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OK, I'm completely lost on this. Define the $p$-adic integers $\mathbb{Z}_p$ as the projective limit $$\lim_{\leftarrow} \mathbb{Z}/p^n \mathbb{Z}.$$ So, if $a \in \mathbb{Z}_p$, then $a$ can be represented by an infinite sequence of numbers $(a_n)_{n \in \mathbb{Z}_{>0}}$, right? So if $a=(a_1, a_2, \cdots) \in \mathbb{Z}_p$, then $$a \in \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p^2\mathbb{Z} \times \cdots,$$ with $a_{n+1} \equiv a_n \mod p^n$. So $a_2 \equiv a_1 \mod p$, $a_3 \equiv a_2 \mod p^2,$ so $a_3 \equiv a_1 \mod p,$ and so on (right?). So if I understand it correctly, when you determine $a_1$, all the other $a_i's$ are fixed. So my first question is: how do you determine this $a_1$?

Also, I don't grasp the concept of the $p$-adic extension (now with $\mathbb{Q}_p$ as an example).Let $p$ be prime, $x \in \mathbb{Q}_p$ and $v = \nu_p(x)$, the $p$-adic valuation of $x$. Then there exists a sequence of integers $0 \leq a_i \leq p-1$ for $i \geq v$ such that $$ x = \sum_{i=v}^{\infty} a_ip^i.$$ This really doesn't make any sense to me, as an element $x \in \mathbb{Q}_p$ is defined to be an infinite sequence of numbers, while $\sum_{i=v}^{\infty} a_ip^i$ clearly has dimension $1$, i.e. it's just a 'regular' number.

I think my misunderstanding of the latter is also the reason why I don't get this: let $p \neq 2$ be a prime and $a \in \mathbb{Z}_p^{\times}$. We define $\Big(\frac{a}{p}\Big)$ to be the legendre symbol $\Big(\frac{a'}{p}\Big)$, with $a' \in \mathbb{Z}$ and $a' \equiv a \mod p$. Again, as $a \in \mathbb{Z}_p^{\times}$ is an infinte sequence of numbers, how can $a \mod p$ be equal to another integer modulo $p$? Can someone clear up my confusion or tell me where I went wrong? Thanks!

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Btw, you should use a different notation with the first $a_i$'s

Say $x \in \mathbb{Z}_p$ and it's in fact a sequence of $x_i \in \mathbb{Z}/p^i$, $i \ge 1$ so that $x_{i+1} \equiv x_i \mod p^{i}$. OK, for $x_1$ you can choose a representative $a_0$ from $0$ to $p-1$ (it's a residue mod $p$). Now $x_2$ is a residue mod $p^2$ so you can take a representative between $0$ and $p^2-1$. Write it in base $p$. It has to be of the form $a_0+ a_1 p$ ( same $a_0$ since $x_2 \equiv x_1 \mod p$). Now move over to $x_3$, choose a representative between $0$ and $p^3-1$ uniquely, you write it in base $p$, it must have an expression as $a_0 + a_1 p + a_2 p^2$ ( the $a_0$, $a_1$ are the same since $x_3 \equiv x_2 \mod p^2$ ).

In the end, $x$ corresponds to an infinite sum $a_0 + a_1 p + \cdots$, determined by the condition $a_0 + \cdots + a_{k-1} p^{k-1} \equiv x_k$

Obs: We are not using at all that $p$ is prime, it's not required at this stage, and you can even have $p=10$, when things are even easier. There you can see that $-1 = 9 + 9 \cdot 10 + 9 \cdot 100 + \cdots $

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  • $\begingroup$ Why does $x$ corresponds to that sum? What I get from your explanation, is that $x = (a_0, a_0 + a_1p, a_0 +a_1 + a_2p^2, \cdots)$, I don't see why an infinite more-dimensional vector can be equal to a one-dimensional infinite sum. $\endgroup$ – Riley Jun 19 '15 at 10:20
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    $\begingroup$ @Riley: Think of a real number and the infinite decimal expansion. If you stop at a finite stage you get approximations of this number. Or , you can go on forever and get the number as an infinite sum. For $p$-adics the $x_n$'s are approximants of $x$. $\endgroup$ – orangeskid Jun 19 '15 at 10:25
  • $\begingroup$ OK, that I understand. But still, take $\pi$ for example. Then $(x_n)_{n \in \mathbb{Z}_{>0}}= (3, 3.1, 3.14, 3.141, 3.1415,...) $ is an infinite series which converges to $\pi$. So if you would say that for $j \rightarrow \infty$ it holds that the term $x_j \rightarrow \pi$, I completely agree. However, this is not the same as saying $\lim_{n\rightarrow \infty} (x_n)_{n \in \mathbb{Z}_{>0}} = \pi$, right? $\endgroup$ – Riley Jun 19 '15 at 11:44
  • $\begingroup$ For $\pi$, if $x_n = 3.1... $ ($n$ digits or so) then, yes, $\lim x_n = \pi$. However, there could be other sequences that converge to $\pi$. This particular one is the one that defines $\pi$. $\endgroup$ – orangeskid Jun 19 '15 at 18:19
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I’ll just address your first question or two. The way I recommend strongly that you think of $p$-adic numbers is to imagine them as $p$-ary expansions extending (potentially) infinitely to the left. Let’s use $5$ for our typical prime number.

