9
$\begingroup$

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?

$\endgroup$

2 Answers 2

11
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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? $\endgroup$
    – Dave
    Sep 11, 2011 at 19:05
  • 1
    $\begingroup$ Same trick would work. $r \to xy$, $s \to y$. $\endgroup$
    – user325
    Sep 11, 2011 at 19:14
  • 1
    $\begingroup$ @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. $\endgroup$
    – jspecter
    Sep 12, 2011 at 1:31
  • $\begingroup$ @jspecter, ah you are right. Sorry for the mistake. $\endgroup$
    – user325
    Sep 12, 2011 at 2:30
5
$\begingroup$

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 $g, p_1, \cdots, p_k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.