If every left multiple of x is nilpotent then the same is true for the left multiples of yx. In other words, the elements that generate nil left ideals can be scaled.
The sum of all left nil ideals is the same as the sum of all right nil ideals, i.e. the set of all x that generate one sided nil ideals. This set is an ideal, denoted onenil(R). Since onenil(R) contains every nil ideal, it contains upnil(R). The only catch is, nobody knows if it is nil. If it is, onenil(R) = upnil(R). This is equivalent to asserting every one sided left nil ideal lies in upnil(R).
If all left nil ideals lie in upnil(R) then the sum of any set of left nil ideals lies in upnil(R) and is nil.
Conversely, if the sum of finitely many left nil ideals is nil then the same holds for infinitely many left nil ideals, for each element lives in the sum of finitely many left nil ideals. The sum of all left nil ideals is a nil ideal, living in upnil(R). Yet onenil(R) contains upnil(R), so they are equal.
Another equivalent condition states every left nil ideal generates a nil ideal. If true, the sum of two left nil ideals is contained in the sum of the two generated nil ideals, which is nil, hence kothe's conjecture holds.
Conversely, if every left nil ideal is in upnil(R) then the ideal generated is in upnil(R), and is nil.
A ring that satisfies these conditions is called kotherian. Hey, if Noethe can lend her name to an adjective, noetherian, why not Kothe? Note that a commutative ring is kotherian, with lownil(R) = upnil(R) = onenil(R).
If H is a left nil ideal and every finitely generated left ideal inside H is nilpotent then H generates a nil ideal. Represent e as above, and note that e is in the span of a finitely generated nilpotent left ideal. Thus e lives in a nilpotent ideal, en = 0, and the ideal generated by H is nil.
If R has a finite characteristic, and the integer n is nil in R, n generates a nil ideal, which has already been divided out. Therefore the integers do not participate in any of the left nil ideals. The characteristic is not a power of n.
Let H be a left nil ideal. If H contains a left nilpotent ideal it generates a nilpotent ideal, which is a nil ideal, which was divided out. Therefore H does not contain any left nilpotent ideals.
Suppose H intersects every finitely generated subring of R in a left nilpotent ideal. As before, represent e as a linear combination of elements of H, with coefficients from R on the right. Build a subring U using 1, the elements of H extant in e, and the multiples of these elements in e, with the coefficients moved to the left. So if e employs h5x, xh5 contributes to U, along with h5, and 1. Take all finite sums of all finite products and U becomes a subring of R. The intersection of U and H is nilpotent. Raise e to the nth power and expand into a sum of products. Each term is the product of n elements drawn from U∩H, times something on the right. This is 0, hence een = 0, and H generates a nil ideal.
With uplow(R) divided out, there are no nil ideals. Therefore there exists a finitely generated ring U, having a left nil ideal H that is not nilpotent. If this cannot exist, then kothe's conjecture is true.
Change notation, so that R is the finitely generated ring formerly known as U, and H is H∩U. Multiply H by anything on the right and the result lies in H, thus H is a two sided ideal. The quotient R/H is Z, if R has characteristic 0, or Zn, if R has characteristic n. If R has characteristic p, R/H is a field, and H is a maximal ideal. Since H is nil it is in jac(R), and in every maximal ideal. Thus the maximal ideals are H adjoin p, for every prime p dividing the characteristic of R.
Let each of the generators of H span a principal left ideal in H. If these are all nilpotent then their sum, H, is nilpotent, which is a contradiction. Therefore there is a principal left nil ideal in H that is not nilpotent.
I'm not sure where to go from here, but I'm not alone. Last I heard, kothe's conjecture was still open.