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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)

HK Lee
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inequal
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4 Answers4

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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*}

Bernard
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  • thanks, but how is the isomorphism in the middle possible, you just took LHS modulo $2$, how do you know that ? – inequal Feb 20 '15 at 12:52
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    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: $(R/I)/(I+J)/I\simeq R/(I+J)$. – Bernard Feb 20 '15 at 13:30
  • How do you know the first isomorphism $\mathbb{Z}[\sqrt{-5}]/(2,1+\sqrt{-5})\cong \mathbb{Z}[x]/(2,x+1,x^2+5)$? I see the isomorphism just before it, but I'm not sure how you deal with an expression like $(\mathbb{Z}[x]/(x^2+5))/(2,1+\sqrt{-5})$. – Addem Apr 08 '18 at 04:38
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    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. – Bernard Apr 08 '18 at 10:31
  • @Bernard, hi! I wonder, in your last comment, is there any specific reason why you want to keep $x^2 + 5$ in the ideal $(2, 1+x, x^2 + 5)$? I mean, $(2, 1+x, x^2 + 5)/(x^2 + 5) = (2, 1+x)/(x^2 + 5)$, right? Sorry if this question sounds silly, I just wonder if I've missed something in the reasoning. Thanks in advance. – msd15213 Aug 03 '18 at 14:38
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    @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 $(2,1+x, x^2+5)$. – Bernard Aug 03 '18 at 17:50
  • @Bernard I see. I wonder, for questions like this, if there is a general way to determine how one should write the "correct" ideals and eventually arrive at a field? Or is everything more or less ad hoc? Thanks. – msd15213 Aug 03 '18 at 19:18
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    @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). – Bernard Aug 03 '18 at 19:41
  • @Bernard that's really helpful, thanks a lot! – msd15213 Aug 07 '18 at 01:51
  • Dear, Bernard! Thanks a lot for such detailed answer but let me ask you a question: We know that $\mathbb{Z}[\sqrt{-5}]\cong \mathbb{Z}[x]/ (x^2+5)$. Could you explain me please how did you get that $\mathbb{Z}[\sqrt{-5}]/(2,1+\sqrt{-5})\cong \mathbb{Z}[x]/(2,x+1,x^2+5)$? I am trying to prove it rigorously, but no results. – RFZ Dec 08 '18 at 17:11
  • That's simple: by the intial isomorphism $\mathbf Z[\sqrt{-5}\simeq \mathbf Z[x]/(x^2+5)$, the quotienting ideal $(2, 1+\sqrt{-5})$ becomes the ideal generated by $2$ and $1+x$ in $\mathbf Z[x]/(x^2+5)$, i.e. (2,1+x, x^2+5)/(x^2+5)$. Finally, I use the third isomorphism theorem. – Bernard Dec 08 '18 at 17:28
  • I have spent one day and finally I've got everyting in your post! You've got that $(2,1+\sqrt{-5})=\mathbb{Z}[\sqrt{-5}]$. Hence factor-ring by this ideal should isomorphic to zero. On the other hand, you've derived that this factor-ring is isomorphic to the field consisting of two element. This is contradiction. Am i right? – RFZ Dec 08 '18 at 19:06
  • Yes, that is the contradiction, so there can be no element with norm $1$. – Bernard Dec 08 '18 at 19:26
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$(2,1\!+\!w)\,=\, (\alpha) \,\Rightarrow\, (\alpha^2)\, =\, (2)\ \ $ when $\ \ \color{#c00}{w^2} =\, \color{#c00}{4n}\!-\!1\,\ $ [e.g. $\ w^2=-5\ $ if $\ n=-1$], $ $ by

$\smash[b]{(2,1\!+\!\color{#c00}w)^{\color{#c00}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 $\,\ (a,\,b)^2 = (a^2,\ ab,\ b^2)$

Bill Dubuque
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  • 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. – inequal Feb 21 '15 at 10:52
  • @inequal Yes, you can finish using norms. I colored $4n$ to help see that it comes from $,w^2.\ \ $ – Bill Dubuque Feb 21 '15 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 $? – jasmine Aug 28 '19 at 21:45
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    @jasmine We are showing that $1$ is a linear combination of the generators of the ideal. – Bill Dubuque Aug 28 '19 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).

Robert Soupe
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    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 $7$? – inequal Feb 22 '15 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. – Robert Soupe Feb 23 '15 at 01:02
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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!

hunter
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  • (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 $1$). – hunter Feb 20 '15 at 11:41
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    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 $a,b\in \mathbf Z[\sqrt{-5}]$ – inequal Feb 20 '15 at 11:55
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    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? – Timbuc Feb 20 '15 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. – hunter Feb 20 '15 at 14:56
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    @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. – Bill Dubuque Feb 20 '15 at 18:03
  • @BillDubuque whoops! Thanks! – hunter Feb 20 '15 at 19:51