Is ideal prime in ?
is Euclidean doman, so I have to check whether is an irreducible element?
How do I do this?
Thank you for your help.
Here's what I did so far:
I try to find GCD .
Firstly, I find GCD
GCD
Next, GCD
So GCD
Now I check is (-1-i) irreducible.
So in any case, one of them is +-1, which means that (-1-i) is irreducible, so (1+3i, 3+11i, -1-11i) is generated by an irreducible element so it is prime.
I'm pretty sure I made mistakes, could someone please correct me?
Let me find the factorization for with the approach I used. You need to know some stuff about the arithmetic in the Gaussian integers. The first one is that if with is irreducible then it means be the case that is a prime number in . Since is of the form where it means all its factors cannot be real or pure imaginary i.e. they must have the same form where . Write where are irreducibles (and are not necessarily distinct) in which are neither purely real or imaginary. We see that and so . Notice that . Also, each must be a prime number as explained. This forces and the can be relabeled so that .
Now we need to use some more arithmetic knowledge about . The only prime (up to associates) with is . However, with there are two primes (not associates) so that have their norm equal to . This is in fact true in general if where is an odd prime in then both have norm and they are not associates. This does not happen with the even prime because and so the conjugate of is associate with . Conjugates turn out being non-associate if the norm is odd, not even. So we need to find a prime so that . The good news is that if a Gaussian integer has norm a prime it is automatically irreducible so we just need to find such . It is clear that works, which tells us that other non-associate prime that has norm is . Same idea with it is easy to see that works, which tells us the other non-associate prime is .
Therefore, one of these have to hold:
Where is a unit i.e. which corrects the primes to propely associated with .
If we perform the computations (for the first one) we get:
, now we see that .
Thus, .
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This computation is not the most pleasant thing to do but I think it is a lot faster than just dividing out the primes and trying to look for factors because this approach eliminates guesswork. It just reduces the problem into cases.
You try doing !