# Ideal is graded iff it is generated by homogeneous elements

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?

• Are you asking about the equivalence of $I$ being graduated and $I$ being generated by homogeneous elements or about the equivalence of $I$ being generated by homogeneous elements and $I$ being generated by finitely many homogeneous elements or both? Commented Jul 9, 2018 at 16:53
• I'm asking for the first part but the second one is also interesting
– user562724
Commented Jul 9, 2018 at 17:32

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.

• but you have to show that if $I$ is graded then $I$ is generated by homogeneous elements. Where do you use the assumption of $I$ is graded?
– user562724
Commented Jul 11, 2018 at 17:00
• my proof is as folllows. If $I$ is graded then $\bigcup\limits_{i\geq 0}I\cap S_{i}$ is a generating set for $I$ which is homogeneous
– user562724
Commented Jul 11, 2018 at 17:04
• I was under the assumption that the if-part was clear to you. You proof works perfectly. I proved that if $I$ is generated by homogeneous elements then $I$ is graded. And additionally that if $I$ is generated by homogeneous elements, then a finite subset is sufficient. Commented Jul 13, 2018 at 17:52