Giving a hypersurface by higher degree polynomials Suppose a hypersurface in $\mathbb{P}^n$ is given by an equation $F(Z) = 0$. It is easy to show that polynomials $Z_i F$ ($i = 0,\ldots,n$) give the same hypersurface, but I have trouble demonstrating that $K[Z_0,\ldots,Z_n]/I \cong K[Z_0, \ldots, Z_n]/I'$ where $I = (F)$, $I' = (Z_i F \mid i = 0,\ldots,n)$.
I tried to construct an isomorphism explicitly, but I got stuck: intuitively, $\varphi: G \mapsto \sum_{i=0}^n Z_i G$ seems like it could be right, it is defined correctly because if $G = FH$, then $\varphi(G) = \sum_i Z_i F H \in I'$.
But then I have to show that $\varphi$ is surjective, and I'm stuck: I can't find a way to show that for any polynomial $G$ we have $G \equiv \sum_i Z_i F \: (\mathrm{mod}\ I')$.
Am I on the right track? Is there an easier way to show that $K[Z]/I \cong K[Z]/I'$?
 A: The reason you have trouble showing your algebras are isomorphic is that they are not isomorphic!    
For example take $n=1$ and $F=z_0$.
The algebras $K[z_0,z_1]/(z_0)$ and $K[z_0,z_1]/(z_0^2,z_1z_0)$ are not isomorphic because the first one is reduced (meaning its only nilpotent is zero) whereas in the second one the class of $z_0$ is nilpotent.  
In scheme theory, a more advanced version of algebraic geometry, one would even say that the locus of zeros of the ideals $(z_0)$  and$(z_0^2,z_1 z_0)$ are different subschemes of $\mathbb P^1_K$, even though their underlying sets are the same, namely the singleton set $\lbrace [0:1]\rbrace$.
A: The problem is that rings correspond to working with affine schemes.
Inside $\mathbf{A}^{n+1}$, the ideals $I = (F)$ and $J = (x_0,x_1,\ldots,x_n)I$ cut out different schemes, because one has an embedded point at the origin $(0,\ldots,0)$.
Asking that the projective hypersurfaces are the same is equivalent to asking whether
$$\mathrm{Proj}(K[Z]/I) = \mathrm{Proj}(K[Z]/J),$$
 which is weaker than
asking that $K[Z]/I = K[Z]/J$. For the former, one can check equality on
the affines $x_i \ne 0$ which cover $\mathbf{P}^n$, but the latter is false in this case, because $F \notin J$ (just by considering degrees, for example).
