1

My question: In $\mathbb{Z}[\sqrt{-5}]$, why $(3,\sqrt{-5}-1)(3,\sqrt{-5}+1)=(3)$. My computation figure out $(3,\sqrt{-5}-1)(3,\sqrt{-5}+1)=(3)(3,\sqrt{-5}-1,\sqrt{-5}+1,2)$. $3$ and $2$ are prime numbers, $\sqrt{-5} -1$ and $\sqrt{-5}+1$ are irreducible.

This is just a trivial problem for many people, but it takes me to an ambiguity. Thank all for your help!

Minh Nguyễn Hoàng
  • 569
  • 3
  • 10

3 Answers3

3

$ ab=3c\, \Rightarrow\, (3,a)(3,b) = 3(3,a,b,c)\,$ [$ = (3)\,$ if $\,(3,\color{#0a0}{a,b})=1,\,$ as here by $\,2 = b\!-\!a\in (\color{#0a0}{a,b})$]

$\color{#c00}{ab = dc}\,\Rightarrow\, (d,a)(d,b) = (dd,da,db,\color{#c00}{ab}) = \color{#c00}d(d,a,b,\color{#c00}c)\,$ [$= (d)\,$ if $\,(a,b,c,d)= 1$] $ $ generally.

See this post and its links for elaboration on this gcd arithmetic and closely related topics such as Euler's four number theorem (Vierzahlensatz), Riesz interpolation and Schreier refinement.

Bill Dubuque
  • 272,048
1

Note that $(3, 2, a, b) = (1)$ for any $a, b\in \Bbb Z[\sqrt{-5}]$. So $(3)(3, 2, a, b) = (3)$.

Arthur
  • 199,419
1

If an ideal $I$ in a ring $R$ contains $1$, then it contains $r \cdot 1$ for any $r \in R$, so $I$ is the entire ring: $I=R$.

Your ideal $(3,2,\ldots)$ contains $3$ and $2$ so it also contains $3-2=1$, so it is the entire ring $\mathbb{Z}[\sqrt{-5}]$.

If $J$ is an ideal in $R$ then $JR = J$ because for $j \in J$ and $r \in R$ we have

  • $j = j\cdot 1 \in JR$ (so $J \subset JR$),
  • $j \cdot r \in J$ (so $JR \subset J$).
Ricardo Buring
  • 6,323