You can write any ordinary integer in base $5$, so that the expansion $4213_5$ means $3+1\cdot5+2\cdot5^2+4\cdot5^3$. Same as decimal expansion, but with $5$ as the base. Now remember that $p$-adically, the powers of $p$ are smaller and smaller, so that an expansion going to the left, like $\dots4444_5;$ means $4+4\cdot5+4\cdot5^2+\cdots=\sum_{i=0}^\infty4\cdot5^i$, a good $5$-adically convergent geometric series ’cause the common ratio is $5$, which is smaller than $1$. (I like to use a semicolon for the radix point instead of a period to remind myself that I’m working $p$-adically.)Evaluate it with the formula for geometric series, and get $4/(1-5)=-1$.

This says that the successive approximations $4_5$, $44_5$ (equals $20+4$, remember) and $444_5$ (equals $100+20+4$) all have the property that they’re congruent to $-1$ modulo higher and higher powers of $5$.

One very strong advantage of this notation is that you do addition, subtraction, and multiplication of $5$-adic numbers exactly as you learned to do in elementary school for decimally-notated integers. There will be carries proceeding to the left, just as before. All you need that’s different is a $5$-ary multiplication table (and maybe an addition table will help too). Division has to be done rather differently, because you need it to work right to left, but I won’t go into that.

I’ll leave you with one nice fact: to seven $5$-adic places, $\dots2013233_5;$ is a square root of $-1$ in $\Bbb Z_5$. You can check that the sequence $3_5=3_{10}$, $33_5=18_{10}$, $233_5=68_{10}$, etc. are numbers $n$ such that $n^2+1$ is more and more highly divisible by powers of $5$. That means that the number represented by this infinite $5$-adic expansion (I’ve only written down part of it, of course) is a square root of $-1$ in $\Bbb Z_5$.

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When $a_1$ is fixed modulo $p$, then $a_2$ will be represented by an integer between $0$ and $p^2-1$ (both included), furthermore, modulo $p$ $a_2=a_1$ hence if $a_1$ is between $0$ and $p-1$ one will have :

$$a_2=a_1+\alpha\times p $$

Now $\alpha$ is not determined at all by $a_1$, namely there are $p$ different choices which will give different numbers modulo $p^2$ they are $\alpha=0,...,p-1$.

More generally, if you fixed some $a_{i_0}$ then all the $a_i$'s for $i\leq i_0$ are fixed (explicitely $a_i=a_{i_0}$ mod $p^i$) however there are $p$ choices for $a_{i_0+1}$, $p^2$ choices for $a_{i_0+2}$ which will verify $a_{i_0+1}=a_{i_0}$ mod $p^{i_0}$, $a_{i_0+2}=a_{i_0}$ mod $p^{i_0}$...

Now your second question is clear, the dimension 1 is no more true (since $a_1$ does not determine the whole sequence $(a_i)$).

For your last remark if :

$$x=\sum_{i=0}^{\infty}a_ip^i \text{ with } 0\leq i\leq p-1$$

is any $p$-adic integer then $x\in \mathbb{Z}_p^{\times}$ if and only if $a_0\neq 0$. Furthermore $a_0$ is by definition the congruence of $x$ modulo $p$.

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  • $\begingroup$ Why does $x$ corresponds to that sum? What I get from your explanation, is that $x = (a_0, a_0 + a_1p, a_0 +a_1 + a_2p^2, \cdots)$, I don't see why an infinite more-dimensional vector can be equal to a one-dimensional infinite sum. $\endgroup$ – Riley Jun 19 '15 at 10:23
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    $\begingroup$ @Riley A one-dimensional infinite sum does not make much sens. Think that you can approximate any real number by an infinite sum of succesive $10^{-n}$ with coefficients in $\{0,...,9\}$. Is this is kind of sum a "one-dimensional infinite sum"? Of course not, because $\mathbb{R}$ has a dimension over $\mathbb{Q}$ infinite (even uncoutably infinite). $\endgroup$ – Clément Guérin Jun 19 '15 at 10:29
  • $\begingroup$ OK, that was a rather poor formulation, my apologies for that. What I mean is this: how can an infinite sequence $\alpha = (\alpha_1, \alpha_2, \cdots)$ be equal to something of the form $(\cdot)$, which represents only $1$ number. You have infinitely many terms in $\alpha$ and only one in the other. The dimensions don't add up. It's like saying $(1,2,3,4, \cdots) = a$ for an $a \in \mathbb{Z}$, which is complete nonsense ofcourse. I know that this example is completely off-topic, but I expect that you now understand where my confusion lies. $\endgroup$ – Riley Jun 19 '15 at 12:12
  • $\begingroup$ Perhaps you should start from completions of Q. It can be shown (Ostrowski's theorem) that Q admits only 2 distances which are compatible with + and x : the "archimedaean" one, built from the usual absolute value, and the p-adic one, built from the "p-adic valuation" defined by v_p(a/b) = v_p(a) - v_p(b), where a, b are in Z and v_p (a) = the exact power of p dividing a (same thing for b). Then take completions "à la Cauchy). The archimedean completion gives the reals, R. The p-adic completion gives Q_p. It only remains to check that you recover the definition via inverse limits. $\endgroup$ – nguyen quang do Nov 19 '16 at 9:12

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