Let Z[X] be the ring of polynomials in one variable over Z.

My question: Is every prime ideal of Z[X] one of following types?

If yes, how would you prove this?

(0).

(f(X)), where f(X) is an irreducible polynomial.

(p), where p is a prime number.

(p,f(X)), where p is a prime number and f(X) is an irreducible polynomial modulo p.

**Answer**

Let P be a prime ideal of Z[x]. Then P∩Z is a prime ideal of Z: this holds whenever R⊆S are commutative rings. Indeed, if a,b∈R, ab∈R∩P, then a∈P or b∈P (since P is prime). (More generally, the contraction of a prime ideal is always a prime ideal, and P∩Z is the contraction of P along the embedding Z↪Z[x]).

Thus, we have two possibilities: P∩Z=(0), or P∩Z=(p) for some prime integer p.

**Case 1.** P∩Z=(0). If P=(0), we are done; otherwise, let S=Z−{0}. Then S∩P=∅, S is a multiplicative set, so we can localize Z[x] at S to obtain Q[x]; the ideal S−1P is prime in Q[x], and so is of the form (q(x)) for some irreducible polynomial q(x). Clearing denominators and factoring out content we may assume that q(x) has integer coefficients and the gcd of the coefficients is 1.

I claim that P=(q(x)). Indeed, from the theory of localizations, we know that P consists precisely of the elements of Z[x] which, when considered to be elements of Q[x], lie in S−1P. That is, P consists precisely of the rational multiples of q(x) that have integer coefficients. In particular, every integer multiple of q(x) lies in P, so (q(x))⊆P. But, moreover, if f(x)=rsq(x)∈Z[x], then s divides all coefficients of q(x); since q(x) is primitive, it follows that s∈{1,−1}, so f(x) is actually an *integer* multiple of q(x). Thus, P⊆(q(x)), proving equality.

Thus, if P∩Z=(0), then either P=(0), or P=(q(x)) where q(x)∈Z[x] is irreducible.

**Case 2.** P∩Z=(p).

We can then consider the image of P in Z[x]/(p)≅Fp[x]. The image is prime, since the map is onto; the prime ideals of Fp[x] are (0) and ideals of the form (q(x)) with q(x) monic irreducible over Fp[x]. If the image is (0), then P=(p), and we are done.

Otherwise, let p(x) be a polynomial in Z[x] that reduces to q(x) modulo p and that is monic. Note that p(x) must be irreducible in Z[x], since any nontrivial factorization in Z[x] would induce a nontrivial factorization in Fp[x] (since p(x) and q(x) are both monic).

I claim that P=(p,p(x)). Indeed, the isomorphism theorems guarantee that (p,p(x))⊆P. Conversely, let r(x)∈P(x). Then there exists a polynomial s(x)∈Fp[x] such that s(x)q(x)=¯r(x). If t(x) is any polynomial that reduces to s(x) modulo p, then t(x)p(x)−r(x)∈(p), hence there exists a polynomial u(x)∈Z[x] such that r(x)=t(x)p(x)+pu(x). Therefore, r(x)∈(p,p(x)), hence P⊆(p,p(x)), giving equality.

Thus, if P∩Z[x]=(p) with p a prime, then either P=(p) or P=(p,p(x)) with p(x)∈Z[x] irreducible.

This proves the desired classification.

**Attribution***Source : Link , Question Author : Makoto Kato , Answer Author : Xam*