A proof of the Noether Normalization Lemma

Question

Look at the following proof of the Noether Normalization Lemma taken from Qing Liu's book "Algebraic Geometry and Arithmetic Curves":

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I don't understand the highlighted part. To be more specific, I don't have any idea how to find the vector $m\in\mathbb N^n$ such that $\left<m,\nu_0\right> > \left<m,\nu\right>$ for every $\nu_0\neq \nu$ with $\alpha_\nu\neq 0$.

Thanks in advance.

Answer 1

You could take $m = (d^{n-1},d^{n-2},\ldots,d,1)$ where $d-1$ is the largest digit appearing among the indices $\nu$. For this $m$, the scalar product $\langle m,\nu\rangle$ is the number whose base-$d$ representation is the string $\nu$, and so $\langle m,\nu\rangle>\langle m,\nu'\rangle$ precisely when $\nu$ is greater than $\nu'$ in the lexicographical order.

Answer 2

Induction on $n≥2$. Set $v_i=(v_{i1},…,v_{in})$, $1≤i≤t$ and suppose $v_1>⋯>v_t$ (lexicographical order). Now set $v'_i=(v_{i2},…,v_{in})$. Suppose $v_{11}=⋯=v_{k1}>v_{i1}$ for $i>k$.
If $k=t$ we get $v'_1>⋯>v'_t$ and use induction.
If $k<t$ use induction for $v'_1>⋯>v'_k$ and find $m'∈\mathbb N^{n−1}$ such that $⟨m',v'_1⟩>⟨m',v'_i⟩$ for $i=2,…,k$. For $i>k$ choose $d_i$ such that $$⟨m',v'_1⟩+d_iv_{11}>⟨m',v'_i⟩+d_iv_{i1}.$$ Now set $m_1=\max_{i>k}d_i$, $m=(m_1,m')∈\mathbb N^n$ and we are done.