Remember that nil(R) is the intersection of all prime ideals, and since maximal ideals are prime, jac(R) contains nil(R).
A ring is jacobson semisimple if jac(R) = 0. This implies a reduced ring, with nil(R) = 0.
Let G be any ideal in R, and let H be all polynomials with coefficients in G. Verify that H is an ideal in S, and H is the extension of G into S.
If P is a prime ideal in R containing G, let Q be the extension of P into S. Note that Q contains H. We need to show that Q is a prime ideal in S.
Let u and v be polynomials not in Q, yet uv lies in Q, and u and v have the lowest possible degrees. One of the two lead coefficients lies in P, say it is the lead coefficient of u. Subtract away the first term of u, and u still lies outside of Q, while uv remains in Q. This is a contradiction, hence Q is prime.
If u is a polynomial in rad(H) then it is in each such Q. Its coefficients are in P for each prime P containing G. Thus rad(H) is contained in the extension of rad(G).
Now let Q be an arbitrary prime ideal in S containing H. Contract back to R by looking at the constant terms of Q. This is a prime ideal P in R, containing G. Now P generates a prime ideal inside Q, so we may as well ratchet Q back to the extension of P. If the coefficients of u are in P for every P containing H, then u is in the intersection of each such Q, and u is in rad(H). In other words, the extension of rad(G) is contained in rad(H). The two sets are equal: rad(ext(G)) = ext(rad(G)).
Set G = 0, and ext(nil(R)) = ext(rad(0)) = rad(ext(0)) = nil(S).
We know the jacobson radical always contains the nil radical. We want to show that a polynomial u that is not nilpotent, not in nil(S), is not in jac(S) either. That will prove nil(S) = jac(S).
Let u be a polynomial that is not nilpotent in S, and consider 1-xu, where x is one of the indeterminants of S. In the previous theorem we characterized the units of S. The constant term must be a unit and all other coefficients must be nilpotent. Thus the coefficients of u must be nilpotent. This places u in the extension of nil(R), which is nil(S). Yet u does not lie in nil(S), so 1-xu is not a unit. By characterization 2, u is not in jac(S). This completes the proof; nil(S) = jac(S).
If nil(R) = 0, as when R is jacobson semisimple, then S is jacobson semisimple. But the converse is not true. Let R be a dvr, with one maximal ideal, which is its jacobson radical. It is not jacobson semisimple. Yet nil(R) = 0, hence jac(S) = 0, and S is jacobson semisimple.