# Prime elements in $\mathbb{Z}[\sqrt{5}]$

I need to determine which of the elements $$3+2\sqrt{5}$$, $$9+4\sqrt{5}$$ and $$4-\sqrt{5}$$ are prime elements in $$\mathbb{Z}[\sqrt{5}]$$, respectively which are associated.

My ansatz is as follows:

So let $$x=3+2\sqrt{5}$$ divide $$ab$$ for $$a,b \in R$$. Thus, there are $$u,v \in \mathbb{Z}$$, such that $$ab=(a_1+b_1\sqrt{5})(a_2+b_2\sqrt{5})=(a_1a_2+5b_1b_2)+(a_1b_2+a_2b_1\sqrt{5})=(u+v\sqrt{5})(3+2\sqrt{5})=(3u+10v)+(3v+2u)\sqrt{5}.$$ How do I go from here in order to check whether $$x|a$$ or $$x|b$$?

• I deleted my comment because it was about irreducible elements, not prime elements, which need not be irreducible in non-UFDs. Commented Nov 10, 2020 at 8:56
• @HermeticallySealedHalibut So the norm doesn't help to check for prime? Commented Nov 10, 2020 at 9:57
• In general, no, at least not as directly as it does with irreducibles. It is still a useful tool though, see lhf's answer below. Commented Nov 11, 2020 at 10:43

$$3+2\sqrt{5}$$ has norm $$-11$$ and $$4-\sqrt{5}$$ has norm $$11$$. Perhaps they are associates. Indeed $$\frac{3+2\sqrt{5}}{4-\sqrt{5}} = 2+\sqrt{5}$$ and $$2+\sqrt{5}$$ is a unit because it has norm $$-1$$. In fact, $$(2+\sqrt{5})(-2+\sqrt{5})=1$$.

$$9+4\sqrt{5}$$ is a unit because it has norm $$1$$ and so $$(9+4\sqrt{5})(9-4\sqrt{5})=1$$.

It remains to decide whether $$3+2\sqrt{5}$$ is prime. Consider $$\frac{\mathbb Z[\sqrt{5}]}{\langle 3+2\sqrt{5} \rangle} = \frac{\mathbb Z[\sqrt{5}]}{\langle 3+2\sqrt{5},11 \rangle} = \frac{\mathbb Z[X]}{\langle X^2-5,3+2X,11 \rangle} \cong \frac{\mathbb Z_{11}[X]}{\langle X^2-5,3+2X \rangle} = \frac{\mathbb Z_{11}[X]}{\langle X^2-5,2(7+X) \rangle} = \frac{\mathbb Z_{11}[X]}{\langle X^2-5,X+7 \rangle} = \frac{\mathbb Z_{11}[X]}{\langle X+7 \rangle} \cong \mathbb Z_{11}$$ which is a domain. Therefore, $$3+2\sqrt{5}$$ is prime.

• $(1)$ Why $2(7+X)=X+7$ in $\mathbb{Z}_{11}[X]$ in the last line? and $(2)$ How did you abolished the factor $x^2-5$ in the equality $\mathbb{Z}_{11}[X]/ \left\langle X^2-5,X+7 \right\rangle=\mathbb{Z}_{11}[X]/\left\langle X+7 \right\rangle$. Please explain
– MAS
Commented Nov 26, 2020 at 17:16
• @Why: (1) $2$ is invertible mod $11$, (2) $(X-7)(X+7)=X^2-49=X^2-5 \bmod 11$.
– lhf
Commented Nov 26, 2020 at 17:26
• oh,nice. Thank you very much
– MAS
Commented Nov 27, 2020 at 10:09