Primitive Rings, Left and Right Primitive

Left and Right Primitive

If R contains a minimal left (or right) ideal, then R is left primitive iff R is right primitive iff R is prime.

We saw earlier that primitive implies prime, so let R be prime, and let x generate a minimal left ideal in R. We know that Rx is simple; let's show Rx is also faithful. Suppose y kills Rx, whence Ry * Rx = 0. The product of two left ideals lies in 0, which is a prime ideal, hence either y or x is 0. Therefore R is left primitive.

Since prime implies semiprime, use the previous theorem to exhibit the minimal right ideal xR, whence R is right primitive.

If you want a one-sided primitive ring, it can't have a left or right minimal ideal.

Unique Faithful Module

If R is left primitive, and has the aforementioned minimal left ideal Rx, then every simple faithful left module is isomorphic to Rx. Given another simple faithful module M, choose w in M with xw nonzero. Thus Rxw is a nonzero submodule of M, and Rxw = M. Right multiplication by w is a left R homomorphism from one simple module onto another, with a kernel that has to be 0, hence the two modules are isomorphic.

Ring of Endomorphisms is Bi-Primitive

Return to our earlier example, where V is a right K vector space, and R is the left primitive ring of K endomorphisms of V. If we can find a minimal left ideal, then R becomes right primitive as well.

Build a basis for V that starts with the vector v₀. Let e be the endomorphism that projects v₀ onto v₀, and squashes all the other vectors down to 0. Since e² = e, e is an idempotent in R. Write R as the direct product of two left R modules R*e and R*f, where f = 1-e. (This is not a direct product of rings, since e and R do not commute.)

Let q be a left R module homomorphism mapping R into V as follows. If g is an endomorphism in R, let q(g) = g(v₀). Since (g₁+g₂)(v₀) = g₁(v₀) + g₂(v₀), q respects addition in R. Multiply g₁ by g₂ on the left, and we are really applying g₁ to v₀, and then g₂. Before or after q, multiplication in R corresponds to the action of R on V. Thus q is indeed a left R module homomorphism from R into V.

Note that f lies in the kernel of q, as does R*f. If g lies in the kernel of q, write g = g₁+g₂, components taken from Re and Rf. We know q(g₂) = 0, so this means q(g₁) = 0. Now g₁ is a map that extracts v₀ and moves it somewhere else, then q comes along and applies this map only to v₀. The result is suppose to be 0. Therefore g₁ = 0, and Rf is the kernel. The projection q discards Rf, and pushes Re into V.

For any v₁ in V, there is some endomorphism g that maps v₀ to v₁. Therefore Re maps onto V. The kernel of q is Rf, and the image is V, hence V, as a left R module, is isomorphic to R/Rf, which is Re. We already showed V is a simple left R module, hence Re is a minimal left ideal. Therefore R is also right primitive.