Let K be a field and K[x] the polynomial ring over K in an indeterminate x. If P is a prime ideal of K[x], then there exists an irreducible polynomial f in K[x] such that P = K[x]f. This result is quite old and basic; however no corresponding result seems to be known for a polynomial ring in n indeterminates x1, ..., xn over K. Actually, it seems to be very difficult to find some system of generators for a prime ideal of K[xl,...,Xn]. Now, K [x1, ..., xn] is a Noetherian ring and by a converse of the principal ideal theorem for every prime ideal P of K[x1, ..., xn] there exist n polynomials fi, f..n such that P is minimal over (fi, ..., fn), the ideal generated by { fl, ..., f}n ([4], Theorem 153). Also, as a consequence of ([1], Theorem 1) it follows that any prime ideal of K[xl, X.., Xn] is determined by n polynomials. However it is not clear in [1] how to find these polynomials and no converse result is proved. The purpose of this paper is to show a result which in particular implies that every prime ideal of K[xi, ..., xn] is determined by a sequence of n polynomials which is in some sense irreducible, and such that the converse is also true. In general, let R be any prime ring. If P is a nonzero prime ideal of the polynomial ring R[x] with P n R = 0 (an R-disjoint prime ideal), then P = Q[x]fo n R[x] where Q is a ring of right quotients of R and fo E C[x] is an irreducible polynomial, C being the extended centroid of R ([2], Corollary 2.7). This characterization has the problem that we have to compute Q to have a prime ideal determined. A better way (an intrinsic one) to determine an R-disjoint prime ideal has been given in [3]. We proved that every R-disjoint prime ideal P of R[x] is determined by just one polynomial in P which is in some sense irreducible ([3], Theorem 1.4).