# Why is $(2, 1+\sqrt{-5})$ not principal?

Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbb{Z}[\sqrt{-5}]$?

Say $(2,1+\sqrt{-5})=(\alpha)$, then since $2\in(2,1+\sqrt{-5})$ we have $2\in (\alpha)$, so $\alpha\mid2$ in $\mathbb Z[\sqrt{-5}]$. Writing $2=\alpha\beta$ in $\mathbb Z[\sqrt{-5}]$ and taking norms, $4=N(\alpha)N(\beta)$ in $\mathbb Z$, so $N(\alpha)\mid4$ in $\mathbb Z$. Similarly, since $\sqrt{-5}\in(\alpha)$ we get $N(\alpha)\mid5$, thus $N(\alpha)$ is a common divisor of $4$ and $5$, therefore $N(\alpha)=(1)$, so $\alpha$ is a unit. But that means $1\in (2,1+\sqrt{-5})=(\alpha)$, so it must be the whole ring, but it cannot to reach a contradiction, so how can I find an element, which is not in $(2, 1+\sqrt{-5})$.

Or is there a simpler method ? (Maybe not UFD would imply not PID)

• This last idea in brackets seems to work. However, it only implies that the ring is not PID, it does not imply anything about the ideal in question. Commented Feb 20, 2015 at 11:38
• Why would $\sqrt{-5}\in \alpha$? Commented Feb 20, 2015 at 12:32

First if the ideal $(2, 1+\sqrt{-5})$ were principal, generated by $\alpha$, $N(\alpha)$ would divide $N(2)=4$ and $N(1+\sqrt{-5})=6$, hence would divide $\operatorname{gcd}(4,6)=2$. There is no element with norm $2$, hence $N(\alpha)=1$, which means $\alpha$ would be a unit; in other words, we would have $$(2, 1+\sqrt{-5})=\mathbf Z[\sqrt{-5}].$$

Now $\mathbf Z[\sqrt{-5}]\simeq \mathbf Z[x]/(x^2+5)$. Hence
\begin{align*}\mathbf Z[\sqrt{-5}]/(2, 1+\sqrt{-5})&\simeq \mathbf Z[x]/(2,x+1,x^2+5)\simeq \mathbf Z_2[x]/(x+1,x^2+1)\\ &=\mathbf Z_2[x]/\bigl(x+1,(x+1)^2\bigr)=\mathbf Z_2[x]/(x+1)\simeq\mathbf Z_2. \end{align*}

• thanks, but how is the isomorphism in the middle possible, you just took LHS modulo $2$, how do you know that ? Commented Feb 20, 2015 at 12:52
• Do you mean $\,\mathbf Z[x]/(2,x^2+5)\simeq \mathbf Z_2[x]/(x^2+1)$? It comes from the third isomorphism theorem: $(R/I)/(I+J)/I\simeq R/(I+J)$. Commented Feb 20, 2015 at 13:30
• That's because it is isomorphic to $\;\bigl(\mathbf Z[x]/(x^2+5)\bigr)\!\!\bigm/\!\!\bigl((2,1+x, x^2+5)/(x^2+5)\bigr)$, and we apply the third isomorphism theorem. Commented Apr 8, 2018 at 10:31
• @michaelshiyu:You can't write $(2,1+x))/(x^2+5)$ because $x^2+5$ is not contained in the ideal $(2,x1+x)$. But the smallest ideal of $\mathbf Z[x]$ which contains both $(2,1+x)$ and $(x^2+5)$ is $(2,1+x, x^2+5)$. Commented Aug 3, 2018 at 17:50
• @michaelshiyu: I simply used the following general more or less obvious fact: the ideal generated by an ideal $I$ in a quotient $R/J$ (where $J$ is another ideal of $R$) is the quotient ideal $I+J/J$ (which is isomorphic to $I/I\cap J$ by the second isomorphism theorem). Commented Aug 3, 2018 at 19:41

Hint $\,\ (2,1\!+\!w)\,=\, (\alpha) \,\Rightarrow\, (\alpha^2)\, =\, (2)\ \$ when $\ \ w^2 =\, \color{#c00}{4n}\!-\!1\,\$ [e.g. $\ w^2=-5\$ if $\ n=-1$]

since $\smash[b]{\,\ (2,1\!+\!w)^2 =\, (4,2\!+\!2w,\color{#c00}{4n}\!+\!2w)\, =\, 2\!\!\!\!\underbrace{(\color{#0a0}2,1\!+\!w,2n\!+\!w)}_{\large\quad \color{#0a0}{2n}+1+w-(2n\,+\,w)\:=\,1}\!\!\!\!\! \!\!=\, (2)}$

