Consider the polynomial ring $\mathbb{F}[x_{1},...,x_{n}]$ in $n$ variables and let $I \subset \mathbb{F}[x_{1},...,x_{n}]$ be an ideal. We call $I$ graded if we can decompose it into it's homogeneous components, i.e.
$$ I = \bigoplus\limits_{i \geq 0}(I \cap R_{i}), $$ where $R_{i} = \text{span}\{x^{a} : a \in \mathbb{N}^{n}, \sum_{j=1}^{n}a_{j} = i\}$, i.e. $R_{i}$ is the set of all homogeneous polynomials of degree $i$.
Now in a book it says that $I$ is graded if and only if $I$ is generated by homogeneous elements. There was no proof. Anyone can help? For the if part we can consider $\bigcup_{i\geq 0}(I\cap R_{i})$ as generating set of homogeneous elements for $I$. But now $\mathbb{F}[x_{1},...,x_{n}]$ is a noetherian ring so every ideal is finitely generated. How can I found a finite generating set of homogeneous elements?
Let me answer the real question first:
Let $S \subseteq \mathbb{F}[x_1,\dots,x_n]$ be a set of homogeneous elements and $I$ the ideal generated by it. We want to show that $I$ is graded, that is, we have to show that $I = \bigoplus_{i \geq 0} (I \cap R_i)$. First of all, we have $I \cap R_i \subseteq I$ for all $i \geq 0$ and so $ \bigoplus_{i \geq 0} (I \cap R_i) \subseteq I$. Also, $ \bigoplus_{i \geq 0} (I \cap R_i)$ is an ideal containing $S$, thus $I \subseteq \bigoplus_{i \geq 0} (I \cap R_i)$ and putting things together, you you get that $I$ is graded.
Now, let me address the issue of being finitely generated:
Since $\Bbb F[x_1,\dots,x_n]$ is noetherian, the set of ideals generated by finite subsets of $S$ has a maximal element $J$ generated by some finite subset $S_0 \subseteq S$. If $J$ were not equal to $I$ then there would exist some $s \in S$ which is not in $J$ but then the ideal generated by $S_0 \cup \{s\}$ would be strictly larger than $J$ in contradiction to the maximality of $J$. Thus, $J = I$ and $I$ is generated by $S_0$, a finite set of homogeneous elements.