# Proving $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$

Could anyone help me prove that $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$?

As $6=2*3=(1+\sqrt{-5})(1-\sqrt{-5})$ so $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. Therefore is not a PID or euclidean domain

• If they were reducible, then what could the norms of the factors be? Commented Nov 28, 2011 at 14:16
• @HenningMakholm: What do you mean 'norm' in this context? we have not covered norms in ring theory yet. Commented Nov 28, 2011 at 14:18
• Your text may call it something else. I'm talking about the function $a+b\sqrt{-5}\mapsto a^2+5b^2$ (which preserves products). Commented Nov 28, 2011 at 14:21
• That hasn't been covered, the lecturer is just shocking. Hence, struggling with the basics. Commented Nov 28, 2011 at 14:25
• Well alternatively prove that $|a|\ge 1$ for all nonzero $a\in \mathbb Z[\sqrt{-5}]$, and since $|ab|=|a|\cdot|b|$, any two factors of, say, $1+\sqrt{-5}$ must have absolute value equal to or less than $|1+\sqrt{-5}|$. Find all possible candidates by means of graph paper and a compass, and show that neither of them works. Commented Nov 28, 2011 at 14:29

The standard method is:

Define a function $N\colon \mathbb{Z}[\sqrt{-5}]\to\mathbb{Z}$ by $N(a+b\sqrt{-5}) = (a+b\sqrt{-5})(a-b\sqrt{-5}) = a^2+5b^2$.

1. Prove that $N(\alpha\beta) = N(\alpha)N(\beta)$ for all $\alpha,\beta\in\mathbb{Z}[\sqrt{-5}]$.
2. Conclude that if $\alpha|\beta$ in $\mathbb{Z}[\sqrt{-5}]$, then $N(\alpha)|N(\beta)$ in $\mathbb{Z}$.
3. Prove that $\alpha\in\mathbb{Z}[\sqrt{-5}]$ is a unit if and only if $N(\alpha)=1$.
4. Show that there are no elements in $\mathbb{Z}[\sqrt{-5}]$ with $N(\alpha)=2$ or $N(\alpha)=3$.
5. Conclude that $2$, $3$, $1+\sqrt{-5}$, and $1-\sqrt{-5}$ are irreducible.

This is a common technique for dealing with rings of the form $\mathbb{Z}[\theta]$, where $\theta$ is an algebraic integer.

It can also be done directly, though it is a bit more laborious. Here's what I came up with on the fly:

Assume that $(a+b\sqrt{-5})(c+d\sqrt{-5}) = 2$. Then $ac+5bd = 2$ and $ad+bc=0$. If $a=0$, then we must have $c=0$ (since $bc=0$ but we cannot have $a=b=0$), but then $5bd=2$ is impossible. Thus, $a\neq0$ and $c\neq 0$. If $b=0$, then $d=0$, so the factorization occurs in $\mathbb{Z}$ and is trivial; symmetrically if $d=0$. So we may assume that all of $a,b,c,d$ are nonzero.

Then $ad=-bc$, so $2d=acd + 5bd^2 = -bc^2 + 5bd^2 = b(5d^2 - c^2)$. If $b$ is odd, then $b|d$, so writing $d=bk$ we get $abk + bc=0$, so $ak+c=0$, thus $c=-ak$. Hence $c+d\sqrt{-5} = -ak+bk\sqrt{-5} = k(-a+b\sqrt{-5})$. But this gives $$2 = (a+b\sqrt{-5})(c+d\sqrt{-5}) = k(a+b\sqrt{-5})(-a+b\sqrt{-5}) = -k(a^2+5b^2).$$ Since $a$ and $b$ are both nonzero, $a^2+5b^2$ is at least 6, which is impossible. So $b$ is even, $b=2b'$.

Then $b'(5d^2-c^2) = d$, so setting $5d^2-c^2 = k$ we have $$a+b\sqrt{-5} = a+2b'\sqrt{-5},\qquad c+d\sqrt{-5} = c+kb'\sqrt{-5}.$$ From $ad=-bc$ we get $ak=-2c$. If $a$ is even, then we have $a=2a'$, so $$2 = (2a'+2b'\sqrt{-5})(c+d\sqrt{-5}) = 2(a'+b'\sqrt{-5})(c+d\sqrt{-5}),$$ which yields that $c+d\sqrt{-5}$ is a unit (in fact, this is impossible with $c$ and $d$ both nonzero, but that doesn't matter). If $a$ is odd, then $k$ is even and $c=ak'$, with $2k'=k$. So now we have \begin{align*} 2 &= (a+b\sqrt{-5})(c+d\sqrt{-5})\\ &= (a + 2b'\sqrt{-5})(ak' + 2b'k'\sqrt{-5})\\ &= k'(a+2b'\sqrt{-5})(a+2b'\sqrt{-5})\\ &= k'(a^2 - 20b'^2) + 4ab'\sqrt{-5} \end{align*} which implies $a=0$ or $b'=0$ (hence $b=0$), contradicting our hypotheses.

