Every ideal of $\mathbb{Z}[x]$ is finitely generated

Question

I am trying to prove that if $I$ is an ideal of $\mathbb{Z}[x]$ then $I$ is generated by $x$ and the least positive integer in $I$. But I need help in finishing the proof.

If there are no positive integers, i.e., constant polynomials (other than $0$, of course) in $I$ then $I=(x)$. Otherwise, let $a$ be the least positive integer present in $I$.

Let $p(x)=a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ be a polynomial in $I$. It is enough to show that $a$ divides $a_0$ to prove my claim. By the division algorithm, there are integers $q,r$ with $0\le r< a$ such that $a_0 = qa + r$. Thus, $p(x)=a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + qa + r$. Then $r= p(x)-a_n x^n - a_{n-1} x^{n-1} - \ldots - a_1 x - qa$.

Now, if $r$ was in $I$ then we would be done because then $r=0$. But how do I show that $r \in I$?

Does this answer your question? Hilbert's Basis Theorem - Clever Proof? and Show that Z is a PID — Rushabh Mehta, Feb 23 '20 at 05:34
You are wrong that every ideal is generated by $x$ and the least positive integer in $I$ (if any). For example. $I=(x^2+1)$ is not. — Arturo Magidin, Feb 23 '20 at 05:36
@ArturoMagidin Thanks. My course instructor will cover Hilbert's Basis Theorem in the upcoming lecture. He gave $Z[x]$ as an example for a Noetherian ring but left it as an exercise to prove the above claim that I've made. He seems wrong. — ashK, Feb 23 '20 at 05:40
@AshishK He is not wrong; it’s just that not every ideal is of the type you think it is. There’s plenty more ideals. — Arturo Magidin, Feb 23 '20 at 06:32
@ArturoMagidin the claim was made by him and was left by him as an exercise — ashK, Feb 23 '20 at 06:34

score 1 · Answer 1 · answered Feb 23 '20 at 06:09

Extended hints. Let $I$ be an ideal of $\Bbb{Z}[x]$.

Let $J$ be the set of leading coefficients of polynomials in $I$. Prove that $J$ is an ideal of $\Bbb{Z}$.
Because $\Bbb{Z}$ is a PID we have $J=m\Bbb{Z}$ for some integer $m$. Let $p(x)\in I$ be a polynomial with leading coefficient $m$. Assume that $\deg p(x)=n$. Show that every polynomial $f(x)\in I$ can be written in the form $$f(x)=q(x)p(x)+r(x),$$ where $\deg r(x)<n.$
Show that the set $$R=\{r(x)\in I\mid \deg r(x)<n\}$$ is a finitely generated abelian group.
Conclude that $I$ is a finitely generated ideal.

I couldn't find a proof of this with a quick search. Surely we have done this on our site already, hence CW. It is essentially the standard proof of Hlibert basis theorem. Because $\Bbb{Z}$ is a PID we can cut a few corners. — Jyrki Lahtonen, Feb 23 '20 at 06:10

Every ideal of $\mathbb{Z}[x]$ is finitely generated

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