Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question

1 Answer 1

up vote 3 down vote accepted

Lam (A first course in noncommutative rings, 2ed) does it for (unital) rings $R$ in Lemma 11.28 (page 186):

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 $B$.

Here, the statement equivalent to the Axiom of Choice is Zorn's Lemma; it says that:

Every partially ordered set in which every chain (i.e. totally ordered subset) has an upper bound contains at least one maximal element.

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.