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?

• Your chain of equalities claims that $[\mathbf{Q}(i, \sqrt[n]{p}) \, : \, \mathbf{Q}(\sqrt[n]{p})] = 2$, but you haven't shown that to be true (using the argument here, that claim is actually equivalent to the claim in the title). Aug 2 '20 at 22:19
• @BrandonCarter the field is real so it cannot have $i$, so the degree is $2$. Aug 2 '20 at 22:20
• You can still use Eisenstein's criterion even if $p$ itself is no longer prime. Aug 2 '20 at 22:21
• @Sorfosh: Sure, but you should mention that in the argument. Aug 2 '20 at 22:21
• @hardmath yes i know, i just do not know what prime ideal to pick here. Aug 2 '20 at 22:21

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.