counter examples for alleged sub-algebras This is a homework question, in which we've got a bunch of kinds of subsets of a given Lie-algebra, and needed to decide wether these are sub-algebras, ideals, or non of the above. 
I have managed to find all besides 2 counter examples:
Given a commutative ring $k$ and $k$-Lie-algebra $A$, and $I,J\subset A$ are sub-algebras. then define: $\left[I,J\right]=\mbox{span}\left\{ \left[a,b\right]\,|\, a\in I,\, b\in J\right\}$ . this is obviously a $k$-modul. I'm pretty sure that $\left[I,J\right]$ is not necessarily a sub-algebra, but i can't find a counter example.
The other case, is exactly like the above, only this time $I$ is an ideal ($J$ is not an necessarily an ideal). this time i proved that this is a sub-algebra, but i'm pretty sure it is not necessarily an ideal, yet niether for this could i find a counter-example. 
Any ideas for those 2? 
Thanks alot!
 A: You are correct about the first one. To find a counterexample you have to go no further than $3 \times 3$ matrices: $\mathfrak{gl}_3(k)$.
Consider $I = \left\{ \begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix} \;{\Huge|}\; a,b,c \in k \right\}$ and $J = \left\{ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ a & 0 & 0 \end{pmatrix} \;{\Huge|}\; a \in k \right\}$.
It's easy to see that $I$ and $J$ are subalgebras of $\mathfrak{gl}_3(k)$.
Then $[I,J] = \left\{ \begin{pmatrix} a & 0 & 0 \\ b & 0 & 0 \\ 0 & c & -a \end{pmatrix} \;{\Huge|}\; a,b,c \in k \right\}$, but this is not a subalgebra. Just try bracketing elements with $a=c=0$, $b=1$ and $b=c=0$, $a=1$. This will give you a matrix with a non-zero entry in the $(3,1)$-position. (Thus not closed.)
Next, $\mathfrak{gl}_3(k)$ is reductive (almost simple) so it doesn't have any ideals that will help. To find a counterexample for the second situation one needs to look at a different algebra. Consider upper-triangular matrices $U$. Then  the same $I$ is an ideal of $U$ (while not an ideal of $\mathfrak{gl}_3(k)$). 
Now let $J = \left\{ \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & a \end{pmatrix} \;{\Huge|}\; a,b \in k \right\}$. This is (an abelian) subalgebra of $U$. 
Then $[I,J] = \left\{ \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix} \;{\Huge|}\; a,b \in k \right\}$ is not even a subalgebra. Hmm...actually this works for the first one too. Oh well, I'll leave up both. :)
