I am trying to answer this question:
Let $k$ be a field and $k[x,y].$ Define the subring $A \subset k[x,y]$ by $A = k[x, xy, xy^2, xy^3, ...].$ Show that $A$ is not Noetherian.
And I got the following hint:
Hint: Consider the ideal $I = (x, xy, xy^2, xy^3, ...).$ Assume $xy^{n+1}= f_{0}x + f_1xy + \dots + f_n xy^n$ for $f_i \in A.$ Divide by $x$ and evaluate at $x=0.$}
But I have the following question:
Is this form of the elements in $I$ correct (If $I$ is assumed to be a finitely generated ideal):
Do we have that each element is a polynomial(in the picture $f_i$) in those variables $(x, xy, xy^2, xy^3, ...)$ with coefficients (in the picture $g_{i_{s_i}}$) also polynomials in $(x, xy, xy^2, xy^3, ...)$?
Could anyone clarify this to me please?
There is an important piece I think you are missing here: you begin by supposing $I$ has some finite generating set $\{f_1,\ldots, f_t\}$. However, in a Noetherian ring, any generating set of any ideal contains a finite generating set. The hint suggests you should begin with $\{x,xy,xy^2,\ldots\}$ as a generating set, and suppose that $I$ is actually generated by $\{x,xy,\ldots, xy^n\}$. Then $xy^{n+1}$ must be in the span of these generators, so for some $f_0,\ldots, f_n$, $xy^{n+1}=f_0x+f_1xy+\cdots+f_nxy^n$. These ``coefficients'' $f_i$ are simply elements of $A$, i.e., certain polynomials in $x$ and $y$.