In this blog, I found the following lemma-
Lemma 4: Coefficients of corresponding powers of $(α - 1)$ must be congruent mod $λ$ provided all powers are less than the $(λ - 1)$ st.
if:
$a_0 + a_1(α - 1) + a_2(α-1)^2 + ... + a_{λ-2}(α-1)^{λ-2} ≡ b_0 + b_1(α - 1) + b_2(α-1)^2 + ... + b_{λ-2}(α-1)^{λ-2} (\mod λ)$
then:
$a_0 ≡ b_0 (\mod λ)$
$a_1 ≡ b_1 (\mod λ)$
...
$a_{λ-2} ≡ b_{λ-2} (\mod λ)$
Proof:
(1) If $a_0 + a_1(α - 1) + ... + a_{λ-2}(α - 1)^{λ-2} ≡ $ $b_0 + b_1(α -1) + ... + b_{λ-2}(α-1)^{λ-2} (\mod (α - 1)^{λ-1}$, then:
$c_0 + c_1(α-1) + ... ≡ 0 (\mod (α - 1)^{λ-1})$ where $c_i = a_i - bi$
(2) Then $c_0 ≡ 0 (\mod α - 1)$ which also gives us that $c_0 ≡ 0 (\mod λ)$
(3) Then $c_1(α-1) ≡ 0 (\mod (α-1)^2)$ so that $(α-1)$ divides $c_1$ which means that $c_1 ≡ 0 (\mod λ)$.
(4) We can use the same logic to show that all $c_i ≡ 0 (\mod λ)$
(5) So, $a_i - b_i = 0$, this implies that $a_i = b_i$
I couldn't understand the second line-
(2) Then $c_0 ≡ 0 (\mod α - 1)$ which also gives us that $c_0 ≡ 0 (\mod λ)$
How it is derived that $c_0 ≡ 0 (\mod α - 1)$ and how it implies $c_0 ≡ 0 (\mod λ)$ ?
Thanks for your question. It has been 14 years since I wrote that blog post.
I was using the blog to take notes on this book by Harold Edwards.
Here's the argument:
(1) From Step 1:
$$c_0 + c_1(α-1) + \dots + c_{\lambda-2}(α-1)^{\lambda-2} \equiv 0 \pmod { (α - 1)^{λ-1} } \text{ where } c_i = a_i - b_i$$
(2) Since $(α-1)$ divides $(α-1)^{\lambda-2}$, it follows that:
$$c_0 + c_1(α-1) + \dots + c_{\lambda-2}(α-1)^{\lambda-2} \equiv 0 \pmod { (α - 1) }$$
(3) This wouldn't be true if $c_0 \not\equiv 0 \pmod {\alpha - 1}$ so we can conclude that $c_0 \equiv 0 \pmod {\alpha - 1}$
(4) From Corollary 3.2, here, it follows that:
$$ (α - 1)^{λ-1} = λ \times \text{ unit}$$
(5) Since $\lambda$ divides $(α - 1)^{λ-1}$, it follows that:
$$c_0 + c_1(α-1) + \dots + c_{\lambda-2}(α-1)^{\lambda-2} \equiv 0 \pmod { \lambda }$$