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

let be an A-algebra and let then there are 3 equivalent conditions:

i) is integral over A

ii)the subring generated by is finite over A (as an A-module?)

ii) an A-algebra such that and C is finite over A.

Proof:

since y is integral then we know that there exists a monic polynomial s.t (*) but then the proof says is generated by , as an A-module.

To show this let

if then substracting times the relation (*) kills the leading term......I cannot get to this.... please help

how to show ?

can I just set ?? then and (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 .

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