Suppose P is nonzero. Choose a nonzero polynomial f(x) from P having the smallest possible degree. Since P contains no nonzero integers, it follows that deg(f) > 1.

We may assume that f(x) is primitive. That is, f(x) has content one [c(f) = 1]. The reason: write f(x) = c(f)f₁(x) with f₁(x) primitive. Since P is prime and contains no nonzero integers, it follows that f₁(x) is in P.

We claim that P = (f):

Let g(x) be in P. Factor into irreducibles:

g = g₁g₂g₃^{. . .}g_n

Since P is prime, it follows that g_i is in P for some i = 1, 2, 3, ..., n.

By the division algorithm in Q[x], there exist q(x) and r(x) in Q[x] such that

g_i(x) = f(x)q(x) + r(x),

where r(x) = 0 or r(x) is nonzero and deg(r) < deg(f).

Suppose that q(x) is a polynomial in Z[x]. Then since r(x) = g_i(x) - f(x)q(x), it follows that r(x) is in P. Thus, r(x) = 0 since f(x) has minimum degree. This gives g_i(x) = f(x)q(x) and so g is in the principal ideal (f).

On the other hand, suppose that q(x) is not a polynomial in Z[x]. Taking the LCD of the coefficients of q(x), we obtain a nonzero integer n such that nq(x) is in Z[x]. Thus, nr(x) = ng_i(x) - nf(x)q(x) and so nr(x) is in P. Whence, r(x) = 0 and so g_i(x) = f(x)q(x).

Let m = c(nq) [i.e., the content of nq(x)]. Then ⁿ/_mq(x) has integer coefficients and is primitive.

We have ⁿ/_mg_i(x) = ⁿ/_mq(x)f(x). Taking contents of both sides, ⁿ/_mc(g_i(x)) = c(ⁿ/_mq(x))c(f) = (1)(1) = 1.

Since g_i is irreducible and noninteger, it follows that c(g_i) = 1. So ⁿ/_m = 1. Hence q(x) is in Z[x], contradiction.

REMARK: This is rather like the proof that every Euclidean domain is a PID.