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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?

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Assume that $xy^n$ lies in the ideal. Consider $k[x,y]$ as polynomial ring in $x$ over $k[y]$, and compare the coefficient of $x$. –  user27126 Jan 9 '13 at 17:09
    
I think it should be reverse; consider $k[x,y]$ as a ring in $y$ over $k[x]$. –  Bombyx mori Jan 9 '13 at 17:12
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Dear @QiL: why did you suppress your fine answer? –  Georges Elencwajg Jan 10 '13 at 11:27
    
@GeorgesElencwajg: it doesn't add anything to your answer, and YACP's comment is more simpler. –  user18119 Jan 10 '13 at 15:35

2 Answers 2

up vote 3 down vote accepted

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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@YACP: what is the precise statement and how to prove it ? –  user18119 Jan 9 '13 at 19:32
    
@QiL A monomial $m\in K[X_1,\dots,X_n]$ belongs to a monomial ideal $I=(m_1,\dots, m_t)$ ($m_1,\dots,m_t$ are also monomials) iff $m_i\mid m$ for some $i$. (If $m\in I$, then $m=\sum_{i=1}^t a_im_i$, $a_i$ polynomials. Now consider a monomial order on the set of monomials. By looking to the highest monomial in the left hand side we get exactly what we want.) –  user26857 Jan 9 '13 at 19:37
    
@YACP: very clear. Thanks ! –  user18119 Jan 9 '13 at 20:50

Define $I:=(x, xy, ..., xy^{n-1})$. Any monomial in $I$ must contain one of the generators of $I$ as a factor. Say $xy^n \in I$. Write $xy^n=pxy^i$, where $0 \le i<n$ and $p \in R$. Since $R$ is a domain, we may cancel $xy^i$ from both sides, so that we have $y^{n-i} =p \in R$.

Now, take a general polynomial $f \in R$. We may write is as $c_0 + c_1x + c_2xy +\cdots$, where the $c_i$ are in $k$, and all but finitely many of them are $0$. If $f(x, 0)=0$, then $c_0=c_1=0$, which means either $f$ contains $x$ as a factor or $f=0$. Thus, $y^{n-1} \notin R$.

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