I have a problem with an exercise from Tom Marley which is:
Let $R$ be a $\mathbb Z$-graded ring and $I$ an ideal of $R_0$. Prove that $IR \cap R_0 = I$.
For $I \subset IR \cap R_0 $, we can write $x = x·1 \in IR$. Since $I$ is an ideal of $R_0$, $I \subset R_0$, then we have $x \in R_0$, so $x \in IR \cap R_0$.
But for $IR \cap R_0 \subset I$, I have some problems. Any comments and guidance would be highly appreciated.
Let $f\in IR\cap R_0$ and write $f=a\cdot g$ with $a\in I$ and $g\in R$. We can decompose $g=\sum_{d\in\Bbb Z} g_d$ with $g_d\in R_d$ and since $I\subseteq R_0$, we have $a\in R_0$. It follows that $ag_d$ is the homogeneous degree $d$ component of $f$. But it is also true that $f\in R_0$, so $ag_d=0$ for all $d\ne 0$. This means that $f=ag_0\in I\cdot R_0=I$.