I'm in the middle of trying to prove something at the moment and am looking for a decomposition of the Lie algebra $\mathfrak{su}(3)$ into a tensor product of some algebra $A$, and another $B$ containing $\mathfrak{su}(2)$, or some such result. Does anyone know of anything?
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$\mathfrak{su}(3)$ cannot be decomposed as a tensor product, since there are no nonabelian simple Lie algebras of dimension 4, 2 or 1 (thus, any 8d Lie algebra which is a tensor product has a proper ideal, which $\mathfrak{su}(3)$ doesn't). 


$\mathfrak{su}(2)$ is contained in $\mathfrak{su}(3)$, so you can take $A=k$ and $B=\mathfrak{su}(3)$ to answer the question in your title. 


There is no such nontrivial algebra, because $\dim \mathfrak{su}(2) = 3$ and $\dim \mathfrak{su}(3) = 8$. As Mariano pointed out (and I missed), considering $\mathfrak{su}(2) \subset \mathfrak{su}(3)$ works trivially. 

