"Multiplying" by denominator of quotient module

Question

Let $R$ be a ring, $A$ be an $R$-module, and $B, C$ be submodules such that $A = B \oplus C$. Then we can say that $\dfrac{A}{B} = \dfrac{B \oplus C}{B} \cong \dfrac{C}{B \cap C} = \dfrac{C}{0} \cong C$ (using the second isomorphism theorem). My question is: is the converse true?

If we have $\dfrac{A}{B} \cong C$, can we somehow deduce that $A \cong B \oplus C$?

Answer 1

While this is true for vector spaces (that is, if your ring $R$ is a field), the property you consider is false in general. For instance, taking $R=\mathbb{Z}$, $A=R$ and $B=2\mathbb{Z}$, you get that the quotient is the torsion group $C=\mathbb{Z}/2\mathbb{Z}$. No subgroup of $\mathbb{Z}$ can be isomorphic to $C$, so $A$ is not isomorphic to $B\oplus C$.

Rings having the property you consider are called semisimple rings.

Answer 2

No. For example, let $R = A = \mathbb{Z}$, and let $B = 2\mathbb{Z}$. Then $\frac{A}{B} \cong \mathbb{Z}/\mathbb{2Z}$, so $B \oplus C \cong \mathbb{Z} \oplus \mathbb{Z/2Z}$, which cannot be isomorphic to $A$ as $B \oplus C$ has an element which is annihilated by $2$, which $A$ does not.