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Could anyone give me an example of an invertible maximal ideal of some integral domain which is not generated by one element?

hxhxhx88
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3 Answers3

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In the coordinate ring $A=\mathbb R[x,y]\stackrel {def}{=}\mathbb R[X,Y]/(X^2+Y^2-1)$ of the circle the maximal ideal $\mathfrak m=(x-1,y)\subset A$ cannot be generated by a single element, as Möbius so successfully showed us with his band (no, no, he didn't conduct an orchestra).

Complements
1) The square $$\mathfrak m^2=((x-1)^2,(x-1)y, y^2)=((x-1)^2,(x-1)y,1-x^2)=(x-1)(x-1, y,-x-1))=(1-x)(-2)=(1-x)$$ of the above ideal $\mathfrak m$ is generated by a single element, namely $1-x$.

2) Every maximal ideal in $A\otimes_\mathbb R \mathbb C=\mathbb C[X,Y]/(X^2+Y^2-1)$ can be generated by a single element since that complexified ring is a PID.

Edit
At Rodrigo's request in the comment, I'll explain why $A_\mathbb C=\mathbb C[X,Y]/(X^2+Y^2-1)$ is a PID.
It is just the change of variables $U=X+iY, V=X-iY$ which shows that $A_\mathbb C\cong \mathbb C[U,V]/(UV-1)=\mathbb C[U,\frac 1U]$, which is a PID like any ring of fractions $\\$ of the PID $\mathbb C[U]$ .

Georges Elencwajg
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    Dear Georges, +1 for your answer (and the pun). I am curious to know what exactly you have in mind when you speak of the Möbius strip (no, no, he kept all of his clothes on)... – Bruno Joyal Nov 08 '13 at 01:54
  • Dear @Bruno, non-zero ideals in a Dedekind domain are projective modules of rank one. Now, there is a dictionary due to Serre and Swan which in particular associates line bundles to such projective modules and the non-principal ideal $\mathfrak m$ corresponds to the only nontrivial line bundle on the circle, the Möbius bundle. And you made me laugh with your remark that the respectable Herr Professor Doktor Möbius kept his clothes on... – Georges Elencwajg Nov 08 '13 at 07:53
  • Dear Georges: Ah, I see it now! In fact, I knew about Serre-Swan, but for some reason I couldn't fathom that the Möbius strip could arise from something that seems to be happening at a single point. Thanks for clearing it up and I'm happy to have made you laugh! – Bruno Joyal Nov 08 '13 at 11:55
  • Dear Georges, in what area of mathematics does one learn about serre-swan theorem? Does one need to know K-theory? Or can one find it in a not so sophisticated algebraic geometry text book? 2) Also, does it follow from something like Hilbert's Nullstellensatz that the complexified ring you mentioned is a PID or is it by some hands-on computation? – Rodrigo Nov 10 '13 at 01:23
  • Dear @Rodrigo: 1) no, you don't need any $K$-theory. You might check Swan's nice original article. 2) I wrote a hands-on calculation to show PIDness in an edit to my answer. – Georges Elencwajg Nov 10 '13 at 08:48
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$$(2, 1+\sqrt{-5}) \subseteq \mathbf Z[\sqrt{-5}]$$

Bruno Joyal
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Going by standard examples in alg. num. theory, $(2,1+\sqrt{-5})$ in the number ring $\Bbb Z[\sqrt{-5}]$.

anon
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