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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)

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    $\begingroup$ 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. $\endgroup$ Commented Feb 20, 2015 at 11:38
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    $\begingroup$ Why would $\sqrt{-5}\in \alpha$? $\endgroup$
    – Bernard
    Commented Feb 20, 2015 at 12:32

4 Answers 4

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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*}

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  • $\begingroup$ thanks, but how is the isomorphism in the middle possible, you just took LHS modulo $2$, how do you know that ? $\endgroup$
    – inequal
    Commented Feb 20, 2015 at 12:52
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    $\begingroup$ 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: $(R/I)/(I+J)/I\simeq R/(I+J)$. $\endgroup$
    – Bernard
    Commented Feb 20, 2015 at 13:30
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    $\begingroup$ 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. $\endgroup$
    – Bernard
    Commented Apr 8, 2018 at 10:31
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    $\begingroup$ @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 $(2,1+x, x^2+5)$. $\endgroup$
    – Bernard
    Commented Aug 3, 2018 at 17:50
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    $\begingroup$ @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). $\endgroup$
    – Bernard
    Commented Aug 3, 2018 at 19:41
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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 $\,\ (a,\,b)^2 =\, (a^2,\ ab,\ b^2)$

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  • $\begingroup$ 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. $\endgroup$
    – inequal
    Commented Feb 21, 2015 at 10:52
  • $\begingroup$ @inequal Yes, you can finish using norms. I colored $4n$ to help see that it comes from $\,w^2.\ \ $ $\endgroup$ Commented Feb 21, 2015 at 14:41
  • $\begingroup$ +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 $? $\endgroup$
    – jasmine
    Commented Aug 28, 2019 at 21:45
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    $\begingroup$ @jasmine We are showing that $1$ is a linear combination of the generators of the ideal. $\endgroup$ Commented Aug 28, 2019 at 22:16
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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).

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    $\begingroup$ 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 $7$? $\endgroup$
    – inequal
    Commented Feb 22, 2015 at 22:18
  • $\begingroup$ 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. $\endgroup$ Commented Feb 23, 2015 at 1:02
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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!

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  • $\begingroup$ (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 $1$). $\endgroup$
    – hunter
    Commented Feb 20, 2015 at 11:41
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    $\begingroup$ 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 $a,b\in \mathbf Z[\sqrt{-5}]$ $\endgroup$
    – inequal
    Commented Feb 20, 2015 at 11:55
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    $\begingroup$ 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? $\endgroup$
    – Timbuc
    Commented Feb 20, 2015 at 12:27
  • $\begingroup$ The post proves that if the ideal is principal, then it is the whole ring. This answer proves it's not the whole ring. $\endgroup$
    – hunter
    Commented Feb 20, 2015 at 14:56
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    $\begingroup$ @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. $\endgroup$ Commented Feb 20, 2015 at 18:03

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