Proving rigorously that $(x)\subset(x,xy)\subset(x,xy,xy^2)\subset\cdots$ is an ascending chain of ideals of $A$? I was reading that question here How to prove that $k[x, xy, xy^2, \dotsc]$ is not noetherian? :
The question was:
Consider the subring $R=k[x,xy,xy^2,\ldots]$ of $k[x,y]$. I want to prove that $R$ is not noetherian.
An ascending chain of ideals is the following: $$(x)\subset(x,xy)\subset(x,xy,xy^2)\subset\cdots$$ It is intuitively clearly to me that this is an ascending chain of ideals. But how do I prove it rigorously that $$xy^n \notin (x,xy,xy^2,\ldots,xy^{n-1})$$ or that this chain of ideals can never stabillize?
And one of the answers was this:
Preliminary remark
Given a polynomial $f(x,y)\in k[x,y]$, write $[f]_{i,j }$ for its term $ax^iy^j$ in the monomial $x^iy^j$.
Then for a finite sum $f=\sum_lf_l$ of polynomials we have $[f]_{i,j }=\sum [f_l]_{i,j}$
Proof that $xy^n \notin (x,xy,xy^2,\ldots,xy^{n-1})$
Suppose that $xy^n=\sum _{l=0}^{n-1}g_lxy^l$ with $g_l\in R$.
From the preliminary remark we get $$  xy^n=[xy^n]_{1,n}=[\sum _{l=0}^{n-1}g_lxy^l]_{1,n}\stackrel{prel.rem.}{=} \sum _{l=0}^{n-1}[g_lxy^l]_{1,n}=\sum _{l=0}^{n-1}[g_l]_{0,n-l}xy^l$$  But all $[g_l]_{0,n-l}=0$ since $n-l\gt0$  and all non-constant terms of a polynomial in $R$ involve a positive power of $x$.
Contradiction.
1-My question is that I do not understand how this answer proves that $(x)\subset(x,xy)\subset(x,xy,xy^2)\subset\cdots$ is an ascending chain of ideals of $R$? could anyone explain this to me please ? or is this was an answer to the OP second question "a proof that it can not stabilize"
2-Also, why we need in the finite sum $f$ of polynomials in the preliminary remark in this answer?
EDIT: I specifically do not understand what contradiction that we actually get? could anyone clarify this to me please?
 A: Regarding your question 1, you ask precisely one literal question:

But how do I prove it rigorously that
$xy^n \not\in (x,xy,xy^2,\dots,xy^{n−1})$
or that this chain of ideals can never stabillize?

The theorem you recite starts

Proof that $xy^n \not\in (x,xy,xy^2,\dots,xy^{n−1})$...

So evidently, the proof is precisely responding to your question.  Since you argue that this is sufficient, you should already know how the proof of this fact resolves your question.
Regarding your second question.  If one wishes to show that some object is not an element of generated set, one way is to take a fully generic generated element and shows that equating it to your object leads to a contradiction.  So a way to show that $xy^n$ is not an element of the ideal is to equate it to a generic element of the ideal and discover that this leads to a contradiction.
A generic element of this ideal is not the form that is written.  (I was briefly concerned that the proof was wrong for this reason.)  However, it is the generic element with $x$-degree $1$.  The function of the preliminary remark is this : in any sum of element of $k[x,y]$, terms of specific multidegree in the result are exactly the sum of the terms of that multidegree in the summands.  So the output term of $(x,y)$-multidegree $(1,n)$ is the sum of the $(1,n)$-multidegree terms in the summands.  This means we can restrict to terms with $x$-degree $1$ in the generic form.  (And so the generic form of an element is not an error in the proof.)
Since the product of any two (non-scalar, equiv. non-$k$, equiv. positive degree) elements of the ideal has $x$-degree greater than $1$, we need not consider products of terms in the generic form of an element of the ideal that is (for purpose of contradiction) set equal to $xy^n$.  Since $xy^n$ has no terms of multidegree other than $(1,n)$, all terms of other multidegrees in the sum must cancel.  We may therefore assume that all terms in the sum with nonzero coefficients have multidegree $(1,n)$.  But all the terms in the sum have multidegree $(1,\ell)$ with $\ell < n$, so the coefficients must have nonzero degree, which is a contradiction -- the coefficients are scalars (i.e., elements of $k$), having multidegree $(0,0)$.
