The K/C criterion is necessary. If K is a field, equal to its center, then K[x] is commutative, and a commutative primitive ring is a field. Yet K[x] is not a field, illustrating the need for the transcendental element over C.
Synthetic division shows K[x] is a left euclidian domain. Any two polynomials in R have a right gcd, and that gcd is a linear combination of the two polynomials. This result extends to finitely many polynomials. They have a common right gcd, and they generate that gcd. Apply Hilbert's basis theorem, and R is noetherian. Every left ideal is finitely generated. Its generators span, and are generated by, the gcd. Therefore H is principal, and R is a left pid.
A linear polynomial generates a maximal left ideal. Let M be the quotient moduleR/(x-a), where a ∈ K is transcendental over C. Because the kernel is maximal, M is a simple left module. Show M is faithful, and R is left primitive.
Suppose H is a nonzero ideal that kills M from the left, hence H is contained in R*(x-a). Naturally H is a left ideal, and it is principal, generated by g.
Since H is two sided, gR lies in Rg. Note that g has the least degree of all polynomials in H, and there is no harm in making g monic. Given u in R, what v in R satisfies ug = gv? Note that u and v have the same degree. Since g is monic, the lead coefficients of u and v must agree. Subtract the leading terms from u and from v, and we still have ug = gv. Again, the lead coefficients agree. Repeat, and conclude u = v. Thus g is in the center of R. This means all the coefficients of g lie in C.
Since H lies in the maximal left ideal generated by x-a, g = p*(x-a) for some polynomial p. It would be nice to substitute x = a, hence g(a) = 0, and a is algebraic over C, but substitution and multiplication are not always compatible in noncommutative rings. Let me illustrate with p = x-b. Now g = x2 - (b+a)x + ba. Do you see what I did? I replaced xa with ax, since K and x comute. If x is replaced with an arbitrary element w from K, w and a might not commute. Thus g(w) might not equal p(w)*(w-a). But a and a do commute, so it works for a. That means g(a) = p(a)*(a-a) = 0, and a is algebraic over C. This contradicts the selection of a. Therefore H does not exist, M is faithful, and R is left primitive.
Let M be R mod (x-a)R, and R is right primitive.
To see if this is an R isomorphism, we need to examine right multiplication in the ring R. That is, R acts on R, and on M2 through the quotient map. Let's see what happens when x passes from M1 to M2, and is multiplied by b on the right. We have xb = bx = bx-ba2+ba2 = b*(x-a2)+ba2 = ba2.
An R homomorphism has to respect scaling by x. That is, we can apply x before or after the homomorphism. Now x*1 in M1 is a1. This is multiplied on the right by b, and that has to agree with x acting on M2, which is xb, or ba2. Therefore a1b = ba2, and a1 and a2 are conjugate courtesy of b.
One can construct a division ring K where this does not happen. Start with a field C and adjoin two noncommuting indeterminants, I may as well call them a1 and a2, and embed this ring into a division ring K. Yes, this can be done, though that is not obvious. Both a1 and a2 are transcendental over C, and they produce simple faithful modules M1 and M2. Are a1 and a2 conjugate? An element of K looks like an infinite sum of terms, where each term is a word in the free group generated by a1 and a2, with a coefficient from C. Words can be ordered lexicographically, so that we can talk about the least term in the series. If b is such a series, then the least term of a1b corresponds to the least term of ba2. The other terms line up thereafter. So think of b as a single term, a word in the free group on two letters. Prepending one letter, or appending the other, gives the same result. By looking at lengths, a1 cancels the first letter of b iff a2 cancels the last letter. Look at these two cases, and derive a contradiction in each case. Since the least terms cannot agree, a1 and a2 are not conjugate by anything in K, and M1 ≠ M2.