In a right artinian semiprime ring $R$ a finite intersection of maximal ideals is $0$.
The definitions I am working with is as follows:
A ring $R$ (associative with $1$) is called right Artinian if every decreasing sequence of ideals eventually stabilizes, or equivalently, if every subset of ideals of $R$ has a minimal element.
A ring is called semiprime if $0$ is the only nilpotent ideal, or equivalently, the intersection of the prime ideals of $R$ is zero.
A basic result is that if $R$ is right Artinian, then the Jacobson radical $\mathcal{J}(R)$ of $R$ (that is the intersection of all right maximal ideals) is nilpotent ideal (some power of it is zero). From this it follows that in right Artinian ring, the prime radical (the intersection of all prime ideals of $R$) coincides with the Jacobson radical of $R$.
The problem I am having is the following:
Given a right Artinian semiprimitive ring $R$, is it true that there are finite number of maximal right ideals $M_1,\dots,M_n$ with $M_1\cap\dots\cap M_n=(0)$.
From the definition of semiprime it follows that the prime radical is zero, and because $\mathcal{J}(R)$ coincides with the prime radical, it is also zero. That is, the intersection of all right prime ideals of $R$ is zero.
How to get to finite intersections? If we take a look at the set of all finite intersections of maximal right ideals of $R$, this set has a minimal element (because $R$ is right Artinian). Is it true that this element is $(0)$?
Best Answer
As the question stipulates, we are talking about right Artinian semiprime rings, which are precisely the semisimple Artinian rings.
It is true that the zero ideal is a finite intersection of maximal right ideals. To see this, note that the ring is a finite direct sum of minimal right ideals, and that the complements of these are maximal right ideals which intersect to zero.
It is not true that any finite intersection of maximal right ideals is zero. For example, if it is not a division ring, then a single maximal ideal does not create a zero intersection. It is also easy to find counterexamples using two or more right ideals of a square matrix ring.
In case you cannot assume this characterization of semisimple rings, then it can also be solved another elementary way (since you mentioned it has to be right Artinian.)
Let's accept the fact you alluded to that the Jacobson radical and prime radical coincide in a right Artinian semiprime ring.
Then we know the intersection of all maximal right ideals is zero. Suppose for a moment there is no finite collection of maximal right ideals equal to zero. Under this assumption, any finite intersection is nonzero. Furthermore, for any such collection, one can always add another maximal right ideal which eliminates elements when adding it to the intersection. (Explain why.). By induction, one can define a sequence of maximal right ideals $T_n$ such that $A_n=\cap_{i=0}^n T_n$ is a strictly decreasing sequence of submodules of $R$. That's a contradiction! (Explain why.)
You may then conclude that the original hypothesis (that no finite collection intersects to 0) was false.