A subring of a Noetherian ring need not be Noetherian-problem

I have a problem with understanding the following solution to the problem:
A subring of a Noetherian ring is Noetherian? True or false?

Solution
let $\displaystyle A=k[X,Y]$ where k is a field, then A is Noetherian by Hilbert's basis theorem.
Now we want to construct a subring and then show that a sequence of ideals of this subring never becomes stationary thus the subring will not be Noetherian.

1)We start with letting $\displaystyle A_{1}=k+I$ where $\displaystyle I=(X)=AX$ then $\displaystyle A_{1}$ is a subring of $\displaystyle A=k[X,Y]$

as $\displaystyle \alpha,\alpha^{\prime}\in k$, $\displaystyle g,g^{\prime}\in I$ we have then that
$\displaystyle (\alpha+g)+(\alpha^{\prime}+g^{\prime})\in A_{1}$ and
$\displaystyle (\alpha+g)\cdot(\alpha^{\prime}+g^{\prime})\in A_{1}$

2) Important observation. If $\displaystyle f=\alpha+g\in k+I=A_{1}$ then $\displaystyle f(0,y)=\alpha=(0,0)$ as $\displaystyle g(0,y)=0$ and $\displaystyle y,y^{2},.....\not\in A_{1}$

3) Hence we may construct an infinite chain of ideals
$\displaystyle (X)\subset (X,XY)\subset (X,XY^2)\subset......$
this chain never stops $\displaystyle XY^{k+1}\not\in (X,XY^2,......XY^k)$
as
$\displaystyle XY^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(X,Y)XY^{ i}$ where all $\displaystyle f_{i}\in A_{1}$

$\displaystyle \Rightarrow Y^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(0,Y)Y^{i}$
$\displaystyle \Rightarrow Y^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(0,0)Y^{i}$ (contradition if they were contained).


Here are my questions and doubts....

Part 2) How do elements of $\displaystyle I=(X)=AX$ look like?
are they polynomials generated by element $\displaystyle X$ and then multiplied by any element from the ring? f ex let's take $\displaystyle p(X,Y)=3X-4XY\in k[X,Y]$ then $\displaystyle g=X(3X-4XY)=3X^2-4X^2Y\in I$ Right?

If we further assume that $\displaystyle k=\mathbb{R}$ we can take for ex $\displaystyle \alpha=5$ then if $\displaystyle f=\alpha+g=5+3X^2-4X^2Y$ then $\displaystyle f(0,y)=5=\alpha=f(0,0)$ and $\displaystyle g(0,y)=0$

but based on what do they conclude that $\displaystyle y,y^2,y^3,......\not\in A_{1}$??? is it because of the def of $\displaystyle A_{1}$ that it is a ring all polynomials that are all divisible by $\displaystyle X$??? and these ''alone'' terms with powers of y are not divisible by $\displaystyle X$

3) what does the last part of the solution say?

Could someone explain this solution to me step by step please?
Thanks

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 A₁ is add constant terms (from k) to these polynomials. it should be clear for f(X,Y) in A₁, 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 A₁, 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 A₁. 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 A₁.

the same sort of reasoning shows XY² 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 A₁.

but maybe XY² = f₁(X,Y)X + f₂(X,Y)XY for some f₁,f₂ in A₁? we need to rule this out, too.

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

Y² = f₁(X,Y) + f₂(X,Y)Y

now the left hand side doesn't have any X's at all that is if: h(X,Y) = Y², then h(0,Y) = Y² as well. so we evaluate both sides at (0,Y):

Y² = f₁(0,Y) + f₂(0,Y)Y = a₁ + a₂Y (for some a₁, a₂ in k).

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

thank you very much,that really helped me :)