I try to understand the proof of Nakayama Lemma, graded version:
$S$ is the polynomial ring and $m$ is its maximal homogeneous ideal.
Let $M$ be a finitely generated graded $S$-module and $U$ a graded $S$-submodule of $M$. If $M=U+\mathfrak{m}M$ then $M=U$.
''Proof: Obviously, $U\subseteq M$.
Let $m\in M$, $m=u+f\cdot x$, $u\in U$, $f\in\mathfrak{m}, x\in M$.
But $M$ is graded. Therefore $\deg(m)=\deg(u)=\deg(f)\cdot \deg(x)$.....''
Why is true the last line?
My idea: $M=(m_1,...,m_r) $ where $m_i\in M_{a_i},\forall i.$
$m=\alpha_1\cdot m_1+...+\alpha_r\cdot m_r$. Now $\deg(m)=\min\{\deg(\alpha_1)+\deg(m_1),...,\deg(\alpha_r)+\deg(m_r)\}$.
Let $m \in M$ be a homogeneous element.
We will show that $m\in U$.
To prove this we proceed by induction on the degree of $m$.
We write $m = u + fn$ with homogeneous elements $u \in U, f \in \mathfrak{m}$ and $n \in M$, and such that $\deg(m) = \deg(u) = \deg(fn)$.
If the degree of $m$ coincides with the initial degree $\alpha(M)$, then $m = u$.
If $\deg(u) > α(M)$, then either $n = 0$ and $m = u$, or $\deg(n) < \deg(m)$.
In the second case we may assume by induction that $n \in U$, so that $m\in U$, as well.