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

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.

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Are you certain it is irreducible? What if you divide by $\sqrt{-2}$? –  trb456 Mar 5 '13 at 16:03
Just checked with my lecturer - I was wrong, it is reducible. I'll edit the question title accordingly. Thanks anyway! –  user49394 Mar 5 '13 at 16:09
@user49394, I posted an answer that shows how you could prove irreducibility .. but it goes wrong because the thing can be factored: but the way it goes wrong produces a factorization! –  user58512 Mar 5 '13 at 19:10

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 $x^2+2y^2$:

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

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

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

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I was trying to prove it was irreducible .. but then I noticed it factored.. I left this answer because it shows how to prove irreducible. –  user58512 Mar 5 '13 at 16:14
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.
Hint $\rm\,\ ad+b\sqrt{d}\, =\, \sqrt{d}\,(a\sqrt{d}+b)$