lemma from Milne

Question

I am reading Milne book on algebraic geometry. I understood the proof presented below. I don't understand why use A[x] ? Why not proceed with the proof by using A[X]/(1 - hX)? x isn't defined anywhere.

Answer 1

Milne's phrasing in the second sentence of the proof is a little bit informal with notation, but for clarity, you could insert the following sentence immediately before that:

Let $x$ be the class of $X$ in the quotient ring $A[X]/(1 - hX)$.

So, in particular, $A[x]$ isn't a polynomial ring over $A$ in one indeterminate $x$ (i.e., the ring over $A$ generated by an element $x$ with no relations other than those implied by the commutative ring axioms), but rather it's the ring over $A$ generated by an element $x$ that's subject to the relation $hx = 1$.