# Prove that all ideals in $\mathbb{Z}[x]$ are generated by two elements.

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.

• What do you mean by the least positive integer in $\mathbb Z[x]$? Apr 17, 2020 at 4:56
• 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. Apr 17, 2020 at 5:03
• @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. Apr 17, 2020 at 5:23
• What if $a(x)=2x$ and $b(x)=x^2$? What would be your $q(x)$ then? Apr 17, 2020 at 5:44
• Your "feeling" is wrong and this is why: math.stackexchange.com/questions/2466547 Apr 17, 2020 at 7:34

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 $$?
4. Consider the set $$I_n=\{a(x)\in I\mid \deg a(x) 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 $$\Bbb{Z}[x]$$.
• 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. Apr 17, 2020 at 16:06
• 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 :-) . Apr 17, 2020 at 17:13

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.