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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 $c_0 ≡ 0 (\mod λ)$ ?

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1 Answer 1

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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:

$$c_0 + c_1(α-1) + \dots + c_{\lambda-2}(α-1)^{\lambda-2} \equiv 0 \pmod { \lambda }$$

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  • $\begingroup$ I understand $c_0 \equiv 0 \pmod {\alpha - 1}$ but from $ (α - 1)^{λ-1} = λ \times \text{ unit}$ how can we write $c_0 \equiv 0 \pmod { λ }$? because $ (α - 1)^{λ-1} = λ \times \text{ unit}$, not $ (α - 1)= λ \times \text{ unit}$, plz explain. $\endgroup$ Commented Apr 5, 2020 at 19:16
  • $\begingroup$ It comes from step(1). Since $C(\alpha) \equiv 0 \pmod {(\alpha - 1)^{\lambda-1}}$, it follows that $(\alpha-1)^{\lambda - 1}$ divides $C(\alpha)$. Since $\lambda$ divides $(\alpha-1)^{\lambda-1}$, it follows that $\lambda$ likewise divides $C(\alpha)$ $\endgroup$ Commented Apr 5, 2020 at 20:05
  • $\begingroup$ Is $C(\alpha)= c_0 + c_1(α-1) + \dots + c_{\lambda-2}(α-1)^{\lambda-2}$? in this case, it is clear that $C(\alpha) \equiv 0 \pmod {(\alpha - 1)^{\lambda-1}}$, but my issue is with a constant term $c_0$, to be precise, how we write $c_0 \equiv 0 \pmod {λ}$, it is not clear that $C(\alpha) \equiv 0 \pmod {λ}$ implies $c_0 \equiv 0 \pmod {λ}$, because $c_0$ is a constant term of the polynomial $C(\alpha)$ and so far we managed to prove, $c_0 \equiv 0 \pmod {\alpha - 1}$, note that you are proving $c_0 \equiv 0 \pmod {λ}$ from $c_0 \equiv 0 \pmod {\alpha - 1}$. $\endgroup$ Commented Apr 5, 2020 at 20:24
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    $\begingroup$ I did the steps based on how I think. Sometimes, conclusions were not obvious to me so I included additional steps to make it superclear. If you look at the original proof by Edwards, it is very terse and well-done. This was my attempt to make it very easy to process. The result is that most of my steps appear as overkill to professional mathematicians. :-) $\endgroup$ Commented Apr 5, 2020 at 20:42
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    $\begingroup$ If you post another question with any details that you feel are missing, I will be glad to work on it. I've been thinking about doing a reboot on the blog. I've gotten a bit better at math over the years. :-) $\endgroup$ Commented Apr 5, 2020 at 20:49

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