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Show that the ideal generated by 3 and $1+\sqrt{-5}$ is not a principal ideal in the ring $\Bbb Z[\sqrt{-5}]$.

I fail to understand how can 3 and $1+\sqrt{-5}$ generate an ideal.

$$(3,1+\sqrt{-5}) = (a+b\sqrt{-5}) \quad (1)$$

I think the proof would start like so:

It follows from $(1)$ that $$3 = (a+b\sqrt{-5})(c+d\sqrt{-5})$$

But I have no clue why...

Justin D.
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    If I understand your question correctly, you are trying to show that your equation (1) cannot hold, i.e., that $(3,1+\sqrt{-5})$ is not principal. If equation (1) holds, then since $3\in (3,1+\sqrt{-5}) = (a+b\sqrt{-5})$, it must be the case that $3$ is a multiple of $a+b\sqrt{-5}$ (since that is what the ideal $(a+b\sqrt{-5})$ consists of), so that there must be some $c+d\sqrt{-5}$ such that $3 = (a+b\sqrt{-5})(c+d\sqrt{-5})$. – rogerl Nov 03 '13 at 01:15
  • What am I missing here? How can we have $1 + \sqrt{-5} \in Z[\sqrt 5]$? – Robert Lewis Nov 03 '13 at 01:31
  • @rogerl, I understand your explanation. However, I am a little confused by the notation $(3, 1+\sqrt{-5})$ and what it implies for the ideal... – Justin D. Nov 03 '13 at 01:32
  • @RobertLewis, that was a typo. It should be corrected in my original question. – Justin D. Nov 03 '13 at 01:33
  • @JustinD. Recall that the notation $(a,b)$ for an ideal of the ring $R$, and $(a)$ for a principal ideal, mean $(a,b)={ar+bs:r,s\in R}$ and $(a)={ar:r\in R}$. Thus it is always true that $a\in(a)$ and $a\in(a,b)$. – Mario Carneiro Nov 03 '13 at 01:34
  • @Alan Showing that $\mathbb{Z}[\sqrt{-5}]$ is not a PID is not the same as showing that $(3, 1 + \sqrt{-5})$ is not principal. – tylerc0816 Nov 03 '13 at 01:35
  • @MarioCarneiro, so $(3,1+\sqrt{-5}) = {3 + (1+\sqrt{-5})(\sqrt{-5}) }$ in $Z[\sqrt{-5}]$? That does not make sense... – Justin D. Nov 03 '13 at 01:39
  • @JustinD $(3,1+\sqrt{-5}) = {3(c+d\sqrt{-5}) + (1+\sqrt{-5})(e+f\sqrt{-5}),\mid, c, d, e, f\in\mathbb{Z}}$ – rogerl Nov 03 '13 at 01:41
  • I guess what I'm asking is just what $Z[\sqrt 5]$ is? I thought it was all $a + b\sqrt 5$ with $a, b \in Z$. – Robert Lewis Nov 03 '13 at 01:42
  • @JustinD. Rather, $(3,1+\sqrt{-5})={3(a+b\sqrt{-5})+(1+\sqrt{-5})(c+d\sqrt{-5}):a,b,c,d\in\Bbb Z}$, where I have also applied the definition $\Bbb Z[\sqrt{-5}]={a+b\sqrt{-5}:a,b\in\Bbb Z}$. – Mario Carneiro Nov 03 '13 at 01:42
  • Oh I see now. $r, s \in R$. I am still not used to working with $R$ as a ring. I always confuse it with $\mathbb{R}$. – Justin D. Nov 03 '13 at 01:42
  • @JustinD. Do you know how to proceed, given $3=(a+b\sqrt{-5})(c+d\sqrt{-5})$? – Mario Carneiro Nov 03 '13 at 01:45
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    @JustinD Remember how, in the integers, if $d = \gcd(a,b)$ then we can find $r, s\in\mathbb{Z}$ such that $ar+bs = d$, and that in fact for any choice of integers $r, s$ we get $ar+bs$ being some multiple of $d$? So ${ar+bs}$ is the set of all multiples of $d$, which is just $(d)$, the ideal generated by $d$. The idea here is exactly the same, except that the ideal you get is not the set of multiples of any single element of $\mathbb{Z}[\sqrt{-5}]$ (of course, that's what you're trying to prove). – rogerl Nov 03 '13 at 01:45
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    @ Justin D.: a typo? Praise what gods there be! My faith in algebra, number theory, and humanity have all been simultaneously restored! – Robert Lewis Nov 03 '13 at 01:47

