As the title says, I'm trying to find ideals of $\mathbb{Z}[x]$ generated by $n$ elements and no fewer. I suspect $(2^k, 2^{k-1} x, 2^{k-2} x^2, ..., x^k)$ is generated by no fewer than $n=k+1$ elements, but I haven't been able to prove it. I've tried successively replacing single elements in a finite generating set in an attempt to zero out the constant order coefficients of all but one generator, but I'm not able to ensure I've done so while preserving the ideal. If I could do this, an inductive proof would follow immediately. I've also tried to make other examples but none seemed as promising as this one. This one also generalizes the neat example $(2, x)$ (generated by at least 2 elements) which shows explicitly that $\mathbb{Z}[x]$ is not a PID. I've searched and found nothing useful.

This is problem 3.7 from D.J.H. Garling's A Course in Galois Theory. The chapter itself seems to be a standard introduction to commutative algebra with an eye towards polynomial rings over fields. I hope the question isn't too basic for this site; it's my first post. I've gone through the book and done every problem except for (parts of) around a dozen, including this one. He gives questions that appear to rely on more advanced material than was presented in the text from time to time, so perhaps this question is easy for someone more experienced in algebra, or maybe I'm just missing something.

Any help is appreciated!

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    $\begingroup$ This problem is reasonable to do in an elementary way; I hope you won't mind if I give you a hint instead of a solution. Let $\mathcal{m}$ be the ideal $(2, x)$ and let $I$ be your ideal. What is $I/\mathcal{m}I$? What would it be, if your ideal could be generated by $k$ or fewer elements? $\endgroup$ Mar 22, 2011 at 18:12
  • $\begingroup$ The answers to the question mathoverflow.net/questions/12969/…, and in particular the link to the paper by Matlis, may be interesting to you after you have solved your problem in the ring Z[x]. $\endgroup$
    – KConrad
    Mar 23, 2011 at 1:34
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    $\begingroup$ You don't need a nudge, you already going in the right direction. It is in fact true that, if $I=(a_1, \ldots, a_m)$ then $I/mI$ is spanned (as a vector space) by the $a_i$. Now go forth and prove it! $\endgroup$ Mar 23, 2011 at 18:08
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    $\begingroup$ Done. I'm not sure how I could have missed that $\mathbb{Z}[x]$-linear combinations turn into $\mathbb{Z}/2\mathbb{Z}$-linear combinations in the quotient space. It's so obvious now. Thank you again! $\endgroup$ Mar 24, 2011 at 6:23
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    $\begingroup$ Related: this question on M.SE. $\endgroup$
    – Watson
    Aug 16, 2016 at 18:06


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