Thus, the only factorizations of $2$ in $\mathbb{Z}[\sqrt{-5}]$ are trivial.

(And you can probably see why the "standard method" is so much better....)

• Thank you so much for writing this all out, it has really made it clear :) This is a very neat way of doing it Commented Nov 28, 2011 at 20:14
• Hi, I am doing a similar problem to the OP, and was also not introduced to this norm concept. After reading your post, I definitely want to become comfortable with the "standard method", but is there any chance you could explain where it comes from? Commented Sep 16, 2017 at 22:41

Let $A = \mathbb{Z}[\sqrt{-5}]$. For every element $\alpha$ of $A$ there exist unique $a, b \in \mathbb{Z}$ such that $\alpha = a + b \sqrt{-5}$. Consider the function $N \colon A \to \mathbb{N}$ defined by $N(a + b \sqrt{-5}) = a^2 + 5 b^2$. (If you know some field theory, $N$ is the restriction to $A$ of the norm of the field $\mathbb{Q}(\sqrt{-5})$ over $\mathbb{Q}$.) Prove the following facts:

1. $N$ is multiplicative, i.e. $N(\alpha \beta) = N(\alpha) N(\beta)$ for all $\alpha, \beta \in A$.
2. If $\alpha \in A$, then $\alpha$ is invertible in $A$ if and only if $N(\alpha) = 1$.

Do there exist element in $A$ with norm equal to $2$ or $3$? Now, $N(2) = 4$, $N(3) = 9$, $N(1 + \sqrt{-5}) = N(1- \sqrt{-5}) = 6$. As in Henning Makholm's comment, if they were irreducible then what could the norms of the factors be?

• This is great thanks, much nicer than how it is in my notes Commented Nov 28, 2011 at 20:15

Since $$\sqrt{-5} \cdot \sqrt{-5}=-5 \in \mathbb{Z}$$ holds, we have $$\mathbb{Z}[\sqrt{-5}]=\{a+b \sqrt{-5} \mid a, b \in \mathbb{Z}\}.$$ We use $$(a+b \sqrt{-5})(c+d \sqrt{-5})=a c-5 b d+\left(a d+bc\right) \sqrt{5}$$ and $$|a+b \sqrt{-5}|^2=a^2+5 b^2$$.
$$1+\sqrt{-5}$$ is not a unit: it is valid that $$(1+\sqrt{-5})(a+b \sqrt{-5})=a-5 b+(a+b) \sqrt{5} \neq 1,$$ since otherwise $$a=-b$$ and $$-6b=0$$.
If $$(1+\sqrt{-5})=(a+b \sqrt{-5})(c+d \sqrt{-5})$$, then $$6=1+5=|1+\sqrt{-5}|^2=\left(a^2+5 b^2\right)\left(c^2+5 d^2\right)$$ holds. There is no $$a, b \in \mathbb{Z}$$ with $$a^2+5 b^2=2$$ and $$a^2+5 b^2=3$$, since $$a^2 \in\overline{0}, \overline{1}, \overline{4}\rangle \pmod{5}$$ and if $$a^2+5 b^2=1$$, then $$a+b \sqrt{-5}$$ would be invertible with $$(a-b \sqrt{-5})^{-1}=(a-b \sqrt{-5}),$$ since $$(a+b \sqrt{-5})(a-b \sqrt{-5})=a^2+5 b^2. \quad\square$$

We’ll do $$2$$ the rest are similar.

Suppose $$2$$ is reducible, then

$$2 =ab$$

$$a,b$$ non units. Taking norms we have

$$4 = N(a)N(b)$$

So either they’re both $$2$$ or one is $$1$$ and one is $$4$$ and in the later case we are done. but the norm of no element of your ring is $$2$$. If it were we’d have

$$a^2+5b^2=2$$

Which has no integer solutions. Thus WLOG $$N(a)=1, N(b) =4$$ thus $$a$$ is a unit.

• Please don't duplicate prior answers. Commented Aug 13, 2023 at 1:09
• @BillDubuque it was a clearer more concise answer. Commented Aug 13, 2023 at 21:17