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In $\mathbb Z[\sqrt{5}]$, $2$ and $1+\sqrt{5}$ are irreducible but not prime.

To show irreducible

I tried that there exists $\alpha$ and $\beta$ such that

$$2=\alpha\cdot \beta $$ $$N(2)=N(\alpha)\cdot N(\beta)$$ $$4=N(\alpha)\cdot N(\beta)$$

then there are 3 possibilities (namely $1.4 , 4.1 , 2.2$)

Now,I have to show that $N(\alpha)=2$ is not possible

that is $|a^2-5b^2|=2$

$a^2-5b^2=\pm 2$

How can I show that this is not possible? and how can i show that $2$ is not prime in $Z[\sqrt{5}]$?

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You haven't actually defined $N$, but I'll assume from context that

$$N(a + b \sqrt{5}) = a^2 - 5b^2$$

Consider $a^2 - 5b^2 = \pm 2$ modulo $5$; I'll consider the $+2$ case, as the $-2$ case is identical. The equation then reduces to

$$a^2 \equiv 2 \mod{5}$$

But $0^2 \equiv 0$, $1^2 \equiv 4^2 \equiv 1$, and $2^2 \equiv 3^2 \equiv 4$. We conclude that there is no $a$ satisfying the equation, so $N(\cdot) \neq 2$.


In order to show that $2$ is not prime, you must find two numbers $\alpha$ and $\beta$ such that $2 | \alpha\beta$ but $2\nmid \alpha$ and $2 \nmid \beta$. Andre Nicolas' answer suffices here.

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  • $\begingroup$ this eq. has no solution in $Z_5$ it implies that it has no solution in $Z[\sqrt{5}]$? $\endgroup$ – Siddhant Trivedi Sep 4 '13 at 3:48
  • $\begingroup$ @SiddhantTrivedi If there were a solution in $\Bbb{Z}$, then there would necessarily be a solution in $\Bbb{Z}_5$. Keep in mind that $N$ is a function into the integers. $\endgroup$ – user61527 Sep 4 '13 at 3:49
  • $\begingroup$ you are saying that to find the solution in $Z$ is equal to the find the solution in $Z[\sqrt{5}]$? because of norm function.because norm maps on $\mathbb{N}\cup${0}$ $\endgroup$ – Siddhant Trivedi Sep 4 '13 at 3:56
  • $\begingroup$ @SiddhantTrivedi Not quite. I'm saying that if we suppose that there are $a$ and $b$ satisfying $a^2 - 5b^2 = 2$, then $a^2 = 2 \mod{5}$. $\endgroup$ – user61527 Sep 4 '13 at 3:58
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To show that $2$ is not prime, observe that $2$ divides $(\sqrt{5}+1)^2$ but $2$ does not divide $\sqrt{5}+1$. Or else we can use $(\sqrt{5}-1)(\sqrt{5}+1)$.

Similarly, we can show that $\sqrt{5}+1$ is not prime. For $(\sqrt{5}+1)(\sqrt{5}-1)=4$. So $\sqrt{5}+1$ divides $(20(2)$, but $\sqrt{5}+1$ does not divide $2$.

To show that $a^2-5b^2$ cannot be equal to $\pm 2$, observe that any square is congruent to $0$, $1$, or $-1$ modulo $5$. Since $a^2-5b^2\equiv a^2\pmod{5}$, we cannot have $a^2-5b^2$ equal to anything congruent to $\pm 2\pmod{5}$. In particular, it cannot be equal to $2$ or $-2$.

Remark: It turns out that this flaw can be fixed. If we consider the numbers of the form $a+b\sqrt{5}$, where $a$ and $b$ are integers, together with numbers of the form $\frac{a+b\sqrt{5}}{2}$, where $a$ and $b$ are odd integers, we get a structure in which every non-unit irreducible is prime.

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  • $\begingroup$ 2 divides $(1+\sqrt{5})^2$? $\endgroup$ – Siddhant Trivedi Sep 4 '13 at 4:04
  • $\begingroup$ @SiddhantTrivedi $(1 + \sqrt{5})^2 = 1 + 5 + 2 \sqrt{5} = 6 + 2 \sqrt{5} = 2 (3 + \sqrt{5})$ $\endgroup$ – user61527 Sep 4 '13 at 4:11
  • $\begingroup$ Sure it does. Note that $(1+\sqrt{5})^2=6+2\sqrt{5}=2(3+\sqrt{5})$. $\endgroup$ – André Nicolas Sep 4 '13 at 4:12

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