Primitive ideals

 

Our aim will be to state a classification theorem of primitive ideals. So let us first define such ideals:

Definition 6726

An ideal I is said to be primitive if it is the annihilator of a simple left R-module. The set of primitive ideals of R is denoted by Prim(R). The intersection of the primitive ideals of R is called the radical of R.

Since the annihilator of a module is the intersection of the annihilators of its elements, the radical of R is also the intersection of the maximal left ideals of R. We have th following implication:

$$I \mbox{ primitive } \Rightarrow I \mbox{ prime } \Rightarrow I \mbox{ semiprime. }$$

Let us prove the first one: Assume that I is the annihilator of a simple left R-module M. Let I₁ and I₂ be ideals of R containing I and distinct from I. Then $I_2M\neq 0$ and $RI_2M\subseteq I_2M$, hence I₂M=M, and consequently $(I_1I_2)M=I_1M\neq 0$. So we get that $I_1I_2\not\subseteq I$, so I is a prime ideal as requested.

On the other hand, all maximal ideals are primitive.

Note that the converse of the above implications are not true. To give an example of a prime ideal that is not primitive, we can simply take the ring of integers $R=\mathbf{Z}$ and the ideal I=0.

Proposition 6759

  Assume that a ring R is Noetherian. Let I be an ideal of R, and $\mathcal{I}$ the set of ideals of R, a power of which is contained in I. Then: noetherr

• $\mathcal{I}$ has a largest element J.
• If $I\neq R,\ J$ is the smallest semiprime ideal of R containing I. We then call J the root of I.

Proof:Let J be the sum of the elements of $\mathcal{I}$. Since R is Noetherian, there exists $I_1,\ldots,I_r\in\mathcal{I}$ such that $J=I_1+\ldots+I_r$. A sufficiently large power of $I_1+\ldots +I_r$ is contained in I, an hence $J\in \mathcal{I}$. Then J is the largest element of $\mathcal{I}$. If $I\neq R$, then J is clearly semiprime. If J' is a semiprime ideal containing I, the image of J in R/J' is nilpotent, and hence $J\subseteq J'$.

$\mathbf{Q.E.D.}$

There is one more interesting proposition, which will be left unproven here:

Proposition 6782

Let I be a proper ideal of $\mathcal{U}(\EuFrak{g})$. Then the following are equivalent: semiprimenumlist

• I is semiprime.
• I is an intersection of primitive ideals.

In particular, the ideal of $\mathcal{U}(\EuFrak{g})$ is an intersection of primitive ideals. For every subset T of Prim(R), let I(T) be the intersection of the elements of T. This set is an ideal of R. If R is a quotient algebra of $\mathcal{U}(\EuFrak{g})$, we can obtain a bijection between the set of non-empty closed subsets of Prim(R) and the set of semiprime ideals of R.