Proving that $ 2 + 3\sqrt{-2} $ is reducible in $ \mathbb{Z}[\sqrt{-2}] $

Question

Prove that $ 2 + 3\sqrt{-2} $ is irreducible in $ \mathbb{Z}[\sqrt{-2}] $

So far, I have let $ 2 + 3\sqrt{-2} = (a + b\sqrt{-2})(c+ d\sqrt{-2}) $

I then took the norm and got $\mathbf{N}(2 + 3\sqrt{-2}) = 22 = (a^2 + 2 b^2)(c^2 + 2 d^2) $

I think I must then split 22 into $ (2)(11) $ but I don't know how to proceed from there.

Help is much appreciated!

Note: I originally posed the question as proving it was *ir*reducible. Apologies if I sent people down the wrong track in the answers below! Thank you again for the help.

Answer 1

Let $ 2 + 3\sqrt{-2} = (a + b\sqrt{-2})(c+ d\sqrt{-2}) $ then $\mathbf{N}(2 + 3\sqrt{-2}) = 22 = (a^2 + 2 b^2)(c^2 + 2 d^2)$

That is a good way to start!

Now we just need to show $a^2+2b^2 = 2$ and $c^2+2d^2 = 11$ is impossible, but the first part is possible so we need to show $c^2+2d^2 = 11$ is impossible:

This is easy, let's just write out all numbers of the form $x^2+2y^2$:

$$\begin{array}{|c|c|c|} \hline 0 & 2 & 8 & 18 \\ \hline 1 & 3 & 9 & 19 \\ \hline 4 & 6 & 12 & \\ \hline 9 & \color{red}{11} & & \\ \hline \end{array}$$

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So we have a factorization from the $x=1,y=3$ box which is $3^2 + 2\cdot 1^2 = 11$.

$$2(3+\sqrt{-2})(3-\sqrt{-2})$$

Answer 2

Find the elements in $\mathbb{Z}[\sqrt{-2}]$ of norm $2$ (they are $\pm i \sqrt{2}$) and $11$ (they are $\pm 3 \pm i \sqrt{2}$), and check whether one of the possibile products will give you $2 + 3\sqrt{-2}$. (Of course @trb456 has already give you a hint.)

Then if you want you may use the fact that if the norm of an element is a prime integer, then the element is irreducible. This will show that the two factors that you have found are irreducible, so $2 + 3\sqrt{-2}$ is the product of two irreducibles.

Answer 3

Hint $\rm\,\ ad+b\sqrt{d}\, =\, \sqrt{d}\,(a\sqrt{d}+b)$