1$).  Take $A = kQ$, and take $M$ and $N$ the representations you drew.  Then $M$ and $N$ are right and left $A$-modules, respectively (I think you've permuted left and right in your statement).
In this case, $M$ is the projective right-module $e_1A$.  That makes our life easy: then $M\otimes_A N = e_1A\otimes_A N = e_1N \cong k$.
