Left-ideals in subalgebras

Suppose $A$ and $B$ are (complex) unital associative algebras (with not necessarily the same units) and $B\subseteq A$. Also, let $\mathcal{L}$ be a maximal left ideal in $A$. Is it true that either $B\subseteq\mathcal{L}$ or $B\cap \mathcal{L}$ is a maximal left ideal in $B$?

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1 Answer

The answer is no. Let $A = \mathbb{C}(x)$ and $B = \mathbb{C}[x] \subset A$, i.e. the field of rational functions in one variable and the subring of polynomials. Consider $\mathcal{L} = 0$, the zero ideal in $A$.

If $A$ and $B$ are commutative, finitely generated, and have the same unity element, the answer is yes: this is nontrivial and closely related to the Nullstellensatz. This counterexample shows that the finitely generated hypothesis is necessary, but I am unsure about the rest.

Edit: You may be interested in the notion of PI ring (short for polynomial identity ring). Some Googling and a Springerlink search turned up a 1966 paper of Amitsur and Procesi, "Jacobson rings and Hilbert algebras with polynomial identities," which contains a closely related theorem (I think it implies an affirmative answer to your question) for finitely generated PI algebras over a field.

Before spending too much time investigating these topics in noncommutative algebra, I would suggest getting comfortable with the relevant commutative algebra, if you aren't already. Look up "Jacobson ring" if you are unfamiliar with this notion.

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Thank you. Do you think it might work in finitely generated, non-commutative algebras? – SteelplantFailure Dec 18 '11 at 12:40