Tensor product representations of $\text{sl}(2;\mathbb{C})$ Define $V_m$ as the space of all homogeneous polynomials in two complex variables of degree $n$. 
Then we can define a representation of $SU(2)$ on the space $V_m$ by the formula
$$[\Pi_m(U)f](z) = f(U^{-1}z).$$
We can then define the Lie algebra representaion of $\text{su}(2)$ by the formula
$$\pi_m(X) = \left. \frac{d}{dt}\Pi_m(e^{tX}) \right |_{t=0}$$
which works out to be
$$\pi_m(X)f = -\frac{\partial f}{\partial z_1}(X_{11} z_1+ X_{12} z_2) - \frac{\partial f}{\partial z_2}(X_{21}z_2 + X_{22}z_2).$$
It turns out that this representation extends to a representation of $\text{sl}(2;\mathbb{C})$
Now we have two ways of interpreting the tensor product representation $\pi_m \otimes \pi_m$. 
1) We can define $\pi_m \otimes \pi_m$ as a representation of $\text{sl}(2;\mathbb{C}) \oplus \text{sl}(2;\mathbb{C})$ acting on $V_m \otimes V_m$ by 
$$\pi_m \otimes \pi_m (X,Y) = \pi_m(X) \otimes I + I \otimes \pi_m(Y)$$
2) We also have $\pi_m \otimes \pi_m$ as a representation of $\text{sl}(2;\mathbb{C})$ acting on $V_m \otimes V_m$ defined by 
$$\pi_m \otimes \pi_m (X) = \pi_m(X) \otimes I + I \otimes \pi_m(X)$$
The exercise is to show that $V_1 \otimes V_1$ as a representaion of $\text{sl}(2;\mathbb{C})$ (our second interpretation) is reducible, whilst as a representation of $\text{sl}(2;\mathbb{C}) \oplus \text{sl}(2;\mathbb{C})$(the first interpretation) it is irreducible. 
I think, in the first interpretation, that $\pi_1(X) \otimes \pi_2(Y)$ is irreducible if and only if $\pi_1(X)$ and $\pi_2(Y)$ are irreducible. Is this true?
I can't immediately proof that the second is reducible. I presume it is because we have $\pi_1(X) \otimes I$ and $I \otimes \pi_1(X)$ (as opposed to $X$ and $Y$). Any hints?
 A: The results you are trying to show actually hold for any semi-simple Lie algebra.  One can prove the first fact using character theory, see e.g. theorem 3.9 here: http://books.google.com/books?id=F3NgD_25OOsC&lpg=PR1&dq=compact%20lie%20groups&pg=PA50#v=onepage&q&f=false .
The second one is easier to prove and in fact works for reps of any group.  Here is a hint: there is a canonical decomposition of $V\otimes V$ as $T_1 \oplus T_2$.  Then its easy to show (for this particular decomposition) that any tensor product rep will preserve the summands.
A: Here is a direct computation for (2) to show that it is reducible. Let me denote by $H,X,Y$ the usual basis for the real Lie algebra $\mathfrak{sl}_2(\Bbb{C})$, that is
$$H = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array}\right), X =  \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right), Y = \left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right).$$ 
Now we know for example that $\pi_1(X)$ is the linear operator $-z_2\frac{\partial}{\partial z_1}$ acting on $V_1$. So we find in general that in the basis $z_1,z_2$ of $V_1$ we find the matrices for $\pi_1(X),\pi_1(Y)$ and $\pi_1(H)$ to be
$$\begin{eqnarray*} \pi_1(X) &=& \left(\begin{array}{cc} -1 & 0 \\ 0 & 0 \end{array}\right),\\
\pi_1(Y) &=& \left(\begin{array}{cc} 0 & -1 \\ 0 & 0 \end{array}\right),\\
\pi_1(H) &=& \left(\begin{array}{cc} -1 & 1 \\ 0 & 0 \end{array}\right).
\end{eqnarray*}$$
Then we find that the matrices for $\pi_1\otimes \pi_1$ of $H,X,Y$ in the basis $e_1 \otimes e_1, e_1\otimes e_2,e_2\otimes e_1, e_2 \otimes e_2$ are
$$\begin{eqnarray*} (\pi_1 \otimes \pi_1)(X) &=& \left(\begin{array}{cccc} - 2 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0\end{array}\right), \\
(\pi_1 \otimes \pi_1)(Y) &=& \left(\begin{array}{cccc} 0 & -1 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 &0 & -1 \\ 0 & 0 & 0 & 0\end{array}\right), \\
(\pi_1 \otimes \pi_1)(H) &=& \left(\begin{array}{cccc} -2 & 1 & 1 & 0 \\ 0 & -1 & 0 & 1 \\ 0 & 0 &-1 & 0 \\ 0 & 0 & 0 & 0\end{array}\right). \\
\end{eqnarray*}$$
You can see that span of the vector $e_1\otimes e_1$ is invariant under the action of every element of $\mathfrak{sl}_2(\Bbb{C})$, showing that the representation $V_1 \otimes V_1$ is reducible.
