Suppose that $p,q \in \mathbb{N}$ are prime numbers and $\alpha \in \mathbb{Z[i]}$ satisfy $N(\alpha) = pq$, $p$$3$ mod $4$ and $q$$3$ mod $4$. Using the fact that $p$ and $q$ remain irreducible in $\mathbb{Z[i]}$, prove that $p = q$.

Proof attempt: Assume false, that is, $p \neq q$. As $N(\alpha) = \alpha\alpha^* = pq$, we have $p \mid \alpha\alpha^*$ in $\mathbb{Z[i]}$. Given that $p$ is congruent to $3$ modulo $4$, we have $p$ is irreducible in $\mathbb{Z[i]}$. Then $p$ is prime implying $p \mid \alpha$ or $p \mid \alpha^*$ in $\mathbb{Z[i]}$. Taking norms, we get in both cases that $p^2 \mid pq$ in $\mathbb{Z}$. As $p$ and $q$ are primes in $\mathbb{Z}$, we get $p = q$. Hence a contradiction so $p = q$.

This is my attempt. I'm not certain it's correct. The reason being I haven't used the fact that $q$ is congruent to $3$ modulo $4$.


1 Answer 1


I think a proof is okay and a more general fact holds: since the norm is multiplicative, then $s | x$ in $\mathbb{Z}[i]$ implies $x = sy$ for some $y$ and

$$ N(x) = N(s)N(y), $$

so that $N(x)| N(y)$.

If $p$ is a Gauß prime, then $p \mid N(x)= xx^\ast$ implies $p \mid x$ and thus $p^2 = N(p) \mid N(x)$. In your case, we have $p^2 \mid N(x) = pq$, hence $q = p$ by uniqueness of prime factorization (in $\mathbb{Z}$).

In general, gaussian primes are of the form $pi$ and $p$ with $p \in \mathbb{4\mathbb{N}_0}+3$ or $a+ib$ such that $ab \neq 0$ and $N(a+ib)$ is prime in $\mathbb{Z}$. Thus, if $x$ has a certain prime factorization

$$ x = p_1^{n_1} \cdots p_s^{n_s}f_1^{m_1} \cdots f_r^{m_r} $$

in $\mathbb{Z}[i]$, with each $p_j \in 4\mathbb{Z}_0 +3$ and $q_i = f_i^2 \in \mathbb{N}$ primes, taking norms we get

$$ N(x) = N(p_1)^{n_1} \cdots N(p_s)^{n_s}N(q_1^{m_1})\cdots N(q_r^{m_r}) = p_1^{2m_1} \cdots p_s^{2n_s}f_1^{m_1} \cdots f_r^{m_r}. $$

This shows that gaussian primes appearing in the factorization of $N(x)$ for some $x \in \mathbb{Z}[i]$ always have even exponent.

Going back to the problem of the question, since $N(x) = pq$, either $p = q$ or no Gaussian primes divide $x$. But since $p \equiv 3 \pmod{4}$, the former is true. No extra hypotheses on $q$ are needed.


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