TITLE: Is this ring Noetherian?
QUESTION [9 upvotes]: 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?

REPLY [11 votes]: 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.