Module given by generators and relations Let $\frak G$ be a Lie algebra and let $M$ be a $\frak G$-module generated by a vector $v$ satisfying some set of defining relations denoted by $R$. I mean, $M = U(\frak G)/\langle R \rangle$, where $\langle R \rangle$ is the $U(\frak G)$-submodule generated by $R$ and $U(\frak G)$ denotes the universal enveloping algebra of $\frak G$. In this case, $v = \overline 1$.
Suppose that $\frak B \subset \frak G$ is an ideal such that ${\frak B} M = 0$. So, we can naturally regard $M$ as a $\frak G/\frak B$-module.
Can we say that $M$ is a $\frak G/\frak B$-module given by one generator and relations coming from the defining relations of $M$? How to formalize this fact?
I feel that it is some sort of consequence of the Theorem of Isomorphisms, but I don't know how to write it in a formal way.
 A: If I understand you correctly, you want to write $M$ as $U(\frak G/\frak B)/\langle R'\rangle$ where $R'$ is somehow determined by $S$.
This can indeed be done. There is a natural map $U(\frak G) \to U (\frak G/\frak B)$, the unique ring homomorphism that sends elements of $G$ to the corresponding elements of $G/B$. Or alternately, you can see this map as coming from the universal property. It is clear that this map is surjective, as every element of $U(\frak G/\frak B)$ can be written in terms of elements of $\frak G/\frak B$ which can themselves be written in terms of elements of $\frak G$.
Thus, it is the quotient by some ideal $I$. If we can show that $I$ is contained in $\langle R\rangle$  then we are done because $M=U(\frak G )/\langle R\rangle=U(\frak G )/\langle I\rangle / \langle R'\rangle=U(\frak G/\frak B)/\langle R'\rangle$, where $R'$ is the image of $R$ in $U(\frak G)/\langle I\rangle=U(\frak G/\frak B)$.
But it is easy to check that $I$ is just the ideal of $U(\frak G)$ generated by $\frak B$. Take a basis for $\frak G$ which includes a basis for $\frak B$ and use this to define a basis for $U(\frak G)$, then the map to $U(\frak G/\frak B)$ is just removing terms which contain an element of the basis of $\frak B$, which is the same as taking the quotient by the ideal generated by $\frak B$.
So we are done.
