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I was reading Proposition 2.2 in chapter I of Neukirch (page 6 in my edition), which states the following for an extension of rings $A\subseteq B$:

(2.2) Proposition. Finitely many elements $b_1,\dots, b_n\in B$ are all integral over $A$ if and only if the ring $A[b_1,\dots,b_n]$ viewed as an $A$-module is finitely generated.

Neukirch begins the proof by showing that if $b\in B$ is integral over $A$ then $A[b]$ is a finitely generated $A$-module. To do this, he notes that $b$ integral means there is some monic $f(x)\in A[x]$ of degree $n\geq 1$ such that $f(b)=0$. The claim is that $\{1,b,\dots,b^{n-1}\}$ form a generating set for $A[b]$. Neukirch proceeds to take a polynomial $g(x)\in A[x]$ (so that $g(b)$ is an arbritary element in $A[b])$ and states that "we may then write $$ g(x)=q(x)f(x)+r(x) $$ for some $q(x),r(x)\in A[x]$ with $\deg(r(x))<n$".

Here is my problem: $A[x]$ is not a Euclidean domain in general. If $A$ is a field then sure, but if $A=\mathbb{Z}$ then $\mathbb{Z}[x]$ is not Euclidean so it would seem this step in the proof is not justified. What am I missing here?

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    $\begingroup$ You can always perform polynomial division as long as the leading coefficient of the divisor is a unit in the ring. In this example, your $f$ is monic so you can divide as usual. $\endgroup$ – André 3000 May 14 '18 at 18:15
  • $\begingroup$ @Quasicoherent, this is news to me (nice news!). If you'd like to expand this a bit more I'd be willing to accept it as a solution. $\endgroup$ – Arbutus May 16 '18 at 13:17
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This has nothing to do with $\mathbf A[x]$ being Euclidean, nor even $A$$ being a domain.

By induction, you can suppose $B=A[b]$ for a single integral element $b\in B$.

Indeed, if $\;b^n+a_{n-1}b^{n-1}+\dots +a_1b+a_0=0$ is a monic equation for $b$, then $\;b^n\in \langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle$.

We'll prove $b^m\in \langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle$ for all $m\ge n$.

To set the inductive step, suppose $b^n,\dots,b^m\in \langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle$ for some $m$. Then \begin{align} b^{m+1}&=b\cdot b^m\in b\,\langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle =\langle \mkern1.5mu b,b^2,\dots b^{n-1}, b^n\mkern 1.5mu\rangle =\langle \mkern1.5mu b,b^2,\dots b^{n-1}\mkern 1.5mu\rangle+\langle \mkern1.5mu b^n\mkern 1.5mu\rangle \\ &\subseteq\langle \mkern1.5mu b,b^2,\dots b^{n-1}\mkern 1.5mu\rangle+\langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle =\langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle. \end{align}

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  • $\begingroup$ Correct me if I'm mistaken, but it looks like you've provided a separate proof of the fact that if $b$ is integral over $A$ then $A[b]$ is finitely generated. Although I appreciate this, it doesn't quite answer my question as to what Neukirch was doing in his proof. I guess I'm looking for something more along the lines of Quasicoherent's comment to my original post. $\endgroup$ – Arbutus May 16 '18 at 13:13
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    $\begingroup$ You're quite right, I intended to give another proof (more exactly a variant) , not mentioning Euclidean division, since it seemed to find it somewhat unsettling. Quasicoherent is right, you always can perform a polynomial division by a monic polynomial, or more generally if the divisor leading coefficient is a unit. $\endgroup$ – Bernard May 16 '18 at 13:53

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