Prove that $(2,1+\sqrt{-5})^2 = (2)\,$ in $\,\Bbb Z[\sqrt{-5}]$

Question

Let $p$ be the ideal $\{2a+(1+\sqrt{-5})b\mid a,b\in\mathbb{Z}[\sqrt{-5}]\}$ in $\mathbb{Z}[\sqrt{-5}]$. Prove that $p^2$ is the principal ideal $(2)$.

I tried multiplying the ideal with itself and tried to simplify to see if I could relate it to the principal ideal $(2)$ but I can't seem to get it. This is what I have so far.

$(2a+(1+\sqrt{-5})b)^2=4a^2+4(1+\sqrt{-5})ab+(1+\sqrt{-5})^2b^2$

$=4a^2+(4+4\sqrt{-5})ab+(2\sqrt{-5}-4)b^2$

$=2(2a^2+(2+2\sqrt{-5})ab+(\sqrt{-5}-2)b^2)$

If $x\in p^2$, then $x$ is a multiple of $2$, so $x\in(2)$. Thus $p^2\subseteq (2)$.

I am unsure of how to show that $(2)\subseteq p^2$.

Answer 1

Here’s another approach:
To multiply two ideals, say $I=(a_1,a_2,\cdots,a_m)$ and $J=(b_1,\cdots,b_n)$, where what’s in each pair of parentheses is a list of generators of the ideal, all you need to do is write down the products $a_ib_j$, all of them, and see what ideal they generate.

In the present case, we do this: \begin{align} \left(2,1+\sqrt{-5}\,\right)\left(2,1+\sqrt{-5}\,\right)&=\left(4,2+2\sqrt{-5},-4+2\sqrt{-5}\,\right)\\ &=\left(4,2+2\sqrt{-5},2\sqrt{-5}\,\right)\\ &=\left(4,2,2\sqrt{-5}\,\right)\\ &=\left(2,2\sqrt{-5}\,\right)=(2) \end{align} Notice that each simplification is reversible: you can go from left to right, as written, but also from right to left.

Answer 2

Let $\,w = 1\!+\!\sqrt{-5}.\,$ As in $\,{\small \rm 4NT} =$ Euler's Four Number Theorem, using $\,(2,3)\!=\!(1),\,$ we have $$\,\underbrace{2\cdot 3 = \color{0a0}w\!\:\color{c00}{\bar w}\, \Rightarrow\, (2) = (2,\color{#c00}w)(\color{#90f}2,\color{#0a0}{\bar w})}_{\textstyle a\,d\, =\, b\,c\ \underset{\small \rm 4NT}\Rightarrow\, (a) = (a,b)\,(a,c)} = (2,\color{#c00}w)^2\ \ {\rm by}\ \ \underbrace{\color{#0a0}{\bar w}\equiv -\color{#c00}w\!\!\!\pmod{\!\color{#90f}2}}_{\textstyle \bar w + w \,=\, 2}\ \ \ \ \small\bf QED\qquad$$

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$\small \bf 4NT$ Proof $ $ As for gcds, $\rm\small 4NT$ constructs a nontrivial (ideal) factorization of $\,2\,$ from a factorization of any multiple $\,w\bar w\,$ of $\,2\,$ that is nontrivial (i.e. $\,2\nmid w,\bar w),\,$ i.e. a witness that $\,2\,$ is composite (not prime). To do so we simply take the ideal gcd (= ideal sum) of $\,2\,$ with each factor $\:\!w,\:\!\bar w\,$ (assuming any common factor of $\color{#0a0}{a,b,c,d}$ has been cancelled). The proof is the same as that linked for gcds:

$$ \bbox[1px,border:3px solid #c60]{\bbox[10px,border:1px solid #c00]{\begin{align}&(\color{#0a0}{a,b,c,d})\!=\!(1),\ \color{#c00}{ad = bc} \ \Rightarrow\ (a,b)\,(a,c) = (aa,ab,ac,\color{#c00}{bc}) = (\color{#c00}a)(\color{#0a0}{a,b,c,}\color{#c00}d) \:\!=\:\! (a)\\[.2em] &\rm\small \color{#0a0}{A\!+\!B\!+\!C\!+\!D}\!=\!(1),\, \color{#c00}{AD\!=\!BC}\Rightarrow(A\!+\!B)(A\!+\!C)=A^2\!+\!AB\!+\!AC\!+\!\color{#c00}{BC}= \color{#c00}A(\color{#0a0}{A\!+\!B\!+\!C}\!+\!\color{#c00}D) = A\end{align}}} \qquad$$

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Alternatively, we can repeat the proof of $\small \rm 4NT$ in this special case as below, but doing that obfuscates the key arithmetical idea (factorization refinement) that $\small\rm 4NT$ conveys.

${w^2} =\, 2w\,-\,6\,$ for $\,w = 1+\sqrt{-5},\,$ so apply below with $\,a=2$

$\!\begin{align}\color{darkorange}{w^2} = ab w + ac,\ \color{#c00}{(a,c)\!=\!(1)} \Rightarrow\ (a,w)^2\! &= (a^2,aw,\color{darkorange}{w^2})\\ &=\, (a^2,aw,ac+abw)\\ &=\, a(a,\ \,\color{#0a0}w,\ \,c\:\!+\ \,b\color{#0a0}w)\\ &=\, a(\color{#c00}a,\ \ w,\ \,\color{#c00}c)\\ &=\, a(\color{#c00}1) \end{align}$

The ideal arithmetic in Lubin's answer is a special case of the above. For an introduction to this simple ideal (and gcd) arithmetic see here and its links.

