Question: Is the ring $ R = \{ f \in \mathbb{C}[x,y] \mid {\nabla f}(0,0) = (0,0) \} $ Noetherian?
I guess it isn’t Noetherian as I suspect that $$ (x y + y^{2}), \quad (x y + y^{2},x^{2} y + y^{3}), \quad (x y + y^{2},x^{2} y + y^{3},x^{3} y + y^{4}), \quad \ldots $$ is an ascending chain of ideals. I don’t know whether this is right or not.
Thanks!
Your ring $R$ is generated as an algebra over $\mathbb C$ by the monomials of $\mathbb C[x,y]$ of degree $2$ and $3$, of which there are 7. That means there is a surjection from $\mathbb C[z_1,\dots,z_7]$ onto your ring, which is therefore noetherian.
Let us check my claim about $R$ being generated by those monomials. If a polynomial is in $R$ then its linear terms are zero, and therefore all the monomials it has also belong to $R$. It follows that $R$ is generated as a complex vector space by all the monomials it contains. Now you can easily check that a monomial $x^iy^j$ is in $R$ if $i+j>1$ or $i+j=0$. Such a monomial can we written as a product of monomials of degree $2$ and $3$.