Consider the first point, \((0,0)\) and let the first break occur at \((i,m i)\), so that the first slope is \(m\). Then by construction no points lie below this line; that is \((j, \nu (j))\) lies above the line for all \(j\). In other words, this means \(\nu (a_j) \geq m j\) for every j. But, if the second slope was smaller we would have to have \((j, \nu (a_j))\) lying below this line, which contradicts above.

Need to write up.

We need to show a Cauchy sequence in \(\mathcal{A}_c\) converges with respect to the norm \(\| \cdot \| _c\). Therefore, consider a sequence of power series

Saying this sequence is Cauchy means

Using the definition of the Gauss norm, this means that

and certainly,

for each \(n\) whenever \(i,j {\gt} M.\) That is each of the sequences \((a_{i,n})_{i \in \mathbb {N}}\) forms a Cauchy sequence. Since \(K\) is complete, this means they are convergent. Now consider

where \(a_n = \lim _{i \to \infty } a_{i,n}\). We want to show this is the limit of the \(f_n\) and is also in \(\mathcal{A}_c\).

For the first part, we know that if \(i,j {\gt} M\) we have

and letting \(j \to \infty \) it follows that if \(i {\gt} M\)

Therefore, we have \(\forall \; i {\gt} M\)

and so \(f_i(x) \to g(x)\) with respect to \(\| \cdot \| _c.\)

Finally, since \(f_i \in \mathcal{A}_c\), there exists \(N\) such that for all \(n {\gt} N\), \(|a_{i,n} | c^n \leq \varepsilon .\) Thus for all \(n {\gt} N\)

and therefore \(|a_n|c^n \to 0\), which by definition means \(g(x) \in \mathcal{A}_c\). With this we conclude the proof.

We need to show that any power series is the limit of a sequence of polynomials. Towards this let

be a power series in \(\mathcal{A}_c\). Consider the sequence of truncations of \(f(x)\),

we want to show \(f\) is the limit of this sequence. Now

which tends to zero as \(k \to \infty \) since \(\lim _{n \to \infty } |a_n|c^n = 0\). Thus \(f_k (x) \to f(x)\) with respect to the Gauss norm and we conclude.

Need to write up.

The idea is to start with an approximate factorisation, and then improve it. If we prove by induction this is always possible, it will produce convergent sequences of polynomials and power series which in the limit gives the factorisation we want.

Base case: To start the induction we need to find a \(\delta {\lt} 1\) and an initial pair \(g_1 (x)\) and \(h_1 (x)\). Towards this consider

Then, since the assumption is that \(\| f(x)\| _c = |a_N |c^N\) and the terms of higher degree are smaller, we have

and let \(\delta \) be a measure of how much smaller:

clearly \(0 {\lt} \delta {\lt} 1\). Then we can easily check that all our properties are satisfied.

Inductive step: Let \(\delta \) be a fixed real number, \(0 {\lt} \delta {\lt} 1\). Suppose at stage \(n\) we have a polynomial \(g_n (x)\) and a power series \(h_n (x)\) such that

1. \(\| f(x) - g_n (x) h_n(x) \| _c \leq \delta ^n \| f(x)\| _c\);

2. \(\deg g_n (x) = N\);

3. \(\| g_n(x) \| _c = |b_{n,N} | c^N\);

4. \(h_n (x) \in \mathcal{A}_c\);

5. \(\| h_n(x) - 1 \| _c \leq \delta ^n\); and

6. \(\| f (x) - g_n (x) \| _c {\lt} \delta \| f(x)\| _c.\)

Firstly, \(\| f(x) - g_n (x) \| _c \leq \delta \| f(x) \| \) implies \( \| f(x) \| _c = \| g(x)_n \| _c\), using the fact \(\| \cdot \| _c\) is non-archimedean and that \(0 {\lt} \delta {\lt} 1\).

By Lemma 4.11 we can find a polynomial \(r(x)\) of degree less than \(N\) and a power series \(q(x) \in \mathcal{A}_c\) such that

and

Now consider

We claim these are better approximations and satisfy the wanted properties.

Firstly, since \(\deg r(x) {\lt} N\) we must have \(\deg g_{n+1} (x) = N = \deg g_n (x)\) and \(g_{n+1}\) will also have leading term \(b_{n,N}\). Moreover, since \(\mathcal{A}_c\) forms a ring, \(h_{n+1} \in \mathcal{A}_c\) (polynomials are trivially restricted power series).

Next, we have

which means by the same argument as for \(g_n\)

We also have

Now we want to show it indeed gives a better approximation:

and therefore

Since, \(0 {\lt} \delta {\lt} 1\) this implies we do indeed have a better approximation.

Therefore \(g_{n+1}\) and \(h_{n+1}\) satisfies all our properties with \(\delta ^n\) replaced by \(\delta ^{n+1}\).

To conclude, we want to show these functions actually form the sequence we want. Firstly, they are Cauchy as

and

Therefore, as \(\mathcal{A}_c\) is complete, we have they converge to a \(g(x)\) and \(h(x)\). It is then immediate via taking limits that these have the wanted properties.

We prove this by induction on \(k\):
If \(k=1\), write the break point as \((N, m_1N)\). We know all points are on or above the line \(y = m x\), and the ones where \(j {\gt} N\) are strictly above it. This translates to

• \(|a_j | (p^j)^{m_1} \leq 1\) for all \(j\),

• \(|a_N | (p^N)^{m_1} = 1\), and

• \(|a_j | (p^{m_1})^j\) < 1 if \(j {\gt} N.\)

This says that \(\| f(x) \| _{p^{m_1}} = 1\) and that the maximum is last realised at the degree \(N\). Thus, we can apply the Weierstrass preparation theorem to conclude that there is a polynomial of degree \(N\) and a power series \(h(x)\) satisfying

Combining the middle equation with \(\| f(x) \| _{p^{m_1}} {\lt} 1\) means \(\| f(x) - g(x) \| _{p^{m_1}} {\lt} 1\). Now, using facts about Newton polygons of polynomials, it follows that the zeros of \(f(x)\) in the closed ball of radius \(p^m\) coincide with the those of \(g(x)\). That is, we must have the Newton polygon of \(g(x)\) coincides with the Newton polygon of \(f(x)\) on \(0 \leq x \leq N\). Finally, we have \(f(x)\) converges on the closed ball of radius \(p^{m_1}\), by Theorem 4.5, and thus so does \(h(x).\)
By induction, assume we have \(f(x) = g_{k-1}(x)h_{k-1}(x)\) for a polynomial \(g_{k-1}(x)\) and power series \(h_{k-1}(x)\), such that the Newton polygon of \(g_{k-1}(x)\) coincides with the first \(k-1\) segments of the polygon for \(f(x)\) and \(h_{k-1}(x)\) has no zeroes in the closed ball of radius \(p^{m_{k-1}}\).
Let us now consider the \(k\)-th segment. By definition, the \(k\)-th segment ending at \(x = N\) means that for any \(i {\gt} N\), \((i, \nu _p (a_i))\) lies above the line of slope \(m_k\) through the point \((N, \nu _p (a_N))\). This means that, writing \(c = p^{m_k}\), \(\| f(x) \| _{p^{c}} = \lvert a_N \rvert c^N\) and \(\lvert a_n \rvert c^N {\gt} \lvert a_i \rvert c^i \) for any \(i {\gt} N\). Thus, we can apply the \(p\)-adic Weierstrass preparation theorem to get a polynomial \(g(x)\). It is then easy to see \(g_1 (x) \mid g(x)\), and that its Newton polygon coincides with the relevant portion of the Newton polygon of \(f(x).\)