Is \(\mathbb{Q}(\sqrt{5})\) a Unique Factorization Domain?

[math_grl]

So you see it all over the place, $$\mathbb{Q}(\sqrt{-5})$$ is not a UFD by finding an element such that it has two distinct prime factorizations...but what about showing that $$\mathbb{Q}(\sqrt{5})$$ is a UFD?

I'm only concerned with this particular example, I might have questions later on regarding a general method.

 

[MathematicalPhysicist]

Isn't the proof goes like the proof of the fundamental theorem of arithmetic? the uniqueness part.

 

[math_grl]

I guess it could be. So you are saying that you take some random element $$a + b\sqrt{5} \in \mathbb{Q}(\sqrt{5})$$ and claim there are two distinct prime factorizations and show they actually differ by a unit?

 

[MathematicalPhysicist]

Well if you show it for the integers coeffiecients then it will obviously follow for rational coeffiecnts (I am referring to a,b), by taking common denominator in both sides and then multiply by the lcm of the two denominators of both sides we reducing the problem to integer coeffiecints.

I believe that because Q(\sqrt 5) is spanned by 1 and \sqrt 5
It's enough to show that $$(a+b\sqrt 5) c \sqrt 5=d+e\sqrt 5$$ and $$(a+b\sqrt 5) c =d+e\sqrt 5$$, where all the parameters are integers, is uniquely factorised (this obviously follows from the uniquness in simple integers).
Obviously in the proof we should impose the condition that gcd(a,b,c)=1 otherwise we can divide both sides by gcd(a,b,c).

I am just not sure how this will follow for negative integers, maybe this works upto a sign in the whole integers.