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I was trying to prove that $\mathbb{Z}[x]$ is noetherian, so every ideal in $\mathbb{Z}[x]$ is finitely generated.

I feel that all ideals in $\mathbb{Z}[x]$ are essentially generated by two elements - a polynomial and the smallest integer belonging to the ideal.

Let $a(x) \in I$, where $I$ is an ideal in $\mathbb{Z}[x]$, be a polynomial whose degree is the least. Let $b(x)$ be another polynomial whose degree is more than $a(x)$ then $r(x)=a(x)-b(x)q(x) \in I$ becomes the polynomial of the smallest degree (we first assume that $r(x)$ is a non constant polynomial). So $r(x)$ has to be zero.

If $r(x)$ is a constant in $\mathbb{Z}$ and let $r $ be the least positive integer in $\mathbb{Z}[x]$. If $r(x) \in (r)$ then we are done, or let $d=(r(x),r)$ then I will be generated by $(a(x),d)$. What I think is that I am going wrong in the last paragraph. Can someone point out my mistake.

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  • $\begingroup$ What do you mean by the least positive integer in $\mathbb Z[x]$? $\endgroup$
    – awllower
    Apr 17, 2020 at 4:56
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    $\begingroup$ That $\mathbb{Z}[X]$ is Noetherian follows from the fact that $\mathbb{Z}$ is Noetherian via the Hilbert Basis Theorem. I'm afraid that your claim is not true, if I'm not mistaken. Let $p$ be a prime, and let $n$ be a positive integer. One may show that the ideal $\langle p^{n}, p^{n-1}X, \ldots, pX^{n-1}, X^{n} \rangle$ may not be generated by fewer than $n+1$ elements, for instance. It is true that every prime ideal can be generated by two elements, however. $\endgroup$ Apr 17, 2020 at 5:03
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    $\begingroup$ @smita: I don't have an answer for that, because I don't know why someone would avoid the Hilbert Basis Theorem. It is literally the exact tool for this kind of problem, and most modern proofs are about a paragraph long and don't use anything outside of basic commutative algebra. If you are taking a class where it hasn't been introduced, you could look at its proof, and then imitate its proof for $\mathbb{Z}$ specifically. But I'd just as soon write down a general proof - writing a proof for just $\mathbb{Z}$ strikes me as silly. $\endgroup$ Apr 17, 2020 at 5:23
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    $\begingroup$ What if $a(x)=2x$ and $b(x)=x^2$? What would be your $q(x)$ then? $\endgroup$ Apr 17, 2020 at 5:44
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    $\begingroup$ Your "feeling" is wrong and this is why: math.stackexchange.com/questions/2466547 $\endgroup$
    – user26857
    Apr 17, 2020 at 7:34

2 Answers 2

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As Angina Seng pointed out, your mistake is that you think you can do polynomial division as if you were working over a field. You cannot reduce $a(x)$ modulo $b(x)$ and expect a low degree remainder. If the leading coefficient of $a(x)$ is not divisible by that of $b(x)$ you are dismounted at the first obstacle: what choice of $q(x)$ would give a low degree $r(x)$ when for example $$ r(x)=(4x^5+7)-q(x)(3x^3+5)? $$ You see what the problem is? The leading coefficient of the product $q(x)(3x^3+5)$ will be divisible by three, and hence cannot be equal to four, which is what you would need to cancel the term $4x^5$.


You can (and arguably should) use a general proof of Hilbert basis theorem. The following minor shortcut is available in $\Bbb{Z}[x]$. Leaving the steps as exercises :-)

Let $I$ be a non-zero ideal of $\Bbb{Z}[x]$.

  1. Let $J\subseteq\Bbb{Z}$ be the set of leading coefficients of polynomials of $I$. Prove that $J$ is an ideal of $\Bbb{Z}$. Warning: Proving that $J$ is closed under addition requires a bit of care.
  2. Why does there exist an integer $m$ such that $J=m\Bbb{Z}$? Why does there exist a polynomial $b(x)\in I$ such that the leading coefficient of $b(x)$ is equal to $m$?
  3. Fix a polynomial $b(x)\in I$ as in the previous step. Let's denote $n=\deg b(x)$. If $a(x)\in I$ is arbitrary, why does the polynomial division work well enough to allow us to conclude that there exists a polynomial $q(x)\in\Bbb{Z}[x]$ such that $$r(x)=a(x)-q(x)b(x)$$ has degree $<n$?
  4. Consider the set $$I_n=\{a(x)\in I\mid \deg a(x)<n\}.$$ Why is it a finitely generated free abelian group?
  5. Prove that $b(x)$ together with a $\Bbb{Z}$-basis of $I_n$ generates $I$ as an ideal of $\Bbb{Z}[x]$.
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    $\begingroup$ The "shortcut" is that $\Bbb{Z}$ is not a random noetherian ring but a PID. We can use the theory of f.g. modules over PIDs to simplify the key components of the standard proof of Hilbert basis theorem by a tiny amount. $\endgroup$ Apr 17, 2020 at 16:06
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    $\begingroup$ My recollection of having answered this earlier turned out to be correct. Switching to CW for unlike those calculus or elementary number theory homework answering machines, I believe that we should not try and get paid twice (or in few cases, a hundred times) for the same work. My excuse for not finding that answer before posting is that surprisingly the older post did not contain the buzzword Hilbert basis theorem :-) . $\endgroup$ Apr 17, 2020 at 17:13
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Here is the basic fact about Noetherian rings you need :

Let $R$ be a Noetherian ring, then $R[X]$ is Noetherian.

Since $\mathbb{Z}$ is principal, it is Noetherian, hence $\mathbb{Z}[X]$ is Noetherian.

As @user26857 pointed out in the comments above, one can find ideals in $\mathbb{Z}[X]$ whose minimal number of generators is arbitrarily high.

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