Question

is not a principal ideal domain ​

Answer 1

Answer:

pls mark as brainlist answer

Step-by-step explanation:

Prove that Z[x] is not a principal ideal domain. ... Then we must have x = p(x)f(x) and 2 = p(x)g(x) for some f(x),g(x) ∈ Z[x]. But the second implies that p(x) must be a constant polynomial, specifically p(x) = −2, −1, 1 or 2. We can't have p(x) = ±1 because then 〈p(x)〉 = Z[x] so p(x) = ±2.