Answer 3

Enough to show that $(2, 1+ \sqrt{-5})= (\sqrt{2})$, or $$(\sqrt{2}, \frac{1 + \sqrt{-5}}{\sqrt{2}})=(1)$$

Clearly we have the inclusion LHS $\subset$ RHS, since both of the generators are integral. Moreover, the generators have relatively prime norms ($2$ and $9$), so we get the equality.

$\bf{Added:}$ Here we are using tacitly the following fact: If $K\subset L$ are number fields, $I$, $J$ (fractional) ideals in $K$ then $I \subset J$ if and only if $I^e \subset J^{e}$, where $I^e$ is the extension of $I$ in $L$, that is $B\cdot I$

So from $(\sqrt{2}, \frac{1 + \sqrt{-5}}{\sqrt{2}})=(1)$ as fractional ideals in $\mathbb{Q}(\sqrt{2}, \frac{1+ \sqrt{-5}}{\sqrt{2}})$, we conclude that $(\sqrt{2}, \frac{1 + \sqrt{-5}}{\sqrt{2}})^2=(1)^2 = (1)$, or

$$\left(\frac{1}{\sqrt{2}}\right)^2\cdot (2, 1+ \sqrt{-5})^2 = (1)$$ so $$(2, 1+ \sqrt{-5})^2 = (2)$$

as (fractional) ideals of $\mathbb{Q}(\sqrt{2}, \frac{1+ \sqrt{-5}}{\sqrt{2}})$. But this means ( note that the extension of ideals is multiplicative) that

$$(2, 1+ \sqrt{-5})^2 = (2)$$

as ideals in $\mathbb{Q}(\sqrt{-5})$.

$\bf{Added:}$ Let us mention that if $A\subset B$ is an integral extension of rings, and $I$ is an ideal of $A$, then its extension $I^{e}$ in $B$ consists of all $b\in B$ with an integral equation "from $I$" that is

$$I^e= \{ j \in B\ | \ j^m + i_1 j^{m-1} + \cdots + i_m \textrm{ for some } i_1, \ldots, i_m \in I\}$$

Indeed, there is a canonical way to form integral equations for the sum of elements, and all works great.

We get from the above

$$I^{e c} = I$$

for any ideal $I$ of $A$, if $A\subset B$ is an integral extension of Dedekind rings. Indeed, consider $I$ an ideal in $A$. Then $I^{ece} = I^e$ (this is basic). Now, if we can show that the map $I\mapsto I^e$ is injective, we conclude $I^{ec} = I$. So let $I$, $J$ ideals of $A$ ( nonzero) with $I^e = J^e$. Now this implies $I^e= (I+J)^e= J^e$. So let us show that $I = I+J= J$. Indeed, we have $I \subset I+J$, so $I = (I+J)\cdot K$ for some ideal $K$ of $A$. We get $I^e = (I+J)^e \cdot K^e$, so using the equality $I^e = (I+J)^e$, we get

$$(I+J)^e = (I+J)^e \cdot K^e$$

Now use that $B$ is also Dedeking (in fact it is enough to have cancellation of non-zero ideals) and get

$$(1) = K^e$$ as ideals of $B$. Now comes the lemma about the extension of ideals in integral extensions. We conclude

$$1^n = k_1 1^{n-1} + \cdots + k_n $$

and so $1 \in K$, and $I = I+J$.

I wonder if the above ( the map $I \mapsto I^e$ injective) holds in other cases.

Note that we do need the hypotheses.

1. $\mathbb{Z}[\sqrt{5}] \subset \mathbb{Z}[\frac{1+ \sqrt{5}}{2}]$ with $(2)\ne (1+\sqrt{5})$ on LHS but $=$ on RHS. Here, LHS is not Dedekind

2. $\mathbb{Z} \subset \mathbb{Q}$, both rings have "good" arithmetic properties, but the extension is not integral.

An important conclusion is: the extension of ideals is injective, what we have used before.

Answer 4

Let's call this ideal $\mathfrak{p}$. Note that $2 \in \mathfrak{p}^2$, since $$ 2 = (1+\sqrt{-5})(1 - \sqrt{-5}) \ - \ 2\cdot2. $$ On the other hand, if $m \in \mathfrak{p}^2$, then $m$ is a sum of terms of the form \begin{align} (2a + (1 + \sqrt{-5})b)&(2a'+(1+\sqrt{-5})b') = \\ & 4aa' + 2ab' + 2a'b + bb' - 5bb' + \bigg(2ab' + 2a'b + bb'+bb'\bigg)\sqrt{-5}. \end{align} and everything is divisible by $2$ when we group like terms.

This is a bit of a miracle proof; you can redo it conceptually by observing that $\mathfrak{p}$ is the ideal of all elements of the form $c + d\sqrt{-5}$ where $c$ and $d$ have the same parity.