I was reading a paper, and it said that the following were equivalent using the Axiom of Choice, but I tried working it out, and I wasn't sure how: an algebra $A$ is primitive; $A$ has a proper left ideal $B$ such that $A = B +C$ for any non-trivial two-sided ideal $C$ of $A$. I've tried reasoning it out, and I'm not sure how - can anyone help? I know it should be really easy, but I seem to be missing a key step. If someone could give me a hint or two to work it out, that would be great. 



Some equivalent statements about primitive algebras
Lam (A first course in noncommutative rings, 2ed) does it for (unital) rings $R$ in Lemma 11.28 (page 186):



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> If such a $B$ exists, we may assume (after an application of Zorn's Lemma) that it is a maximal left ideal. The annihilator of the simple left $R$-module $R/B$ is an ideal in $B$, and so it must be zero. This shows that $R$ is left primitive. Conversely, if $R$ is left primitive, there exists a faithful simple left $R$-module, which we may take to be $R/B$ for some (maximal) left ideal $B \subsetneq R$. A nonzero ideal $C$ cannot lie in $B$ (for otherwise $C$ annihilates $R/B$) and so must be comaximal with $B$.
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Here, the statement equivalent to the Axiom of Choice is [Zorn's Lemma](http://en.wikipedia.org/wiki/Zorn_lemma/); it says that:



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> Every partially ordered set in which every chain (i.e. totally ordered subset) has an upper bound contains at least one maximal element.
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In this proof we get to see one of its most common uses: it assures that any **unital** ring has a maximal ideal (see the Wikipedia page for more information).


