Finding ideals of $\mathbb{Z}[x]$ generated by $n$ elements and no fewer

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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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? – David Speyer Mar 22 '11 at 18:12
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]. – KConrad Mar 23 '11 at 1:34
Thank you for the hint! A hint is certainly fine--I'd in fact prefer to get to the final answer on my own, since I've done nearly all the other problems myself. $I/mI$ can be considered a vector space over $\mathbb{Z}/2\mathbb{Z}$ of dimension $k+1$ with basis $\{\overline{2^i x^{k-i}}\}$. If $I = (a_1, ..., a_m)$, perhaps $\{a_i\}$ spans this vector space, from which the result would follow. I haven't been able to show this. Could you give me another nudge in the right direction? Thank you! Also, thanks for the the link to the other thread. That characterization of Dedekind domains is cool. – Josh Swanson Mar 23 '11 at 11:37
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! – David Speyer Mar 23 '11 at 18:08
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! – Josh Swanson Mar 24 '11 at 6:23