Factorize $ 77 $ in $ \mathbb{Z}[\sqrt{-13}]$ I have to show that there are three different factorization of $ 77 $ in the above ring. It's not hard to find them, i.e. $ 77 = 11 \cdot 7 = (8 - \alpha)(8 + \alpha) = (5 - 2 \alpha)(5 + 2 \alpha) $, where $ \alpha^2 = -13 $. It's also not hard to show that these are different factorization, i.e. cannot be obtained from one another by multiplication of units. However, how do I show that there are not more than three? 
I thought I could use the factorization algorithm to obtain $$ (77) = (7, \alpha + 1)(7, \alpha - 1)(11, \alpha + 3)(11, \alpha - 3). $$
I want to show that e.g. $ (7, \alpha + 1) (11, \alpha + 3) $ is principal, but I don't know how to do this. I got something like $ (1, \alpha) $ but that doesn't seem right. I guess my question is, how do you obtain the principal element generating a product of two ideals. Your help is greatly appreciated.
 A: Ideals are useful and important, but in this case I think it would be much simpler to just use the norms of numbers. As you know, $N(a + b \sqrt{-13}) = a^2 + 13b^2$. So we're looking to solve $a^2 + 13b^2 = 7, 11, 49, 77$ or $121$ so that $(a^2 + 13b^2)(c^2 + 13d^2) = 5929$.
Without loss of generality, temporarily ignore solutions with $a$ or $b$ negative. As it turns out, $N(n) = 7$ is impossible, as is $N(n) = 11$. There is only one solution for $49$, and that's $n = 7$. Looking at the sequence $13, 52, 117$, we find that $(2 - 3 \sqrt{-13})(2 + 3 \sqrt{-13}) = 121$, but $77$ is not divisible by either of those factors. So the only solution with positive real integers is $7 \times 11 = 77$.
That takes care of all solutions with $b = d = 0$. Moving on to $b \neq 0$, we have to look in a much smaller circle. The only solutions are $8^2 + 13 \times 1^2 = 77$ and $5 + 13 \times 4^2 = 77$. Any higher value of $a$ or $b$ will overshoot the mark.
So, you have found all three solutions.
If you still want to look at ideals, how many combinations of principal ideals can you get from $$\langle 7, 1 - \sqrt{-13} \rangle \langle 7, 1 + \sqrt{-13} \rangle \langle 11, 3 - \sqrt{-13} \rangle \langle 11, 3 + \sqrt{-13} \rangle ?$$
A: 
I guess my question is, how do you obtain the principal element generating a product of two ideals.

This is going to sound counter-intuitive, but you can use redundant generators.
Hopefully this contrived example in $\textbf Z$ will make my meaning clear: suppose you want to figure out the principal ideal $\langle 3, -6 \rangle \langle 4, 2 \rangle$. Both factors are principal ideals, but pretend not to know that for now. Using FOIL, we get: $$\langle 3, -6 \rangle \langle 4, 2 \rangle = \langle 12, 6, -24, -12 \rangle.$$ Since $\langle 12 \rangle = \langle -12 \rangle$, we can get rid of one generator right away, leaving us with $\langle 12, 6, -24 \rangle$. Next, by comparing the norms of the generators (I told you this example is contrived), we find this chain of proper containment: $\langle -24 \rangle \subset \langle 12 \rangle \subset \langle 6 \rangle$. Since $\langle 6 \rangle$ is maximal among those, we conclude that $\langle 3, -6 \rangle \langle 4, 2 \rangle = \langle 6 \rangle$.

I got something like $(1, \alpha)$ but that doesn't seem right.

It shouldn't be right. Notational quibbles aside, remember that any time an ideal has a unit for one of the generators, the ideal is the whole ring. But of $7, 1 + \sqrt{-13}, 11, 3 + \sqrt{-13}$, none of those numbers are units, since the only units in $\textbf Z[\sqrt{-13}]$ are $1$ and $-1$.
Using our redundant generators, we obtain $$\langle 7, 1 + \sqrt{-13} \rangle \langle 11, 3 + \sqrt{-13} \rangle = \langle 77, 21 + 7 \sqrt{-13}, 11 + 11 \sqrt{-13}, -10 + 4 \sqrt{-13} \rangle.$$
The norms of the generators are $5929, 1078, 1694, 308$. The greatest common divisor of these numbers is $77$. Notice that all four but the first have even norms, but only the last number is divisible by $2$, giving us $-5 + 2 \sqrt{-13}$. Notice that $77$ is divisible by $-5 + 2 \sqrt{-13}$, so we just have $\langle 21 + 7 \sqrt{-13}, 11 + 11 \sqrt{-13}, -5 + 2 \sqrt{-13} \rangle$ to deal with now.
The further whittling down of that ideal's generators I leave as an exercise if you're so inclined.
A: $$(7,1+\sqrt{-13})(11,3+\sqrt{-13})$$ is principal, say $$=a+b\sqrt{-13}$$ then we know $$a^2+13b^2=77$$ and there are only two solutions, 
$$8\pm\sqrt{-13},5\pm \sqrt{-13}$$
Now the product of the ideals contains $11+11\sqrt{-13}$ and $-10+4\sqrt{-13}$ multiplying the first by $2$, the second by $5$and subtracting we get
$$72+2\sqrt{-13}$$
Subtracting $77$
we have $$-5+2\sqrt{-13}$$ the generator.
