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Let $\mathfrak{m}_1$ and $\mathfrak{m}_2$ be left maximal ideals of a unital ring $A$. Show that the simple modules $A/\mathfrak{m}_1$ and$A/\mathfrak{m}_2$ are isomorphic if and only if there exist $a\in A\setminus \mathfrak{m}_2$ so that $\mathfrak{m}_1 a \subseteq \mathfrak{m}_2$.

I don't really know how to begin, I'd love some help.

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up vote 2 down vote accepted

If such an $a$ exists you can consider the $A$-module map defined by multiplication by $a$ on the right from $A/m_1$ to $A/m_2$, you must use the fact that this modules are simple and the assumption on $a$ to show that it is an isomorphism.

If there exists an isomorphism $\varphi$ between the modules, then take $a$ to be the image of $1$ in $A/m_2$ (actually pick someone in the equivalence class). Now how does $\varphi$ look like? Deduce that it is the $a$ you are looking for

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