We call the set $\mathbf{I}(V) = \{f \in K [X] | f(x)=0,\text{ for all } x \in V\}$ the homogeneous ideal of projective variety $V$. Indeed, if $f$ and $g$ both vanish on $V$, and $r$ is an arbitrary polynomial, then $f+g$ and $rf$ vanish on V. Furthermore, one of the homogeneous components of $f$ vanishes on $V$ and $\mathbf{I}(V)$ is finitely generated when K is Noetherian. Thus $\mathbf{I}(V)$ is a homogeneous ideal.
We know a field has only two ideals, the zero ideals and the unit ideal, so it is obvious that K is Noetherian if K is field. But what if K is a ring, is $\mathbf{I}(V)$ still a homogeneous ideal?