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.

| cite | improve this answer | |
  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.