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

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?)

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.