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I want to show $(3,1+\sqrt{-5})^2=(2-\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$. It's easy to see $(2-\sqrt{-5})\subset (3,1+\sqrt{-5})^2$ since $2-\sqrt{-5}=3-(1+\sqrt{-5})$ is in $(3,1+\sqrt{-5})^2$. But I have difficulty proving the converse inclusion.

$(3,1+\sqrt{-5})^2=(3)(3)+(3)(1+\sqrt{-5})+(1+\sqrt{-5})^2=(3)(3,1+\sqrt{-5})+(2)(2-\sqrt{-5})$. How should I proceed?

Jyrki Lahtonen
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Vladimir
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  • I can't see how $a-b\in(a,b)^2$. So you seem to have difficulties with both containments. – user26857 Mar 16 '15 at 11:39
  • @user26857: True, but his expansion on the next to last line does give that: $2(2-\sqrt{-5})$ is in there as are all multiples of $3$. In particular $3(2-\sqrt{-5})$. As the difference of those two you do get the claim. – Jyrki Lahtonen Mar 16 '15 at 12:46

2 Answers2

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Adding the simplest proof I could come up with here:

The ideal $P=(3,1+\sqrt{-5})$ is one of the prime ideals lying above the rational prime $3$, and of index three in the ring of integers. By multiplicativity of indices of ideals we have $$ [\Bbb{Z}[\sqrt{-5}]:P^2]=9. $$ Combining this with the fact $$[\Bbb{Z}[\sqrt{-5}]:(2-\sqrt{-5})]=N(2-\sqrt{-5})=2^2+5\cdot(-1)^2=9,$$ your observation that $(2-\sqrt{-5})\subseteq P^2$ and multiplicativity of index in a tower of abelian groups: $$ 9=[\Bbb{Z}[\sqrt{-5}]:(2-\sqrt{-5})]= [\Bbb{Z}[\sqrt{-5}]:P^2]\cdot =[P^2:(2-\sqrt{-5})] $$ imply that $[P^2:(2-\sqrt{-5})]=1$, in other words $P^2=(2-\sqrt{-5})$.

Leaving my earlier arguments showing how the approach evolved.

From where you are in your last line you can continue as follows:

First verify that $$3\cdot 3=9=(2+\sqrt{-5})(2-\sqrt{-5})\in(2-\sqrt{-5}).$$

Then check that $$ 3(1+\sqrt{-5})=3+3\sqrt{-5}=(2-\sqrt{-5})(-1+\sqrt{-5})\in(2-\sqrt{-5}). $$

Together with your last line these two observation prove the reverse inclusion $$ (3,1+\sqrt{-5})^2\subseteq(2-\sqrt{-5}) $$ so you are done.

The above (or the approach in Bill Dubuque's answer) is the kind of argument you can use, when first starting with this topic. A little bit later, after you have learned about class groups and such, you can take advantage of the fact the class number of this field is $h=2$. An immediate consequence is that the square of any ideal is principal. Therefore $P^2$ is of index $3^2=9$ in the ring $\Bbb{Z}[\sqrt{-5}]$. This means that - without any further calculations whatsoever - we know that $$ P^2=(a+b\sqrt{-5}), $$ where $a,b$ are rational integers and $$ N(a+b\sqrt{-5})=a^2+5b^2=9. $$ This leaves us the possibilities (up to sign) $a=3,b=0$ or $a=2,b=\pm1$. But you already had the key idea that $2-\sqrt{-5}\in P^2$, implying that $P^2\subseteq (2-\sqrt{-5})$, so the claim follows.

Jyrki Lahtonen
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  • Thanks for the higher viewpoint! Can I ask how you know $P$ is of index 3 in the ring of integers? The only relevant thing I know is the index of a principal ideal in the ring of integers is equal to the norm of the generator over $\mathbb{Q}$. – Vladimir Mar 16 '15 at 14:05
  • Sorry about not replying to you question right away. IIRC one of the early examples of prime factorizations in $\Bbb{Z}[\sqrt{-5}]$ is the fact that $P$ and its conjugate ideal $\overline{P}=(3,1-\sqrt{-5})$ are the primes lying above the rational prime $3$. In other words $$P\overline{P}=(3).$$ Because $(3)$ is of index $9$, and as conjugates $P$ and $\overline{P}$ have the same index, they must both have index three. – Jyrki Lahtonen Mar 18 '15 at 13:39
  • The same conclusion can be reached by observing that any number $a+b\sqrt{-5}$ is congruent to $a-b$ modulo $1+\sqrt{-5}$. And of course $a-b$ is then congruent to one of $0,1,2$ modulo $3$. So modulo $P$ there are at most three cosets. Of course, $P$ does not contain any units, so the number of cosets cannot be less than three. – Jyrki Lahtonen Mar 18 '15 at 13:41
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Below we derive it (vs. check it) using simple Euclidean-like reductions, i.e. by reducing some generators mod another (i.e. ideal unimodular transformations generalizing Euclidean algorithm).

Write $\ \overbrace{w = \sqrt{-5}}^{\large w^2\ =\ -5},\ $ so $\ I = (3,w\!+\!1)^2 = (3^2,\,\overbrace{(w\!+\!1)^2}^{\large \color{blue}{2w\,-\,4}\ \ },\,3(w\!+\!1))$

$ 9\in I\,\Rightarrow\, I = (9,\, I\:\!\ {\rm mod}\:\!\ 9),\ $ and $\,\ {\rm mod}\ 9\!:\ \color{blue}{2(w\!-\!2)}\equiv 0\!\! \overset{\large \times\,5\!}\iff \color{#0a0}{w\!-\!2}\equiv 0$

So $\, I = (\color{#c00}9, \color{#0a0}{w\!-\!2}, 3(w\!+\!1)) = \underbrace{(9,w\!-\!2,\color{#c00} 0)}_{\large \text{as claimed}},\,$ by $\, 3(w\!+\!1)\!\!\underset{\color{#0a0}{\large w\,\equiv\, 2^{\phantom{|^.}\!\!}}}\equiv\!\! 3(\color{#0a0}{3})\equiv\color{#c00}{9\equiv0}$

Bill Dubuque
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  • Deriving a result of a calculation has its purposes. As does the ability to pick up a student's incomplete argument and run with it. All roads lead to Rome and all that :-) – Jyrki Lahtonen Mar 16 '15 at 07:20