Exercise 1.3.47* ($2 + \sqrt{-6}$ is irreducible.)

Question

Prove that $2+\sqrt{-6}$ and $2 - \sqrt{-6}$ are primes in the class $\mathscr{C}$ of numbers $a +b \sqrt{-6}$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} We choose $\sqrt{-6} = i \sqrt{6}$.

Let $\mathscr{C}$ be the ring
$$\mathscr{C} = \{a + ib \sqrt{6} \mid a \in \Z, b \in \Z\}.$$
Then
$$N(2 + i\sqrt{6}) = (2 + i\sqrt{6})(2 - i\sqrt{6}) = 4 + 6 = 10.$$
By (1.2), we know that the norm of any nonreal number in $\mathscr{C}$ is no less that $6$:
$$N(a+ib \sqrt{6}) = a^2 + 6 b^2 \geq 6 \qquad 	ext{if } b 
e 0.$$
If $2+ i \sqrt{6} = (a + ib\sqrt{6})(c+id\sqrt{6})$, then 
\begin{align}
10 = N(a + ib\sqrt{6}) N(c+id\sqrt{6}).
\end{align}
If $b
e 0$ and $d
e 0$, then $10 \geq 6	imes 6 = 36$: this is a contradiction. Therefore $b=0$ or $d = 0$. If $b=0$, then $a^2 \mid 10$. Thus $|a| \leq \sqrt{10}$, so $|a| \leq 3$. But $2^2 
mid 10, 3^2 
mid 10$, hence $a^2 = 1$ and $a+ i b\sqrt{6} = \pm 1$ is a unit. Similarly, if $d = 0$, $c+id\sqrt{6}$ is a unit.

This shows that (1) is possible only if $a + ib\sqrt{6}$ or $c+id\sqrt{6}$ is a unit. Therefore $2 + i\sqrt{6}$ is irreducible (Niven, Zuckerman call such an element a prime.)

The same reasoning shows that $2 - i\sqrt{6}$ is irreducible.
\end{proof}

Note: Write $p =(2 + i\sqrt{6}), a = 2, b = 5$. Since $2 \cdot 5 = (2 + i\sqrt{6})(2 - i\sqrt{6})$, 
$$p \mid ab,\qquad 	ext{but } p
mid a, p
mid b,$$
since $N(p) = 10$ doesn't divide neither $4 = N(a)$ nor $25 = N(b)$.

So $p$, which is not a unit, is not a prime in the sense $p \mid ab \Rightarrow p\mid a 	ext{ or } p \mid b$ \quad ($a,b \in \mathscr{C}$).

Exercise 1.3.47* ($2 + \sqrt{-6}$ is irreducible.)

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Exercise 1.3.47* ($2 + \sqrt{-6}$ is irreducible.)

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