Ring Theory – How to Prove k[x, xy, xy², …] is Not Noetherian

commutative-algebranoetherianring-theory

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?

Best Answer

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.

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