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The subring of $\mathbb{C}[x,y]$ consisting of all polynomials $f(x,y)$ whose gradient vanishes at the point $x=y=0$. Is this ring Noetherian?

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2 Answers 2

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Yes, your ring is noetherian because it is the image of the noetherian ring $ \mathbb C[r,s,t,u,v,w,z]$ under the $\mathbb C$-algebra morphism $f:\mathbb C[r,s,t,u,v,w,z]\to \mathbb C[x,y]$ defined by
$$ f(r)=x^2, f(s)=x^3, f(t)=y^2, f(u)=y^3, f(v)=xy,f(w)=x^2y, f(z)=xy^2 $$

Warning The same technique shows that many $\mathbb C$-subalgebras of $\mathbb C[x,y]$ are noetherian .
Beware however that there exist non-noetherian subalgebras of $\mathbb C[x,y]$.
For example $\mathbb C[y,xy,x^2y,x^3y,...]$ is not noetherian since its ideal $(y,xy,x^2y,x^3y,...)$ is not finitely generated.

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What about the subring of $\mathbb{C}[x,y]$ consisting of all polynomials $f(x,y)$ whose partial derivatives in $x$ vanish when $y=0$? Is this Noetherian? –  Dave Sep 11 '11 at 19:05
    
Same trick would work. $r \to xy$, $s \to y$. –  Soarer Sep 11 '11 at 19:14
    
@Soarer. You are missing some polynomials in your image. For example $x^2y.$ In fact, Robert I believe the ring you describe is exactly that in Georges Elencwajg's Warning. –  jspecter Sep 12 '11 at 1:31
    
@jspecter, ah you are right. Sorry for the mistake. –  Soarer Sep 12 '11 at 2:30

I think the answer is yes. The following proof is very similar to that of Hilbert basis theorem.

Lemma: $S = \{f(x) \in \mathbb{C}[x] : f'(0) = 0 \}$ is Noetherian.

Proof of Lemma: Let $J \subset S$ be an ideal, and let $g_1(x)$ be a polynomial of minimal degree in $J$, of degree $n$. Then by division algorithm, for any $h(x) \in J$, it can be written as a sum of $h_1(x) + h_2(x)$, where $h_1 \in S g_1(x)$, and $h_2(x)$ has degree at most $n+1$. Let $g_2(x)$ be a polynomial of degree $n+1$ in $J$, if exists. Then $h_2$ must be a $\mathbb{C}$-linear combination of $g_1$ and $g_2$ for otherwise we have a polynomial of degree less than $n$ in $J$. Thus $J = (g_1,g_2)$.

Proof of original question

Call your ring $R$. Let $I \subset R$ be a nonzero ideal. Write every element in $I$ as a polynomial in $y$, with coefficient a polynomial in $x$.

Notice that the leading coefficients form an ideal of $\mathbb{C}[x]$, which is principal, say generated by $f(x)$. Suppose that $g(x,y) \in I$ has leading coefficient $f(x)$, and $g$ has $y$-degree $n$. Then by division algorithm, it's clear that any $h(x,y) \in I$ is equal to the sum of $h_1(x,y) + h_2(x,y)$, where $h_1 \in R g(x,y)$ and $h_2(x,y)$ has $y$-degree at most $n+1$.

Consider the elements $T$ in $I$ with $y$-degree at most $n+1$. Note that $T$ is a $S$-module. In fact it's a finite $S$-module, generated by $y^{n+1}, xy^n, y^n, \cdots, 1$. So it is a noetherian $S$-module, meaning that there are finitely many generators $p_1,\cdots,p_k$ of $T$ over $S$. But $S \subset R$, so it's clear that $I$ is generated by $g, p_1, \cdots, p_k$.

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