Let J = jac(R) and consider the quotient ring R/J. Any maximal left ideal in R corresponds to a maximal left ideal in R/J, and these intersect in 0 in the quotient ring, so R/J is jacobson semisimple.
The quotient map induces a group homomorphism from the units of R onto the units of R/J. The kernel is the set of elements that map to 1, which is 1+z for any z in J.
Let H be a proper ideal in R. If x is in every maximal left ideal, then x is certainly in every maximal left ideal containing H. The image of x is in every maximal left ideal of R/H, and jac(R)/h is contained in jac(R/H).
Recall that a simple left R module is isomorphic to the cosets of a maximal left ideal in R. Let H be a maximal left ideal and let J be the jacobson radical, hence H contains J. Let y be the image of H in R/J, which is a left maximal ideal in R/J. Since maximal ideals in R/J correspond to maximal ideals in R that contain J, this map is 1-1 and onto. We can associate each simple module in R with a simple module in R/J. But are they the same modules?
The quotient map R/J induces an R module homomorphism from the cosets of H in R to the cosets of y in R/J. Verify that this is indeed a homomorphism, that the kernel is precisely H (the 0 coset), and that every coset of y is accessible, hence the homomorphism is an isomorphism. Finally, the action of R on the cosets of y is dictated by R/J, so we can make this an R/J module, and nothing changes. Therefore the simple modules of R are the simple modules of R/J.
Use this criteria to find jac(R) when R is the n×n upper triangular matrices over a division ring. Let H be the ideal with zeros down the diagonal. Note that Hn is always 0. Now R/H is the n×n diagonal matrices, Which is isomorphic to a vector space of dimension n. There are many maximal ideals, but we can build at least n, by forcing any one of the diagonal entries to zero and letting the others run free. The intersection of these ideals is 0, hence R/H is jacobson semisimple, and H = jac(R).
Let R be a dedekind domain with infinitely many prime ideals. If x is a nonzero element it generates an ideal with a unique (finite) factorization. Thus x is in a finite number of maximal ideals, and misses others. The intersection of all the maximal ideals is 0.
Assume the units of R, along with 0, produce a division subring of R. We know that J contains no units. If y in J is nonzero then 1-y is a unit by criteria 4, hence y is in our subring of units, which is a contradiction.
Polynomials or power series over a division ring have unit subrings, and are jacobson semisimple. This holds for any number of adjoined indeterminants, whether the indeterminants commute or not.
If the units of R form a division ring, we can mod out by an ideal to produce another jacobson semisimple ring, provided the quotient has no new units. For instance, start with F[x,y], where F is a division ring and x and y do not commute, then mod out by yx = xy+1, giving the weyl algebra. This relation does not destroy indeterminants, it just moves them around, so units must have no indeterminants. In other words, the units have not changed, and the algebra is jacobson semisimple.
A simple ring has only one proper ideal, namely 0, and this must be the jacobson radical.