your ideal of what AX looks like is essentially correct: every p(x,y) in AX factors as X(g(X,Y)) for some g(X,Y) in k[X,Y].

so all we do in A_{1}is add constant terms (from k) to these polynomials. it should be clear for f(X,Y) in A_{1}, f(0,Y) is just an element of k (anything with a Y in it, also has an X with it, so setting X = 0 wipes these all out).

note that the polynomial Y in k[X,Y] does not have a constant term, so for it to be in A_{1}, it would have to be in I = (X). but Y has no X terms, either (neither does any power of Y).

now let's look at why XY is not in (X): because if h(X,Y) is in (X), then h(X,Y) = Xf(X,Y) for some f(X,Y) in A_{1}. but this means for h(X,Y) = XY, that f(X,Y) = Y (we can "cancel the X's" since k is a field, so k[X,Y] is an integral domain). and Y isn't in A_{1}.

the same sort of reasoning shows XY^{2}isn't in (X,XY): we already know it's not in (X), and if it were in (XY), we could cancel a factor of XY to get Y in A_{1}.

but maybe XY^{2}= f_{1}(X,Y)X + f_{2}(X,Y)XY for some f_{1},f_{2}in A_{1}? we need to rule this out, too.

well, first we "cancel the X's" on both sides, and we get:

Y^{2}= f_{1}(X,Y) + f_{2}(X,Y)Y

now the left hand side doesn't have any X's at all that is if: h(X,Y) = Y^{2}, then h(0,Y) = Y^{2}as well. so we evaluate both sides at (0,Y):

Y^{2}= f_{1}(0,Y) + f_{2}(0,Y)Y = a_{1}+ a_{2}Y (for some a_{1}, a_{2}in k).

the LHS has degree 2, the RHS has degree 1.