Prove that $5$ is irreducible in $\mathbb Z[\sqrt{2}]$

Consider the UFD $$\mathbb Z[\sqrt{2}]$$. Prime and irreducible elements there are the same. How do I show that $$5$$ is irreducible?

I tried to write $$5=(a+b\sqrt 2)(c+d\sqrt 2)$$ or $$(2bd+ac-5)+(bc+ad)\sqrt 2=0$$. We have $$bc+ad=0$$. If $$c=0$$, then either $$d=0$$ (in which case $$5=0$$) or $$a=0$$ (in which case $$2bd=5$$). Either case gives a contradiction. Suppose $$c\ne 0$$. Then $$b=-(da)/c$$. So $$-2d^2a/c+ac-5=0$$. But I don't see how to proceed. Maybe it's the wrong path?

• A non-trivial factor of $5$ would have norm $\pm 5$. – Lord Shark the Unknown Nov 30 '18 at 18:29
• Hint: Show that $(5, x^2 - 2)$ is a prime ideal in $\mathbb{Z}[x]$ and apply the Third Isomorphism Theorem. Some examples: 1, 2, 3 – André 3000 Nov 30 '18 at 19:15
• See this duplicate. – Dietrich Burde Nov 30 '18 at 19:50

To answer this question it is best to define a norm on $$\mathbb{Z}[\sqrt2]$$ as this will give you information regarding the arithmetic structure in the given domain. You can define the following norm $$\nu: \mathbb{Z}[\sqrt2] \rightarrow \mathbb{Z}$$ given by $$\nu(a+b\sqrt2)=a^2 - 2b^2$$. It is easy to check that this is a multiplicative norm on the given domain; you just need to check that it satisfies the following properties:

• $$\nu(\alpha) = 0$$ if and only if $$\alpha = 0$$

• $$\nu(\alpha \beta) = \nu(\alpha) \nu(\beta)$$ for all $$\alpha,\beta \in \mathbb{Z}[\sqrt2]$$

Now to determine whether 5 is irreducible, first notice that the norm of 5 under $$\nu$$ is 25 i.e. $$\nu(5)=25$$. Then suppose that 5 factored into two elements $$\alpha, \beta \in \mathbb{Z}[\sqrt2]$$ i.e. $$5=\alpha\beta$$. We must have that $$\nu(\alpha\beta)=\nu(\alpha)\nu(\beta)=25$$.

For this to occur either $$\nu(\alpha)=\nu(\beta)=\pm5$$ or $$\nu(\alpha)=\pm1$$ (or alternatively $$\nu(\beta)=\pm1$$). It is easy to see that there does not exist any element in $$\mathbb{Z}[\sqrt2]$$ that has norm $$\pm5$$; that is we cannot have $$\nu(\alpha)=\nu(\beta)=\pm5$$. But then the only other option is that either $$\nu(\alpha)=\pm1$$ or $$\nu(\beta)=\pm1$$. However, if either $$\nu(\alpha)=\pm1$$ or $$\nu(\beta)=\pm1$$, then 5 must be irreducible since $$\alpha$$ or $$\beta$$ is a unit.

• $\nu(\alpha) = \nu(\beta) = -5$ is also a possibility to eliminate (for example, if $m^2 - 2n^2 = \pm 5$, then since 2 is not a QR mod 5, then $m \equiv n \equiv 0 \pmod{5}$, but then $25 \mid m^2 - 2n^2$, contradiction). And then, the remaining case is actually $\nu(\alpha) = \pm 1$ or $\nu(\beta) = \pm 1$ - where you might want to explain the reason this implies $\alpha$ resp. $\beta$ is a unit is because $N(a + b\sqrt{2}) = (a + b\sqrt{2}) (a - b\sqrt{2})$. – Daniel Schepler Nov 30 '18 at 19:20
• @DanielSchepler Can the case of negative norm, which you point out, be avoided by defining the norm to be the absolute value of $a^2-2b^2$? I thought it's a more standard definition. – user437309 Dec 1 '18 at 20:31

Your equation $$5=(a+b\sqrt 2)(c+d\sqrt 2) = (ac + 2bd) + (ad+bc)\sqrt2$$ gives the system $$\begin{cases} ac+2bd = 5\\ ad+bc = 0 \end{cases}$$ Multiplying the first equation by $$d$$, the second by $$-c$$ and adding them yields $$2bd^2 - bc^2 = 5d \implies b(2d^2-c^2) = 5d$$ Hence $$5 \mid b$$ or $$5 \mid (2d^2 - c^2)$$. However, the second possibility is impossible because $$x^2 \equiv \pm 1 \pmod 5$$. Therefore $$5 \mid b$$.

Similarly, multiplying the first equation by $$c$$, the second by $$-2d$$ and adding them gives $$ac^2-2ad^2 = 5c \implies a(c^2-2d^2) = 5c$$ As above we conclude $$5 \mid a$$.

Therefore $$\exists \hat{a}, \hat{b} \in \mathbb{Z}$$ such that $$a = 5\hat{a}$$ and $$b = 5\hat{b}$$. We have

$$5 = (a+b\sqrt 2)(c+d\sqrt 2) = (5\hat{a}+5\hat{b}\sqrt 2)(c+d\sqrt 2) = 5(\hat{a}+\hat{b}\sqrt 2)(c+d\sqrt 2)$$

Dividing be $$5$$ gives

$$1 = (\hat{a}+\hat{b}\sqrt 2)(c+d\sqrt 2)$$

so $$c + d\sqrt{2}$$ is invertible in $$\mathbb{Z}[\sqrt{2}]$$ with $$(c + d\sqrt{2})^{-1} = \hat{a}+\hat{b}\sqrt 2$$.