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I'm trying to show that:

$a_0p^0+a_1p^1+...+a_kp^{k}=b_0p^0+b_1p^1+...+b_kp^{k}$ imply $\forall \ 0\le i \le k \ a_i=b_i$ where $p$ is prime number and $\forall \ 0\le i \le k \ 0\le a_i,b_i<p$

I think that I got an idea to show that, but it's not elegant and I show some way that use a that {$1, p, p^2, ..., p^k$} can be considered as a basis and therefore all coefficients must be equal (just in a way that instead of combination of prime number $p$ and its powers, we would get combination of the basis of polynomials).

My question is why {$1, p, p^2, ..., p^k$} can be considered as a basis? And does it imply just for only for prime numbers?

My way of proving it (in case I'm right):

The problem is equivalent to

$$(a_0-b_0)p^0+(a_1-b_0)p^1+...+(a_k-b_k)p^{k} = 0$$

Now let's assume without loss of generality that $a_k-b_k$ is positive and it gets its lowest value that it isn't zero, so it will be $a_k-b_k=1$. Therefore,
$$S \equiv (b_0-a_0)p^0+(b_1-a_0)p^1+...(b_{k-1}-a_{k-1})p^{k-1}=p^k$$ It's easy to see that the highest value of $S$ is $p^k-1$, and therefore $a_k=b_k$.

In same manner we can show that rest of the coefficients are equal.

Jyrki Lahtonen
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Mr.O
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  • If i am not mistaken all $a_i$ and $b_i$ verify $1\le a_i<p$, $1\le b_i<p$ if that is then a simple induction $\pmod{p}$ yield the result. – Toni Mhax Apr 28 '20 at 07:43
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    @ToniMhax
    You are probably right, and I think that I showed some way to solve this (in the end). I actually ask about the way of solving it by considering {$1, p, p^2, ..., p^k$} as a basis.     – Mr.O Apr 28 '20 at 08:32
  • If you are working over the field $\Bbb{F}_p$ then $p=0$ and so are $p^j$ for all $j>0$. Therefore ${1,p,p^2,\ldots,p^k}$ is not a basis – Jyrki Lahtonen Apr 28 '20 at 09:09
  • Also $a_ip^i=0$ for all $i>0$ and all $a_i\in\Bbb{F}_p$. – Jyrki Lahtonen Apr 28 '20 at 09:12
  • You can just deal ${1, p, p^2, \dots, p^k}$ as a basis. It's enough to make one-to-one correspondence between $p$-adic numerical system on $\mathbb{N}$ and $\mathbb{F}_p[x] \cong \mathbb{F}_p[p]$. – ChoMedit Apr 28 '20 at 10:41
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    @JyrkiLahtonen I'm not working over $\mathbb{F_p}$. It's just that the coefficients are belong to $\mathbb{F_p}$. I mean they get values from from this range. I maybe need to change the notation to $\forall \ 0\le i \le k \ a_i,b_i\in \mathbb{Z}$ and $0 \le a_i,b_i \le p-1$ – Mr.O Apr 28 '20 at 10:53
  • @ChoMedit Can you please elaborate how I can deal {$1,p,p^2,…,p^k$} as a basis. It's actually my main question. – Mr.O Apr 28 '20 at 10:58
  • @JyrkiLahtonen $p$ is a prime number, so I don't really understand why to give the example of $p=0$ – Mr.O Apr 28 '20 at 11:00
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    What I meant is that as an element of $\Bbb{F}_p$ we have $p=0$. This is true for all the primes. As an element of $\Bbb{F}_2$ we have $2=0$, in $\Bbb{F}_3$ we have $3=0$ et cetera. I see that you are not really working over a finite field at all, but you rather want usual base-$p$ representation of natural numbers. – Jyrki Lahtonen Apr 28 '20 at 11:28
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    Remember that $\Bbb{F}_p$ is NOT a subset of the set of integers. Its elements are $\overline{0},\overline{1},\ldots,\overline{p-1}$. That is, residue classes modulo $p$. You are only allowed to leave out the bars when you know not to confuse the concepts :-) – Jyrki Lahtonen Apr 28 '20 at 11:31
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    Yes, it would be clearer, if you wrote $0\le a_i<p$ and $0\le b_i <p$. – Jyrki Lahtonen Apr 28 '20 at 11:32
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    This seems strange. Do you want $p$ to be a generator of a polynomial algebra of some ring $R$? I can't see what the basis means in here. What kind of basis do you consider? – ChoMedit Apr 28 '20 at 11:55
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    I think OP is not really talking about a basis, but about writing numbers to base $p$, and showing each integer has a unique expression in base $p$ (and of course in that case it doesn't matter whether $p$ is prime). – Gerry Myerson Apr 28 '20 at 13:15
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    We already have many answers on the uniqueness of radix representation, e.g. here. – Bill Dubuque Apr 28 '20 at 16:22
  • @ChoMedit I think that Gerry Myerson is right in its observation. I just show that someone wrote that it's a basis, but without saying much more. Thank you. – Mr.O Apr 29 '20 at 23:25
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    @GerryMyerson you probably right. Thank you. – Mr.O Apr 29 '20 at 23:26
  • OK. Then is your question answered at the link that @Gone gave? If so, can we close this question as a duplicate of that one? – Gerry Myerson Apr 29 '20 at 23:31

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