Let $N(\alpha) = pq$ where $p,q$ are distinct primes. Show that $\alpha$ is the product of exactly two irreducibles in the Gaussian integers

Question

I've managed to prove that $\alpha$ is at most 2 irriducibles, however, I'm really struggling to show that there are exactly two. I know this has been posted before, but could anyone provide some hints?

Answer 1

Consider $I:=\alpha\cdot \Bbb Z[i]+p\cdot\Bbb Z[i]$. As $\Bbb Z[i]$ is a principal ideal domain, we have $I=\beta\cdot\Bbb Z[i]$ for some $\beta$. But then $N(\beta)\mid N(\alpha)=pq$ and $N(\beta)\mid N(p)=p^2$, hence $N(\beta)\mid p$. But $\beta\in I$ also implies $p\mid N(\beta)$ (as $(a\alpha+bp)\cdot\overline{a\alpha+bp}=N(a)pq+(\ldots)\cdot p$. We conclude that $\alpha$ is divisible by the non-unit $\beta$. Thus $\alpha$ is the product of at least two irreducibles. But if $\alpha$ were the product of three or more non-units, so would be $N(\alpha)$. Hence $\alpha$ is the product of exactly two irreducibles.

Answer 2

$p$ is irreducible <=> $p$ is prime.

Suppose $\alpha$ is the product of 1 irreducible.

=> $\alpha$ is a prime.

$$\alpha \mid N(\alpha)$$ $$\alpha \mid pq $$

since $\alpha$ is prime,

$$\alpha \mid p \text{ OR } \alpha \mid q $$

If $\alpha \mid p$,

$$N(\alpha) \mid p^2$$ $$pq \mid p^2$$ So we have a contradiction. Similary with $q$.

So $\alpha$ is not irriducible, so $\alpha$ is the product of at least 2 irreducibles. Proving it is equal to 2 irreducible's is simple.

Answer 3

The ring $A=\mathbf Z[i]$ is a UFD, with units $\pm 1,\pm i$. The decomposition of a rational prime $p$ into a product of irreducibles of $A$ is classically known: $(p)=(\pi)$ (inert), $(p)= (\pi .\pi')$ (split) or $(p)=(\pi^2)$ (ramified) . Note that in the split case, $\pi$ and $\pi'$ must be complex conjugates because of the uniqueness (up to units) of the decomposition, so that these two irreducibles are "attached to the same $p$" (up to units).

For $\alpha \in A$, for any irreducible factor $\pi$ of $\alpha$, $N(\pi)$ must divide $N(\alpha)$. In view of the above, the hypothesis $N(\alpha)=pq$, where $p, q$ are two distinct rational primes, implies obviously that $p, q$ must be unramified, and $\alpha$ is the product of exactly two irreducibles (up to units).