Is $\sqrt{-5}$ a prime in $\mathbb{Z}{[\sqrt{-5}]}$?

Question

I know it is irreducible over $\mathbb{Z}{[\sqrt{-5}]}$ but since it is not even UFD, so we can't conclude primality from irreducibility.

My guess is yes since $N(\sqrt{-5})=5$ is prime. I started with $ab=c\sqrt{-5}$ where a, b and c are in $\mathbb{Z}{[\sqrt{-5}]}$, now we have to show one of $a$ or $b$ is divisible by $\sqrt{-5}$.

I got $N(a)N(b)=5N(c)$, hence $5$ divide either $N(a)$ or $N(b)$. What should be my next step ?

I would argue $\sqrt{-5}$ is no more prime in $Z[ \sqrt{-5}]$ than is 1 — Paul Childs, Mar 03 '20 at 05:55
@PaulChilds Well, $\sqrt{-5}$ satisfies the usual definition of a prime, while $1$ doesn't (since $1$ is a unit). — Wojowu, Mar 06 '20 at 00:09
By quotient reciprocity $,\Bbb Z[\sqrt{-5}]/(\sqrt{-5})\cong \Bbb Z/5,$ is a domain, so $,\sqrt{-5},$ is prime in $\Bbb Z[\sqrt{-5}].,$ See here for a similar example with further detail. $\ \ $ — Bill Dubuque, Feb 18 '24 at 18:32

score 5 · Accepted Answer · answered Mar 02 '20 at 06:23

5

Say $5|N(a)$. Since $N(x+y\sqrt{-5}) = x^2+5y^2$, if $a=x+y\sqrt{-5}$ then $5|x$. So $a=5z+y\sqrt{-5} = \sqrt{-5}(y-z\sqrt{-5})$.

You can do this without invoking the norm, just low-tech: Note that an element $x+y\sqrt{-5}$ of $\mathbb{Z}[\sqrt{-5}]$ is a multiple of $\sqrt{-5}$ if and only if $x$ is a multiple of $5$ in $\mathbb{Z}$. Indeed, $5a+y\sqrt{-5} = \sqrt{-5}(y-a\sqrt{-5})$, and $\sqrt{-5}(m+n\sqrt{-5}) = 5(-n) + m\sqrt{-5}$.

Now say $\sqrt{-5}$ divides $(a+b\sqrt{-5})(x+y\sqrt{-5})$. Then it divides $$(a+b\sqrt{-5})(x+y\sqrt{-5}) = (ax-5by) + (ay+bx)\sqrt{-5}$$ and therefore $ax-5by$ is a multiple of $5$; hence $ax$ is a multiple of $5$, so $5$ divides either $a$ or $x$, hence $\sqrt{-5}$ divides either $a+b\sqrt{-5}$ or $x+y\sqrt{-5}$ as required.

answered Mar 02 '20 at 06:23

Arturo Magidin

398,050

1

The norm is not Euclidean for $\mathbb Z[\sqrt{-5}]$, but that doesn't matter here, right? – J. W. Tanner Mar 02 '20 at 06:30
3

@J.W.Tanner: No, it doesn't matter. The only property of the norm being used is that it is multiplicative, so that if $a,b\in\mathbb{Z}[\sqrt{-5}]$, then $a|b$ in $\mathbb{Z}[\sqrt{-5}]$ implies $N(a)|N(b)$ in $\mathbb{Z}$. – Arturo Magidin Mar 02 '20 at 15:25
Thanks for the explanation; +1 – J. W. Tanner Mar 02 '20 at 16:19

Is $\sqrt{-5}$ a prime in $\mathbb{Z}{[\sqrt{-5}]}$?

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