Prove that $6x^5-55x^3 + 50x^2+15$ is irreducible over $\mathbb{Q}[i]$

Question

We need to prove that $6x^5-55x^3 + 50x^2+15$ is irreducible over $\mathbb{Q}[i]$. We let $f(x) = 6x^5-55x^3 + 50x^2+15$. We use Eisenstein's criterion. We know that $(5)$ is a prime ideal in $\mathbb{Q}[i]$. Then we have that $6 \notin (5)$. And $-55,50,15$ are all in $(5)$. And also $6 \notin (5)^2$. Then according to Eisenstein's criterion, this means that $f(x)$ is irreducible over $\mathbb{Q}[i]$. I'm wondering if my proof is correct, because I think that it's too short.
And if my proof is incorrect, could anyone point out the correct direction for me? Thanks!

Answer 1

Your idea of using Eisenstein's criterion was right. Let's start by showing that the polynomial is irreducible in $\mathbb{Z}[i][x]$.

First we need to show that the polynomial is primitive (i.e. it has no constant divisors). To do this we can consider the factorization of the coefficients. As shown here Gaussian primes are completely characterized so we have $$ 6=3 (1+i)(i-i) $$ $$ -55=- 5 \cdot 11=-(2+i)(2-i)11 $$ $$ 50=(2+i)^2(2-i)^2(1+i)(i-i) $$ $$ 15=3(2+i)(2-i) $$ so $\gcd(6,-55,50,15)=1$. Now it is immediate to see that $(2+i) \mid 15,50,-55$ and $(2+i) \nmid 6$ and $(2+i)^2 \nmid 15$ so by the Eisenstein's criterion the polynomial is irreducible in $\mathbb{Z}[i][x]$, so by Gauss lemma it is irreducible in $F[x]$, $F$ being the field of fraction of $Z[i]$, but such field is isomorphic to $Q[i]$, so the polynomial is irreducible in $Q[i][x]$.