Is $f$ irreducible in $\mathbb Z[x], \mathbb Q [x], \mathbb Z[i][x], \mathbb Q [i][x]$

Question

Let $f=2x^{32}+12x^{23}+18x^9+6 \in \mathbb Z[x]$. Is $f$ irreducible in $\mathbb Z[x], \mathbb Q [x], \mathbb Z[i][x], \mathbb Q [i][x]$

Let $a_i$ - coefficient for $x^i$. Then $a_0=6, a_9=18, a_{23}=12, a_{32}=2$.

For $p=3$ we have that $f$ is irreductible in $\mathbb Q(\mathbb Z)[x]$ from Eisenstein's criterion.

However I don't know how to consider task cases.

Answer 1

Part One: $\Bbb{Z}[x]$

Let $g(x)=\dfrac{f(x)}{2}=x^{32}+6x^{23}+9x^9+3$.

Note that $g(x)$ is irreducible over $\Bbb{Z}[x]$ by Eisenstein's criterion with the prime $p=3$.

The polynomial $2$ is also irreducible over $\Bbb{Z}[x]$ because the units of $\Bbb{Z}[x]$ are $\pm1$, and if you write $2=p(x)\cdot q(x)$ for some $p(x),q(x)\in\Bbb{Z}[x]$, then one of $p(x)$, $q(x)$ must be a unit.

So $f(x)$ is not irreducible over $\Bbb{Z}[x]$, but $f(x)=2\cdot g(x)$, and both $2$ and $g(x)$ are irreducible over $\Bbb{Z}[x]$.

Part Two: $\Bbb{Q}[x]$

Since $g(x)$ is primitive (i.e. the coefficients of $g(x)$ have no common factor) and irreducible over $\Bbb{Z}[x]$, it follows by Gauss's lemma that $g(x)$ is irreducible over $\Bbb{Q}[x]$.

Since $2$ is a unit in $\Bbb{Q}[x]$, it follows that $f(x)=2\cdot g(x)$ is irreducible in $\Bbb{Q}[x]$.

Part Three: $\Bbb{Z}[i][x]$

Part three will be similar to part one, because $3$ is prime in $\Bbb{Z}[i]$. This is based on the following characterization of the primes in $\Bbb{Z}[i]$:

Theorem: Let $a+bi\in\Bbb{Z}[i]$. Then $a+bi$ is a prime in $\Bbb{Z}[i]$ iff one of the following is true:

(1) $a=0$ and $b\equiv3$ modulo $4$,

(2) $b=0$ and $a\equiv3$ modulo $4$,

(3) $a\ne0$, $b\ne0$ and $a^2+b^2$ is a prime in $\Bbb{Z}$.

It follows by Eisenstein with the prime $p=3$ that $g(x)$ is irreducible over $\Bbb{Z}[i][x]$.

Note that in $\Bbb{Z}[i]$, we have that $2=(1-i)\cdot(1+i)$, and that $1-i$ and $1+i$ are prime in $\Bbb{Z}[i]$. Hence in $\Bbb{Z}[i][x]$ we can factor $f(x)$ into irreducibles as follws:

$$f(x)=(1-i)\cdot(1+i)\cdot g(x).$$

Part Four: $\Bbb{Q}(i)[x]$

Since $g(x)$ is primitive and irreducible over $\Bbb{Z}[i][x]$, it follows by Gauss's Lemma that $g(x)$ is irreducible over $\Bbb{Q}(i)[x]$.

Hence $f(x)=2\cdot g(x)$ is also irreducible over $\Bbb{Q}(i)[x]$.