I can not find an answer to this that makes any sense to me. I am hoping someone can dumb it down a little or show me something I am missing.
But I can not seem to understand why the Quadratic Field of
$\Bbb Q [\sqrt{-7}]$ has unique factorization.
For example I feel I can draw this equality.
$$8 = 2 \cdot 2 \cdot 2 = (1 + \sqrt{-7})(1 – \sqrt{-7}) $$
Since $2, (1 + \sqrt{-7}), (1 – \sqrt{-7})$ are irreducible, it seems like they are separate sets of factors?
Best Answer
It doesn't make sense to talk about unique factorization in $\Bbb Q[\sqrt{-7}]$ because it is a field, and so any $\neq 0$ element is a unit, since there is an infinite amount of factorizations for, for instance, $4$: $$4=2\times 2=2\times 2\times 2\times\frac{1}{2}=\dots$$
What is true, is that if $d$ is a Heegner number, then the ring of integers of $\Bbb Q[\sqrt{-d}]$ is a UFD. If $K=\Bbb Q[\sqrt{-7}]$, then the ring of integers $\mathcal O_K$ is (since $d\equiv_4 1$) $$\mathcal O_K=\Bbb Z\left [\frac{-1+\sqrt{-7}}{2}\right]$$ Here we have $2=\frac{1+\sqrt{-7}}{2}\frac{1-\sqrt{-7}}{2}$. You're also right that $\Bbb Z[\sqrt{-7}]$ is not a UFD.
Edit:
We have a theorem that states
Let's try to understand what a ring of integers is. First you start with a field, typically this is the rationals, $\Bbb Q$, plus some more. Let us look at $\Bbb Q[\sqrt 2]$, that is, the field where every element is on the form: $$\alpha=a+b\sqrt 2\mid a,b\in\Bbb Q$$ We have a lot of numbers in this field, for instance $1,\frac{2}{3},3+5\sqrt{2},\frac{1119}{11}-\frac{7}{3}\sqrt{2},\dots$
Let us now pick some elements from this field, and find polynomials whos root is that element.
$1$ is easy, because it is the root of $x-1$. $\frac{1+\sqrt 2}{2}$ is slightly harder, but let's try to do it: $$x=\frac{1+\sqrt 2}{2}\\2x=1+\sqrt 2\\2x-1=\sqrt 2\\(2x-1)^2=2\\4x^2-4x+1=2\\4x^2-4x-1=0$$ So in the end, we see that $\frac{1+\sqrt 2}{2}$ is a root of $4x^2-4x-1$. Now, we ask a really intereseting question, which may seem hard. Notice the polynomials $x-1$ and $4x^2-4x-1$. One difference, among others, is that the leading coefficient of $x-1$ is 1, and the leading coefficient of $4x^2-4x-1$ is $4$.
Let us try to find such elements, we already know that $1$ works, since it is a root of $x-1$, but $\sqrt 2$ also works, since it is a root of $x^2-2$ (which has leading coefficient $1$). What about $1+\sqrt{2}$? Let us check: $$x=1+\sqrt 2\\x-1=\sqrt 2\\(x-1)^2=2\\x^2-2x+1=2\\x^2-2x-1=0$$ Indeed! $1+\sqrt 2$ is a root of $x^2-2x-1$, which have leading coefficient $1$. Now you might suspect, that if $m,n\in\Bbb Z$ then $m+n\sqrt 2$ is always a root of a monic polynomial, let's check! $$x=m+n\sqrt 2\\x-m=n\sqrt 2\\(x-m)^2=2n^2\\x^2-2mx+m^2=n^2\\x^2-2mx+m^2-2n^2=0$$ The leading coefficent is $1$, and all other coefficients are integers, so this works out very well!
Now what is the ring of integers? It is the set (also a ring, but this is hard to prove!) of all elements in $\Bbb Q[\sqrt 2]$ that is a root of a monic polynomial with integers coefficients. We typically denote this ring as $$\mathcal O_{\Bbb Q[\sqrt 2]}$$
We have shown that any element on the form $$m+n\sqrt 2\mid m,n\in\Bbb Z$$ is a root of a monic polynomial with integer coefficients. So we know that $$\Bbb Z[\sqrt 2]\subseteq\mathcal O_{\Bbb Q[\sqrt 2]}$$ but we could still have some element outside $\Bbb Z[\sqrt 2]$ that would be inside $\mathcal O_{\Bbb Q[\sqrt 2]}$, right? Let's find out! Look at all the elements in $\Bbb Q[\sqrt 2]$ $$a+b\sqrt 2\mid a,b\in\Bbb Q$$ and suppose that: $$x=a+b\sqrt 2\\x-a=b\sqrt 2\\(x-a)^2=2b^2\\x^2-2ax+a^2=2b^2\\x^2-2ax+a^2-2b^2=0$$ Sure enough, we have found a monic polynomial! Even when $a,b$ are fractions! But there is a problem. Are all the other coefficients integers? For this to be true, we need:
For the first case, we must have $a=\frac{2m+1}{2}$, because we want $a,b$ to be fractions. Also, since $-2a$ is an integer, $(-2a)^2=4a^2$ is also an integer. This means that since $4(a^2-2b^2)=4a^2-8b^2$ is an integer, also $-8b^2$ is an integer, and so we must have $b=\frac{2n+1}{2}$.
Knowing that $a^2-2b^2$ is an integers, we can replace $a,b$ and get: $$a^2-2b^2=\left (\frac{2m+1}{2}\right)^2-2\left (\frac{2n+1}{2}\right)^2=\frac{4m^2+4m-8n^2-8n-1}{4}\\=m^2+m-2n^2-2n-\frac{1}{2}$$ But this is never an integer! So $$\mathcal O_{\Bbb Q[\sqrt 2]}=\Bbb Z[\sqrt 2]$$
With more or less the same reasoning you can show that $$\Bbb Z[\sqrt{-7}]\subseteq\mathcal O_{\Bbb Q[\sqrt{-7}]}$$ But the problem is that there are elements outside of $\Bbb Z[\sqrt{-7}]$ that are still roots of monic polynomials with integers coefficients. For instance, clearly $\frac{-1+\sqrt{-7}}{2}\not\in\Bbb Z[\sqrt{-7}]$, but $$x=\frac{-1+\sqrt{-7}}{2}\\2x=-1+\sqrt{-7}\\2x+1=\sqrt{-7}\\(2x+1)^2=-7\\4x^2+4x+1=-7\\4x^2+4x+8=0\\x^2+x+2=0$$ And this is a monic polynomial with integer coefficients. So you need a ring that is slightly larger than $\Bbb Z[\sqrt{-7}]$, and it turns out that this ring is $\Bbb Z\left[\frac{-1+\sqrt{-7}}{2}\right]$.