# Irreducible but not prime

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 $Z[\sqrt{5}]$?

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.

• this eq. has no solution in $Z_5$ it implies that it has no solution in $Z[\sqrt{5}]$? – Siddhant Trivedi Sep 4 '13 at 3:48
• @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. – user61527 Sep 4 '13 at 3:49
• 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}$– Siddhant Trivedi Sep 4 '13 at 3:56 • @SiddhantTrivedi Not quite. I'm saying that if we suppose that there are$a$and$b$satisfying$a^2 - 5b^2 = 2$, then$a^2 = 2 \mod{5}$. – user61527 Sep 4 '13 at 3:58 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. • 2 divides$(1+\sqrt{5})^2$? – Siddhant Trivedi Sep 4 '13 at 4:04 • @SiddhantTrivedi$(1 + \sqrt{5})^2 = 1 + 5 + 2 \sqrt{5} = 6 + 2 \sqrt{5} = 2 (3 + \sqrt{5})$– user61527 Sep 4 '13 at 4:11 • Sure it does. Note that$(1+\sqrt{5})^2=6+2\sqrt{5}=2(3+\sqrt{5})\$. – André Nicolas Sep 4 '13 at 4:12