Prove $x^n-p$ is irreducible over $Z[i]$ where $p$ is an odd prime. Prove $x^n-p$ is irreducible over $Z[i]$ where $p$ is an odd prime.
By gausses lemma this is equivalent to irreducability over $\mathbb{Q}(i)$. Using field extensions this is easy. $[\mathbb{Q}(i,\sqrt[n]{p}):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]=[\mathbb{Q}(i,\sqrt[n]{p}):\mathbb{Q}(\sqrt[n]{p})][\mathbb{Q}(\sqrt[n]{p}):\mathbb{Q}]=2n$ Thus $[\mathbb{Q}(i,\sqrt[n]{p}):\mathbb{Q}(i)]=n$ and so $x^n-p$ must be the minimal polynomial, and so it is irreducible. However, the book says you can solve this problem using Eisenstein criterion. That is easy when $x^2+1$ is irreducible mod $p$ as $(p)$ is then prime. What do you do in the other cases?

To prove this via Eisenstein's criterion, use the fact that $\mathbb Z[i]$ is a principal ideal domain. In fact, it is Euclidean. Also, for odd primes $p$ in $\mathbb Z$, $p$ remains prime in $\mathbb Z[i]$ for $p=3\mod 4$, and $p$ factors as a product of two distinct primes $p=p_1p_2$ in $\mathbb Z[i]$ for $p=1\mod 4$. (No, this is not obvious...)
Thus, in either case, there is a prime in $\mathbb Z[i]$ dividing the constant term (and all others except the leading term, since those others are $0$), and whose square does not divide the constant term. So we can apply Eisenstein.
Appropriate proof(s) about the remaining-prime and/or factoring into distinct factors depend on your context... but the explanation may give you some motivation to look at otherwise-technical points.
