I'm trying to understand this part of the proof:
I didn't understand why not all coefficients of $f_2,\ldots,f_n$ can lie in the maximal ideal, maybe I'm forgetting something, it should be a very easy thing, but I couldn't understand why, even though it's an doubt in elementary abstract algebra.
I need help.
Thanks in advance.
On
Edit: Oops, I was completely wrong on the definition of unimodular. Here's a (hopefully) correct answer:
Let $k = \mathfrak{o/m}$ be the residue field. Then the units in $\frac{\mathfrak o[x]}{\mathfrak m[x]} \simeq k[x]$ are the nonzero constant polynomials.
If $f_2, \ldots, f_n \in \mathfrak m[x]$ then reducing the relation to $k[x]$ gives $g_1f_1 = 1$ so $f_1$ is a unit. So $f_1 \in \mathfrak o[x]$ is a constant polynomial, hence $f_1 = 1$ and $f_2 = \cdots = f_n = 0$. In this case it's trivial that $f$ has the extension property.
As for why they conclude that it's not possible for $f_2, \ldots, f_n$ to have coefficients in $\mathfrak m$, I assume there is an implicit assumption that $d > 0$ and under this assumption $f_1$ a unit in $k[x]$ is a contradiction.
Suppose that $f_2,\dots,f_n\in\mathfrak m[x]$. From $\sum_{i=1}^ng_if_i=1$ we get $g_1f_1-1\in\mathfrak m[x]$. Let's write $f_1=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ and $g_1=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_0$. Then we get $$g_1f_1-1=b_mx^{n+m}+(b_{m-1}+b_ma_{n-1})x^{n+m-1}+\cdots+(b_0+b_1a_{n-1}+\cdots+b_ma_{n-m})x^n+\cdots+(b_0a_0-1).$$ (If $n<m$, then the last term in the coefficient of $x^n$ is $b_na_0$.)
This entails that all the coefficients of $g_1$ belong to $\mathfrak m$ and $b_0a_0-1\in\mathfrak m$, a contradiction.