Suppose $A=\bigoplus_{n=0}^\infty A_n$ is a Noetherian graded ring. $A_{+}=\bigoplus_{n>0}A_n$ is an ideal of $A$, hence finitely generated by homogeneous elements $x_1,...,x_s.$ Then, we want to prove that $A=A_0[x_1,...,x_s]$, and it is achived by proving $A_n\subseteq A_0[x_1,...,x_s] \forall n\in \mathbb{N}$ by inducting on $n.$ For $y\in A_n, y=\sum^{s}_{i=1}a_ix_i$, then I have a question with the upcoming line of the proof. It says $a_i$ can be replaced by the homogeneous component of degree $n-\deg(x_i).$ In other words, if we take the $n-\deg(x_i)$ degree homogeneous component of $a_i$ to be $b_i$, then $y=\sum^{s}_{i=1}b_ix_i$. My question is, why does it hold? Which property of graded rings is allowing us to accomplish this step? Any help is appreciated.
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Suppose $a_i=b_i+c_i$ where $b_i$ is the homogeneous component of $a_i$ of degree $n-\deg(x_i)$ and $c_i$ has homogeneous components of degree different than $n-\deg(x_i)$. We then have $$ y=\sum_{i=1}^sb_ix_i+\sum_{i=1}^s c_ix_i. $$ On the one hand, $\sum_{i=1}^s c_ix_i\in\bigoplus_{r\neq n}A_r$, on the other hand, $$ \sum_{i=1}^s c_ix_i=y-\sum_{i=1}^s b_ix_i\in A_n $$ so we must have $\sum_{i=1}^s c_ix_i=0$. This means that $\sum_{i=1}^s a_ix_i=\sum_{i=1}^n b_ix_i$ so we can without loss of generality suppose that $a_i=b_i$.
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