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It is well-known that in $\mathbb{Z}[X]$ we do have non-principal ideals, for example $(2,x)$. This is an ideal with two generators. Now I was wondering if there exists an ideal with three generators, which cannot be generated by two elements. (And of course, if so, if we can find ideals with $n$ generators which cannot be generated by $n-1$ elements).

I do have one suggestion: $(8, 4x, 2x^2)$, found by trial-and-error.

My question is twofold:

  1. Is this an ideal as described above?
  2. Is there a more constructive way to think about this question?
user26857
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Miguel
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1 Answers1

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From the way the example is set the ideal $(2^n,2^{n-1}x,2^{n-2}x^2,...,2x^{n-1},x^n)$ is an ideal with $n$ honest generators in $\Bbb Z[x]$ because we cannot generate any of these generators by using the preceding ones.

user26857
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Adelafif
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  • That's the result I hoped for (in this case with $n+1$ generators) but I do not know how to prove it. Can you explain 'because we cannot generate any of these generators by using the preceding ones.' ? – Miguel May 11 '15 at 08:29
  • http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-qu%E2%80%8C%E2%80%8Bick-reference – Timbuc May 11 '15 at 08:33
  • How can you get $x^n$ from the previous generators, Miguel --- they're all multiples of $2$. How can you get $2x^{n-1}$ from the preceding generators, when they're all multiples of $4$? – Gerry Myerson May 11 '15 at 09:10
  • @GerryMyerson The point here is not that "we cannot generate any of these generators by using the preceding ones". That is, this doesn't prove that our ideal can't be generated by less elements. – user26857 May 11 '15 at 10:45