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?

  • 4
    $\begingroup$ 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

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}

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.