Exercise 15.4.12

Proof. Following the similar proof of Theorem 4.2.3 and Corollary 4.2.1, we reason by contradiction. Suppose that $f$ is not irreducible over $\mathbb{Q}(i)$. Then $f=\mathit{vw}$ where $v,w\in \mathbb{Q}(i)[u]$ have degrees less than $d=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)$. By Gauss’ s Lemma over $\mathbb{Z}[i]$, which holds by Theorem A.5.8 because $\mathbb{Z}[i]$ is a UFD, with quotient field $\mathbb{Q}(i)$, there is $\delta \in \mathbb{Q}(i)$ such that $g=\mathit{\delta v},h={\delta }^{-1}w$ are in $\mathbb{Z}[i][u]$. Then

Now consider the ring homomorphism

where $[b]=b+\mathit{\beta \mathbb{Z}}[i]\in \mathbb{Z}[i]∕\mathit{\beta \mathbb{Z}}[i]$ is the coset of $b\in \mathbb{Z}[i]$ modulo the ideal $\mathit{\beta \mathbb{Z}}[i]$.

Then $f=\mathit{gh}$ implies that $[a_0]u^d=\bar{g}\bar{h}$, since $\beta \mid a_1,\dots ,\beta \mid a_d$. However, $\beta$ being prime, $\mathbb{Z}[i]∕\mathit{\beta \mathbb{Z}}[i]={𝔽}_{N(\beta )}$ is a field, which means that unique factorization holds in ${𝔽}_{N(\beta )}[u]$. Since $\beta \nmid a_0$, it follows that $\bar{g}=[a]u^r$ and $\bar{h}=[b]u^s$, where $[a][b]=[a_0]$ and $r+s=d$.

If $r=0$, then $\bar{g}=[a]$ and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)>0$ would imply that the leading term of $g$ is divisible by $\beta$. Then $f=\mathit{gh}$ would imply that the same is true for the leading term $a_0$ of $f$. Thus $\beta \nmid a_0$ implies that $r>0$, and $s>0$ follows similarly.

But then $\bar{g}=[a]u^r$ for $r>0$ implies that $\beta$ divides the constant term of $g$, and the same is true for the constant term of $h$, since $s>0$. Since the constant term $a_d$ of $f$ is the product of the constant terms of $g$ and $h$, it follows that ${\beta }^2\mid a_d$. This contradicts ${\beta }^2\nmid a_d$ and completes the proof. □