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

let $\displaystyle \varphi A\to B$ be an A-algebra and let $\displaystyle y\in B$ then there are 3 equivalent conditions:

i)$\displaystyle y$ is integral over A

ii)the subring $\displaystyle A^{\prime}[y]\subset A$ generated by $\displaystyle A^{\prime}=\varphi(A)$ is finite over A (as an A-module?)

ii)$\displaystyle \exists$ an A-algebra $\displaystyle C\subset B$ such that $\displaystyle A^{\prime}[y]\subset C$ and C is finite over A.

Proof:

$\displaystyle i\Rightarrow ii $ since y is integral then we know that there exists a monic polynomial $\displaystyle f(Y)=Y^n+a_{n-1}Y^{n-1}+...+aY+a_{0}\in A^{\prime}[Y]$ s.t $\displaystyle f(y)=0$(*) but then the proof says $\displaystyle A^{\prime}[y]$ is generated by $\displaystyle 1,y,y^2,....,y^{n-1}$ , as an A-module.

To show this let

$\displaystyle b_{m}y^{m}+....+b_{1}y+b_{0}\in A^{\prime}[y]$ if $\displaystyle m\ge n$ then substracting $\displaystyle b_{m}y^{n-m}$ times the relation (*) kills the leading term......I cannot get to this.... please help

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

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

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 $\displaystyle C=A'[y]$.

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 $\displaystyle y$ which will annihilate it. See Milne's notes on alg. number theory for a full proof.