1. ## is ideal prime?

Is ideal $(1+3i, 3+11i, -1-11i)$ prime in $\mathbb{Z}[i]$?

$\mathbb{Z}[i]$ is Euclidean doman, so I have to check whether $(1+3i, 3+11i, -1-11i)$ is an irreducible element?
How do I do this?

2. Here's what I did so far:

I try to find GCD $(1+3i, 3+11i, -1-11i)$
.

Firstly, I find GCD $(1+3i, 3+11i)$

$N(1+3i)=10$

$N(3+11i)=130$

$3+11i=(1+3i)*4 -1-i$

$1+3i=(-1-i)*(-2-i)+0$

GCD $(1+3i, 3+11i)=-1-i$

Next,
GCD $(-1-i, -1-11i)$
$-1-11i=(-1-i)*(6-5i)+0$
So
GCD $(-1-i, -1-11i)=-1-i$

Now I check is (-1-i) irreducible.

$N(-1-i)=1+1=2$
$2=(a^2+b^2)(c^2+d^2)$

$(a^2+b^2)\in\{+-1, +-2\}, (c^2+d^2)\in\{+-1, +-2\}$

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.

3. Originally Posted by georgel
Here's what I did so far:

I try to find GCD $(1+3i, 3+11i, -1-11i)$ .

Firstly, I find GCD $(1+3i, 3+11i)$

$N(1+3i)=10$

$N(3+11i)=130$

$3+11i=(1+3i)*4 -1-i$

$1+3i=(-1-i)*(-2-i)+0$

GCD $(1+3i, 3+11i)=-1-i$

Next, GCD $(-1-i, -1-11i)$
$-1-11i=(-1-i)*(6-5i)+0$ (*)
So GCD $(-1-i, -1-11i)=-1-i$

Now I check is (-1-i) irreducible.

$N(-1-i)=1+1=2$
$2=(a^2+b^2)(c^2+d^2)$

$(a^2+b^2)\in\{+-1, +-2\}, (c^2+d^2)\in\{+-1, +-2\}$

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.

it's all correct except for (*): we actually have $-1-11i=(-1-i)(6+5i).$ also it's easier to work with $a+bi$ instead of $-a-bi$ when you're working in an ideal.

also note that if $N(a+bi)$ is a prime number, then $a+bi$ is always irreducible.

4. Originally Posted by georgel
Is ideal $(1+3i, 3+11i, -1-11i)$ prime in $\mathbb{Z}[i]$?

$\mathbb{Z}[i]$ is Euclidean doman, so I have to check whether $(1+3i, 3+11i, -1-11i)$ is an irreducible element?
How do I do this?

We can factor into irreducibles:
$1+3i = (1+i)(2+i)$
$3+11i = -i(1+i)(1+2i)(2+3i)$
$-1-11i = -i(1+i)(5-6i)$

Notice that $1+i$ is a common irreducible factor so $(1+3i,3+11i,-1-11i) = (1+i)$.

If you are unsure how I got those factorizations I can show you.

5. Originally Posted by NonCommAlg
it's all correct except for (*): we actually have $-1-11i=(-1-i)(6+5i).$ also it's easier to work with $a+bi$ instead of $-a-bi$ when you're working in an ideal.

also note that if $N(a+bi)$ is a prime number, then $a+bi$ is always irreducible.
Thank you! I'm pretty happy with the fact that that's all I messed up )

Originally Posted by ThePerfectHacker
We can factor into irreducibles:
$1+3i = (1+i)(2+i)$
$3+11i = -i(1+i)(1+2i)(2+3i)$
$-1-11i = -i(1+i)(5-6i)$

Notice that $1+i$ is a common irreducible factor so $(1+3i,3+11i,-1-11i) = (1+i)$.

If you are unsure how I got those factorizations I can show you.
If it's not a problem, I'd be really grateful! We haven't done it in class, and I wasn't able to Google anything useful.

6. Originally Posted by georgel
If it's not a problem, I'd be really grateful! We haven't done it in class, and I wasn't able to Google anything useful.
Let me find the factorization for $3+11i$ with the approach I used. You need to know some stuff about the arithmetic in the Gaussian integers. The first one is that if $a+bi$ with $a,b\not = 0$ is irreducible then it means be the case that $N(a+bi) = a^2+b^2$ is a prime number in $\mathbb{Z}$. Since $3+11i$ is of the form $a+bi$ where $a,b\not = 0$ it means all its factors cannot be real or pure imaginary i.e. they must have the same form $a+bi$ where $a,b\not = 0$. Write $3+11i = \pi_1 \pi_2 ... \pi_k$ where $\pi_j$ are irreducibles (and are not necessarily distinct) in $\mathbb{Z}[i]$ which are neither purely real or imaginary. We see that $N(3+11i) = N(\pi_1) ... N(\pi_k)$ and so $130 = N(\pi_1) ... N(\pi_k)$. Notice that $130 = 2\cdot \cdot 5\cdot 13$. Also, each $N(\pi_j)$ must be a prime number as explained. This forces $k=3$ and the $\pi_1,\pi_2,\pi_3$ can be relabeled so that $N(\pi_1) = 2, N(\pi_2) = 5, N(\pi_3) = 13$.

Now we need to use some more arithmetic knowledge about $\mathbb{Z}[i]$. The only prime (up to associates) with $N(\pi_1) = 2$ is $\pi_1 = 1+i$. However, with $N(\pi_2)=5$ there are two primes (not associates) so that have their norm equal to $5$. This is in fact true in general if $N(\pi) = p$ where $p$ is an odd prime in $\mathbb{Z}$ then $\pi, \bar \pi$ both have norm $p$ and they are not associates. This does not happen with the even prime $p=2$ because $\overline{1+i} = 1-i = -(-1+i) = -(i^2+i) = -i(1+i)$ and so the conjugate of $1+i$ is associate with $1+i$. Conjugates turn out being non-associate if the norm is odd, not even. So we need to find a prime $\pi_2 = a+bi$ so that $a^2+b^2 = 5$. The good news is that if a Gaussian integer has norm a prime it is automatically irreducible so we just need to find such $a,b$. It is clear that $a+bi = 1+2i$ works, which tells us that other non-associate prime that has norm $5$ is $\overline{a+bi} = 1-2i$. Same idea with $N(\pi_3) = 13$ it is easy to see that $2+3i$ works, which tells us the other non-associate prime is $2-3i$.

Therefore, one of these have to hold:
$3+11i = u(1+i)(1+2i)(2+3i)$
$3+11i = u(1+i)(1-2i)(2+3i)$
$3+11i = u(1+i)(1+2i)(2-3i)$
$3+11i = u(1+i)(1-2i)(2-3i)$
Where $u$ is a unit i.e. $u\in \{1,-1,i,-i\}$ which corrects the primes to propely associated with $3+11i$.

If we perform the computations (for the first one) we get:
$3+11i = u(-11 + 3i)$, now we see that $u=-i$.
Thus, $3+11i = -i(1+i)(1+2i)(2+3i)$.
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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 $1+11i$ !

7. I printed it and am now looking into it.
Thank you so much for your effort!