Quadratic Number Rings

Suppose $4 + \sqrt{2} = \alpha \beta$ for irred. $\alpha, \beta$ then by the multiplicativity of the norm

$N(4+\sqrt{2}) = 14 = N(\alpha)N(\beta)$

So wlog $N(\alpha) = 14$ and $N(\beta) = 1$ or $N(\alpha) = 7$ and $N(\beta) = 2$

For $\gamma = a + b\sqrt{2}$, $N(\gamma) = a^2 -2b^2 = 1$ has no solutions for $a,b \in \mathbb{Z}$ (consider reduction by mod 4)

So in fact we must have $N(\alpha) = 7$ and $N(\beta) = 2$. So you want to find elements $\alpha, \beta$ with norms 7, 2 respectively (which automatically implies they're irreducible, why?) and the element of norm 7 is one you've already seen and there's an obvious element of norm 2.

Thank you so much for this!!

One question though, for: $\gamma = a + b\sqrt{2}$, $N(\gamma) = a^2 -2b^2 = 1$, does this not have solution a=1, b=0?

You're right, and indeed there are infinitely many $a, b$ solutions. The reasoning in this case should be "elements of norm 1 are invertible, and invertible things are by definition not irreducible". Compare the statement that in the integers, 1 is not prime (nor is it composite).

Then I want to do the same thing to $5-\sqrt{-2}$. But then the only option for the norms is 3 and 9. I'm only left with a and b (from the first line) having different signs to each other. But then the method doesn't seem to work from here. Is this because $5-\sqrt{-2}$ is irreducible, or because I have made a mistake in my logic? (The calculations should be right, I've triple checked them lol)

Assuming your calculations are correct:

Norm of 27 can be made only from 3x9. You've found all the things of norm 3 (namely $\pm (1 \pm \sqrt{-2})$): there are only two of them, up to multiplication by -1. You've presumably tried dividing by both of them, and found that neither of them comes out as elements of $\mathbb{Z}[\sqrt{-2}]$. That shows you that in fact there is no factorisation as norm-3 times norm-9, and hence that $5-\sqrt{2}$ is irreducible.

Thank you!

No problem I'm procrastinating as much as anyone else!

I'm really sorry to bother you again, but do you know how to prove this:

I can prove by counterexample because in $\mathbb Z[\sqrt{-3}]$ we have: $4=2 \cdot 2=(1+\sqrt{-3})(1-\sqrt{-3})$
But I get the feeling this isn't what the lecturer is after lol.


Edit: Ok I know how to do it now. You basically have to consider 2|1+e.
Sorry for all of the questions. The exam is at 9:30am tomorrow, so I won't be pestering you again with questions about this.

In fact it does hold for e=163, if you first show that the class number of sqrt(-163) is 1. (And similarly for a few smaller e, but none larger.)

Whaaaat? So what I have to prove is wrong???

Ok so I looked at class number on Wikipedia and it says that Z[sqrt(-3)] should have unique factorisation, but my example above shows it doesn't????


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Whaaaat? So what I have to prove is wrong???

Ok so I looked at class number on Wikipedia and it says that Z[sqrt(-3)] should have unique factorisation, but my example above shows it doesn't????

My reply above is updated: I was talking something close to nonsense. Your factorisation is correct, and Z[sqrt(-3)] is not a UFD. This is because -3 is 1 mod 4, so the ring of integers of Q(sqrt(-3)) is not Z(sqrt(-3)) after all.

Thank god.
Thanks again for all of the help.
Good luck in your exams btw (not that you need it though!)