# Thread: A-algebra, element that is integral and finite A-module, explanation a proof

1. ## An A-algebra, element that is integral and finite A-module, proof explanation

let $\varphi A\to B$ be an A-algebra and let $y\in B$ then there are 3 equivalent conditions:
i) $y$ is integral over A
ii)the subring $A^{\prime}[y]\subset A$ generated by $A^{\prime}=\varphi(A)$ is finite over A (as an A-module?)
ii) $\exists$ an A-algebra $C\subset B$ such that $A^{\prime}[y]\subset C$ and C is finite over A.

Proof:
$i\Rightarrow ii$ since y is integral then we know that there exists a monic polynomial $f(Y)=Y^n+a_{n-1}Y^{n-1}+...+aY+a_{0}\in A^{\prime}[Y]$ s.t $f(y)=0$(*) but then the proof says $A^{\prime}[y]$ is generated by $1,y,y^2,....,y^{n-1}$ , as an A-module.
To show this let
$b_{m}y^{m}+....+b_{1}y+b_{0}\in A^{\prime}[y]$ if $m\ge n$ then substracting $b_{m}y^{n-m}$ times the relation (*) kills the leading term......I cannot get to this.... please help

how to show $(ii)\Rightarrow (i)$?

$ii\Rightarrow iii$ can I just set $C=A^{\prime}[y]$?? then $A^{\prime}[y]\subset B$ and $A^{\prime}[y]\subset A^{\prime}[y]$ (every ring is a subring of itself right?)

2. ## Re: A-algebra, element that is integral and finite A-module, explanation a proof

These are classic proofs.

i implies ii is correct.

ii implies iii Correct, you can just set $C=A'[y]$.

iii implies i This is annoying, but should be able to be found in any book on algebraic number theory. You basically construct the characteristic polynomial of a matrix from $y$ which will annihilate it. See Milne's notes on alg. number theory for a full proof.