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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

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    *Hint:* Prove that if there exists a non-trivial ideal, then $R$ cannot be a domain. Then conclude that $R$ is a field. – 2017-02-19

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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.

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Let $A$ be an ideal of $R$ and $B$ such that $A \oplus B = R, A \cap B = 0$. Take $a \in A, b \in B$. Then $ab \in A \cap B$. Now use that $R$ is a domain to conclude that $A, B$ are trivial.