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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?

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    $\begingroup$ A non-trivial factor of $5$ would have norm $\pm 5$. $\endgroup$ – Lord Shark the Unknown Nov 30 '18 at 18:29
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    $\begingroup$ 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 $\endgroup$ – André 3000 Nov 30 '18 at 19:15
  • $\begingroup$ See this duplicate. $\endgroup$ – Dietrich Burde Nov 30 '18 at 19:50
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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.

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  • $\begingroup$ $\nu(\alpha) = \nu(\beta) = -5$ 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 $N(a + b\sqrt{2}) = (a + b\sqrt{2}) (a - b\sqrt{2})$. $\endgroup$ – Daniel Schepler Nov 30 '18 at 19:20
  • $\begingroup$ @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. $\endgroup$ – user437309 Dec 1 '18 at 20:31
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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 $(c + d\sqrt{2})^{-1} = \hat{a}+\hat{b}\sqrt 2$.

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