2 Answers2

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It’s a theorem in algebraic number theory that if $R$ is the ring of integers of a number field $K$ and $z\in R$, then the field-theoretic norm of $z$ is equal to the cardinality of $R/(z)$. Here, $K=\mathbb Q(\sqrt{-5}\,)$ and $R$ is the $\mathbb Z$-span of $\{1,\sqrt{-5}\,\}$, i.e. $R=\mathbb Z[\sqrt{-5}\,]$. Since $R/(3)$ has cardinality $9$, the larger ideal $(3,1+\sqrt{-5}\,)=I$ has the property that $R/I$ has cardinality $3$. But the norm of $a+b\sqrt{-5}$ is $a^2+5b^2$. This never takes the value $3$, that is, there is no $z$ of norm $3$, so the ideal is not principal.

Lubin
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  • @ Lubin: fascinating, +1! Where can I go to see a proof of this? – Robert Lewis Nov 03 '13 at 02:43
  • @ Lubin: that is to say, proofs of the results you use? – Robert Lewis Nov 03 '13 at 02:44
  • @ Lubin: oh, I see, algebraic number theory . . . – Robert Lewis Nov 03 '13 at 02:45
  • @RobertLewis, for these things, I depend on the book “Algebraic Theory of Numbers”, by P. Samuel. The theorem I quoted is Proposition 1 of Section 3.5, it’s page 52 in my edition. – Lubin Nov 03 '13 at 02:58
  • @ Lubin: thanks so much, I'll try and check it out. By the way, do you know any good (hopefully free) online sourxes for this stuff. Thanks.again! – Robert Lewis Nov 03 '13 at 07:35
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    It is of course good to know that this is the one standard technique one always uses to try and show some ideal of a number ring is not principal. Without this, the calculation of the class group would be much more annoying. Also, proving that $|(a)|=|N_{K/\mathbb{Q}}(a)|$ is slightly more difficult than one might assume. Generally one can actually prove that $|I|=|N_{L/K}(I)|$, where here $N_{L/K}$ denotes the ideal norm. Then, we could recover this fact from the fact (also annoying) that $N_{L/K}((a))=(N_{L/K}(a))$ and the fact that $|(a)|=|a|$ for $a\in\mathbb{Z}$ (obvious). – Alex Youcis Nov 03 '13 at 09:20
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A usual procedure to treat this kind of questions is to consider a "norm" on these rings. For $z=a+b\sqrt{-5}$ with $a,b\in\mathbb Z$ define $N(z)=z(a-b\sqrt{-5})=a^2+5b^2\in\mathbb Z$. Then $N(zw)=N(z)N(w)$ for all $z,w\in\mathbb Z[\sqrt{-5}]$, which implies in particular that $z$ is a unit in $\mathbb Z[\sqrt{-5}]$ iff $N(z)=1$.

Suppose that $I=\langle3,1+\sqrt{-5}\rangle$, the ideal in $\mathbb Z[\sqrt{-5}]$ generated by $3$ and $1+\sqrt{-5}$, is principal, say $I=\langle z\rangle$, with $z=a+b\sqrt{-5}$ and $a,b\in\mathbb Z$. The elements $3$ and $1+\sqrt{-5}$ obviously belong to $I$, so they are multiples in $\boldsymbol{\pmb{\mathbb Z}[\sqrt{-5}]}$ of the generator $z$. Therefore we have that $N(z)$ divides both $N(3)=9$ and $N(1+\sqrt{-5})=6$ in $\pmb{\mathbb Z}$, and so $N(z)$ divides $9-6=3$. In other words, $a^2+5b^2=1$ or $3$, so necessarily $b=0$ (otherwise $a^2+5b^2\geq5$), which in turn implies $a^2=1$ (because $3$ is not a square in $\mathbb Z$), so $z=\pm1$, and in particular $\pm z\in I$, so $1\in I$.

Thus, we can write $1=3(c+d\sqrt{-5})+(1+\sqrt{-5})(e+f\sqrt{-5})$, with $c,d,e,f\in\mathbb Z$, which implies $1=3c+e-5f$ and $0=3d+e+f$. Taking classes modulo $3$ we obtain $1=e+f$ and $0=e+f$, which is impossible. This contradiction shows the result.

Matemáticos Chibchas
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    I think it’s quite clear at the end of your second paragraph already that $1\notin (3,1+\sqrt{-5})$, because one sees by direct computation that $(3)=(3,1+\sqrt{-5})(3,1-\sqrt{-5})$, and if either of these factors were equal to $(1)$, the other would be equal to $(3)$, which it isn’t. I do like your explanation better than my own, because it doesn’t depend on that (definitely nonelementary) theorem I used. – Lubin Nov 03 '13 at 03:05