A two sided ideal of a Noetherian ring $R$ is a left Noetherian ring with a minimal left ideal. Consider the set of minimal left ideals of $R$ ordered by inclusion. Then there is a maximal element $\mathfrak b= \bigoplus_{i\in I} \mathfrak a_i$ of direct sums of such ideals. For  any $j\in I$ and $t\in R$, $\mathfrak a_j t$ is either zero or simple. 
How do I show that $\mathfrak b t \subseteq \mathfrak b$? 
Please help- many thanks! 
 A: That's an odd statement of a more general statement that may be easier to grasp.

In any ring, the sum of all minimal left ideals is a two-sided ideal. (If there are no minimal left ideals, the sum is the empty sum $\{0\}$. This ideal is called the left socle of the ring and is denoted $Soc(_RR)$.

There's no doubt at the start that $Soc(_RR)$ is a left ideal. The main issue is to show it's a right ideal. The case of no minimal left ideals being trivial, we assume there's a minimal left ideal $L$. For any given $t\in R$, $Lt$ is obviously another left ideal of $R$. The map $x\mapsto xt$ of $L$ to $Lt$ is clearly a surjective homomorphism of left modules. The kernel of this homomorphism is either $\{0\}$ or $L$ by simplicity of $L$. In the first case $L$ and $Lt$ are isomorphic, and in the second case $Lt=\{0\}$.
Anyhow, this shows that $Lt\subseteq Soc(_RR)$ for any $t\in R$, and it follows immediately that $Soc(_RR)t\subseteq Soc(_RR)$.
It's also still true that there exists a subset of the set of minimal ideals  $L_i$ such that $\oplus L_i=Soc(_RR)$, and this is true for any ring. When $R$ is left Noetherian, you can additionally say that this direct sum is finite.