by using $\,\ (a,\,b)^2 =\, (a^2,\ ab,\ b^2)$

• For the rest: If it were principal, then $(\alpha^2)=(2)$ and since $\alpha$ is a proper divisor of $\alpha^2$ one can show again using norm that it is not possible, am I right ? and why did you colour $4n$ red in the first line, to shorten the computations or does it have something to do with the unit ideal ? Thanks by the way. Commented Feb 21, 2015 at 10:52
• @inequal Yes, you can finish using norms. I colored $4n$ to help see that it comes from $\,w^2.\ \$ Commented Feb 21, 2015 at 14:41
• +1 @BillDubuque i have some doubts $2n+ 1+w -(2n+w) =1$ why u put $2n$ in $2n+1$ ${\!\!\!\!\underbrace{(\color{#0a0}2,1\!+\!w,2n\!+\!w)}_{\large\quad \color{#0a0}{2n}+1+w-(2n\,+\,w)\:=\,1}\!\!\!\!\! \!\!=\, (2)}$ why u write $2n +1 + w-(2n+w)$ i mean it will be $2+ 1 +w - (2n+w)$ why u put $2n$? Commented Aug 28, 2019 at 21:45
• @jasmine We are showing that $1$ is a linear combination of the generators of the ideal. Commented Aug 28, 2019 at 22:16

Yes, there is a simpler method, not that different from what you've done so far, but you have to keep in mind that there are still principal ideals even if the domain is not a principal ideal domain. For example, $\langle 5, \sqrt{-5} \rangle$ is a principal ideal; a little thought should reveal that it's generated by a single element and is in fact $\langle \sqrt{-5} \rangle$.

So if $\langle 2, 1 + \sqrt{-5} \rangle$ is a principal ideal, we should be able to find a single element $x \in \mathbb{Z}[\sqrt{-5}]$ to generate it, an element that is not a unit. There isn't one: $N(2) = 4$ and $N(1 + \sqrt{-5}) = 6$, as you already know, so we'd need for $x$ to satisfy $N(x) = 2$, which as you also already know, has no solutions.

It should be noted that $\sqrt{-5} \not\in \langle 2, 1 + \sqrt{-5} \rangle$. In fact, this ideal does not contain any numbers with odd norm, which means no purely real odd integers. There is no combination of $r, s \in \mathbb{Z}[\sqrt{-5}]$ that will give you $2r + s + s \sqrt{-5} = 3$, for example, because $N(2r + s + s \sqrt{-5})$ must be even. This confirms that $\langle 2, 1 + \sqrt{-5} \rangle$ is not the whole ring.

It might be helpful to compare a similar ideal in a similar domain: $\langle 2, 1 + \sqrt{5} \rangle$ in $\mathcal{O}_{\mathbb{Q}(\sqrt{5})}$. As before, $N(2) = 4$. But $N(1 + \sqrt{5}) = 4$ as well. This may or may not be in a PID (spoiler alert: it is), but this particular ideal is a principal ideal, and it is in fact $\langle 2 \rangle$ (verify that the familiar number $\frac{1}{2} + \frac{\sqrt{5}}{2}$ is an algebraic integer).

• Thank you. Now I was reading this and noticed a similar argument on page $6$ example $4.3$. and for the additional part in your answer, can we show the equality of the ideals ($(2,1+\sqrt{5})$ and $(2)$) in the same manner as in example $4.8$ on page $7$? Commented Feb 22, 2015 at 22:18
• I think so. But I'm still a long way from understanding quotient rings, to be honest with you. Thank you very much for that Conrad paper link. Commented Feb 23, 2015 at 1:02

We say $$a + b\sqrt{-5} \in \mathbb{Z}[\sqrt{-5}]$$ is "good" if the parity of the integers $a, b$ is the same.

Claim: everything in your ideal is good. It's obvious that good things are closed under taking sums, and it's obvious that multiplying something arbitrary by 2 gives something good, so you just have to calculate that multiplying something arbitrary by $(1+\sqrt{-5})$ gives something good.

Indeed, $$(a + b\sqrt{-5})(1 + \sqrt{-5}) = (a - 5b) + (a + b)\sqrt{-5}.$$ and $(a-5b)$ and $(a+b)$ have the same parity for any $a, b$.

On the other hand, there are bad elements in the ring, so your ideal can't be the whole ring. Bonus: prove your ideal is exactly the set of good elements!

• (by the way, a longer argument, but one that generalizes better, is just to write down an arbitrary element of your ideal and show that its norm can't be $1$). Commented Feb 20, 2015 at 11:41
• I assume you mean with ''good'' that it is prime ideal, but how does an element look like in the ideal. Is it of the form $2a+(\sqrt{-5})b$, with $a,b\in \mathbf Z[\sqrt{-5}]$ Commented Feb 20, 2015 at 11:55
• I don't understand how this answers, or in fact even addresses, the question: why isn't that ideal in the post principal? Am I missing something? Commented Feb 20, 2015 at 12:27
• The post proves that if the ideal is principal, then it is the whole ring. This answer proves it's not the whole ring. Commented Feb 20, 2015 at 14:56
• @hunter The OP's argument that the ideal $\,\ne (1)\,$ is erroneous, being based in the false assumption that the ideal contains $\,\sqrt{-5}.\,$ But it is in fact true, e.g. see the argument in Bernard's answer or mine. Commented Feb 20, 2015 at 18:03