Let R be Z adjoin infinitely many noncommuting indeterminants t1 t2 t3 etc. Let a string be a nonempty product of indeterminants with a coefficient from Z, and let an expression be a sum of strings. Thus the set of expressions is the ideal generated by the indeterminants.
Build a kernel K that is string representable, and look at the quotient ring Q = R/K. As we shall see, the adjoined variables generate upnil(Q), and lownil(Q) = 0.
If tl is the largest indeterminant contained in a string s, and s has length at least l, let s lie in K.
If tl is the largest indeterminant extant in an expression e, el lies in K. This because each string in the expansion of el lies in K. Thus el = 0 in Q. Every expression becomes nilpotent, and t1 t2 t3 etc generates a nil ideal in Q.
If e includes a constant, en includes that constant raised to the nth power. Thus e is not nilpotent in Q. Our variables generate the largest nil ideal in Q, which is upnil(Q).
Since t2 is not in K, upnil(Q) is nonzero. If lownil(Q) = 0, we have found our ring.
Suppose Q is not a semiprime ring. There is some nonzero expression u such that uvu = 0 for all v in Q. Pulling back to R, uvu lies in K for all v in R. If u contains strings in K, throw them out, one by one, which does not change the image in Q. Finally none of the strings in u lie in K.
Take the length of the longest string in u, add 1, double it, and set l to this number. Now utlu must lie in K. Expand the product, and every string contains tl, yet none of the strings are long enough. There is no u with uvu in K, and Q is semiprime. That completes the proof.
As a sanity check, note that Q is not commutative, nor is it noetherian. Keep adjoining variables to build a chain of ideals inside upnil(Q).
If you want an ascending chain of right annihilators, as required by the previous theorem, use the elements t2, t2t4t2, t2t4t2t8t2t4t2, etc. Turning to the annihilators, t1 kills the first element (on either side), t3 kills the second, t7 kills the third, and so on.
For each i > 0, map ti onto xiy. Now every string not in K is uniquely a string in the aforementioned indeterminants, with a power of x appended, and a possible y prepended. Use the same rules as above for folding strings of t1 t2 t3 etc into K. By closure, t3t1t2 in K implies yt3t1t2t8x7 is in K, and so on.
consider the ideal H generated by y in Z[x,y]/K. It includes x2y, the image of t2, and is nontrivial. Consider an expression e where everystring contains a y. Let s be one of these strings. If s = y then s2 = 0 in the quotient ring. If s = yx7 then s8 drops to 0, since it includes t77. Finally s is based on a string from t1 t2 t3 etc. If there are no pre and post modifications, then sl = 0, where tl is the largest variable. If a y is prepended we only need square s. If y is prepended and x7 is appended, we'll be introducing extra factors of t7. If there is no lead y, and s begins with t4 and ends with x7, we'll be introducing extra factors of t11. In every case, some power of s is 0, and some power of e is 0. Therefore H is a nil ideal.
If e contains some s without y, or if e includes a constant, the powers of e never reach K. Therefore H is the largest nil ideal, and the upper nil radical.
As before, the quotient ring is semiprime, and the lower nil radical is 0. If u is a power of x, with a possible y in front, then uxu is not in K. If u = y, then ux2u is not in K.
Let u be based on t1 t2 t3 etc, followed by a (possibly empty) power of x. Choose tl as we did above, and utlu is not in K. If u has a y prepended, use uxlu.
If s is part of a larger expression e, find v as above and expand e*v*e. At least one of the resulting strings does not lie in K, and since K is string representable, e*v*e is not in K. Therefore the quotient ring is semiprime, and the lower nil radical is properly contained in the upper nil radical; and the ring is finitely generated, based on 1 x and y.