Understanding $p$-adic fields

Question

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!

Answer 1

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 $

Answer 2

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 $p$.

Answer 3

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 $\Bbb Z_5$.