Exercise 16, § 7

Let A be a ring. A two-sided ideal $\mathfrak{p}$ of A is called primitive if the quotient ring $A/\mathfrak{p}$ is primitive.

a) Prove that the primitive ideals are the annihilators of the simple A-modules. In particular, the primitive ideals of a commutative ring are its maximal ideals.

b) A two-sided maximal ideal is primitive; a primitive ideal is prime (Exercise 15). c) An element $x$ of A is invertible if and only if its image in every quotient of A by a primitive ideal is invertible (prove that if this is the case, $x$ is not contained in any maximal left ideals).

d) Let $\mathfrak{a}$ be a left ideal of A. Prove that if $\mathfrak{a}+\mathfrak{p}= A$ for every primitive ideal $\mathfrak{p}$ of A, we have $\mathfrak{a}= A$.

e) Let M be a finitely generated A-module such that $\mathfrak{p}M = M$ for every primitive ideal $\mathfrak{p}$ of A. Prove that M is zero (by induction on the smallest number of generators of M, using d)).

f ) Let M be a Noetherian A-module. Prove that the intersection of the submodules $\mathfrak{a}M$, where $\mathfrak{a}$ runs through the set of finite products of primitive ideals, is reduced to 0 (use e)).