Classification of prime ideals of Z[X]\mathbb{Z}[X]

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?

  1. (0).

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

  3. (p), where p is a prime number.

  4. (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 PZ is a prime ideal of Z: this holds whenever RS are commutative rings. Indeed, if a,bR, abRP, then aP or bP (since P is prime). (More generally, the contraction of a prime ideal is always a prime ideal, and PZ is the contraction of P along the embedding ZZ[x]).

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

Case 1. PZ=(0). If P=(0), we are done; otherwise, let S=Z{0}. Then SP=, S is a multiplicative set, so we can localize Z[x] at S to obtain Q[x]; the ideal S1P 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 S1P. 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 PZ=(0), then either P=(0), or P=(q(x)) where q(x)Z[x] is irreducible.

Case 2. PZ=(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 PZ[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

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