How can I prove if $R$ is a domain where every submodule is a summand, then it is a field? Suppose $R$ is a domain with the property that, for $R$-modules, every
submodule is a summand. 
I would like to show $R$ is a field.
Stating the definitions I know that for any submodule $A$ there exsists a summand $B$ where $A \oplus B = R$ and $A \cap B=0$.
Also $\forall a,b \in A,B$, $ab \neq0.$
However I am not sure the first step to take to show that every element of $R$ is a unit.
Thanks

Let $I$ be an ideal of $R$. It is sufficient to show that either $I = 0$ or $I = R$. 
Let us assume $I \neq 0$ and try to get $I = R$. From the assumption submodule (ideal) $I$ of $R$ has a complement $J$:
$$ I + J = R, \quad I \cap J = 0.$$
If $J \neq 0$ then, since $R$ is a domain, we have
$$ 0 \neq IJ \subseteq I \cap J = 0,$$
which yields a contradiction. Hence we have $J = 0$ and $I = R$ as desired. 
