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Could anyone give a specific example where the intersection of two principal ideals in an integral domain is not a principal ideal?

user26857
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Mod.esty
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  • Commutative with identity right? – rschwieb May 01 '20 at 02:12
  • @rschwieb Yes, commutative, identity, no zero divisor. – Mod.esty May 01 '20 at 02:18
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    The duplicate post gives an example in $\mathbb Z[\sqrt 5]$ – rschwieb May 01 '20 at 02:23
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    @rschwieb: I think you meant $(2)\cap(1-\sqrt{-5})$ in $\mathbb Z[\sqrt{\color{red}-5}]$ – J. W. Tanner May 01 '20 at 02:31
  • @J.W.Tanner sorry yes, I missed the minus sign while typing on a phone. What you said! – rschwieb May 01 '20 at 02:34
  • so I imagine $(2)\cap(1-\sqrt{-5}) = (a+b\sqrt{-5},c+d\sqrt{-5})$ for some $a,b,c,d\in \mathbb Z$; can you tell me values for $a,b,c,$ and $d$? I don't think $(2,1-\sqrt{-5})$ is the intersection – J. W. Tanner May 01 '20 at 02:36
  • @algebra.And.analysis: the "Here" link isn't working – J. W. Tanner May 01 '20 at 02:39
  • @J.W.Tanner....alright... – Mod.esty May 01 '20 at 02:41
  • @J.W.Tanner $(2) \cap (1+\sqrt{-5})$ contains $6$ and $2(1+\sqrt{-5})$. However, 6 can only be broken down into $23, -2(-3), (1+\sqrt{-5})*(1-\sqrt{-5})$. $2(1+\sqrt{-5})$ can only be broken down into $2(1+\sqrt{-5})$. – Mod.esty May 01 '20 at 03:46
  • Thanks, @algebra.And.analysis; so is $(2)\cap(1-\sqrt{-5})=(6,2+2\sqrt{-5})$? – J. W. Tanner May 01 '20 at 03:57
  • @J.W.Tanner Yes, it is true. $(6, 1+\sqrt{-5}) \subset (2) \bigcap (1+\sqrt{-5}) ={(1+\sqrt{-5})(c+d\sqrt{-5})=(c-5d)+(c+d)\sqrt{-5}\mid c-5d and c+d are even numbers, c and d are integers}$. According to the equation, $6(c_1+d_1\sqrt{-5})+2(1+\sqrt{-5})((c_2+d_2\sqrt{-5})) = (c-5d)+(c+d)\sqrt{-5}$, there must be integers $c_1, d_1, c_2, d_2$ which let the equation is satisfied for all $c, d$ that have the properties in the upfront set. – Mod.esty May 01 '20 at 06:48

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