# Rings of polynomials whose (partial) derivatives vanish are Noetherian?

I have the following two questions on rings of polynomials. They seemed similar enough that I thought I'd go ahead and group them here as opposed to making separate listings for them. The questions are:

($1$) Determine whether or not the ring of polynomials in $z$ whose first $k$ derivatives vanish at the origin for some fixed positive integer $k$ is Noetherian.

($2$) Determine whether or not the ring of polynomials in $z$ and $w$ whose partial derivatives with respect to $w$ vanish for $z=0$ is Noetherian.

Note that $z$ and $w$ are complex variables, that is, the polynomials are over $\Bbb C$.

My work: It seemed like I could view each of these as subrings of $\Bbb C[z]$ and $\Bbb C[z,w]$, and since both of these are Noetherian, I wanted to conclude that each of the above were Noetherian, but I realized it isn't necessarily the case that a subring of a Noetherian ring is Noetherian. Can anyone offer some help for me? Thanks!

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This is Exercise 7.4 iv) & v) from Atiyah and MacDonald, Commutative Algebra. – user26857 Jan 25 '13 at 2:26
Similar question: math.stackexchange.com/questions/63579/is-this-ring-noetherian – user26857 Feb 4 '13 at 10:48

(1) $R=\mathbb C+X^{k+1}\mathbb C[X]$, and this is a subring of $\mathbb C[X]$. Furthermore, the extension $R\subset \mathbb C[X]$ is finite and $\mathbb C[X]$ is noetherian, hence $R$ is noetherian. (An elementary approach: set $S=\mathbb C[X^{k+1}]$. Then $R$ is a finitely generated $S$-module (generated by $1,X^{k+2},\dots,X^{2k+1}$), and therefore $R$ is a noetherian $S$-module. It follows that $R$ is a noetherian ring.)

(2) $R=\mathbb C[X]+X\mathbb C[X,Y]$. This ring is not noetherian since the ideal $I=X\mathbb C[X,Y]$ is not finitely generated. Assume by contrary that $I$ is finitely generated and let $Xf_1(X,Y),\dots,Xf_n(X,Y)$ be a system of generators. Then every element of $I$, say $Xg(X,Y)$, can be written as follows: $$Xg(X,Y)=Xf_1(X,Y)[a_1(X)+Xb_1(X,Y)]+\cdots+Xf_n(X,Y)[a_n(X)+Xb_n(X,Y)].$$ Then $$g(X,Y)=f_1(X,Y)[a_1(X)+Xb_1(X,Y)]+\cdots+f_n(X,Y)[a_n(X)+Xb_n(X,Y)].$$ Now we send $X$ to $0$ and get $$g(0,Y)=f_1(0,Y)a_1(0)+\cdots+f_n(0,Y)a_n(0).$$ This shows that every polynomial of $\mathbb C[Y]$ can be written as a linear combination of $f_1(0,Y),\dots,f_n(0,Y)$ with complex coefficients, a contradiction.

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This may be a dumb question, but it seems like we're viewing $\mathbb{C}[Y]$ as a $\Bbb C$-module? Is that correct? Is that why it can't be finitely generated? – anon271828 Jan 25 '13 at 22:40
That is why I was asking. Maybe I'm just not grasping the conceptual difference between modules and what you currently have. We can think of the $f_i$'s as a basis for $\Bbb C[Y]$? – anon271828 Jan 26 '13 at 0:10
Then why does it follow that everything in there can be written as a linear combination? Isn't that what a basis is? Err... Almost a basis. I guess it may not be linearly independent, but the linear combination part is what has me hung up, I guess. – anon271828 Jan 26 '13 at 0:44

In question (1) the ring obtained for $k=1$ is $S=\mathbb C+z^2\mathbb C[z]$.
That ring is noetherian because it is isomorphic to $\mathbb C[X,Y]/(Y^2-X^3)$ : can you show that?

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