I have asked many question tonight on $p$-adic and I am still confused. So here is a very basic thing I want to know but nobody has cleared this doubt. It might be very silly, but please answer it.
If $\alpha \in Z_p$, then $\alpha=(\alpha_0,\alpha_1......)$, where $\alpha_i \in Z/p^{i+1}Z$ Right?. Now somebody says that $\alpha \equiv 1 \hspace{.2cm} mod \hspace{.09cm}p$. What does it mean? Should I think of $1$ as $(1,0,0,0\ldots.....)$ as $\alpha-1$ is defined only when $1 \in Z_p$, so by $1$ here, book means that it is $(1,0,0\ldots...)$? And then $\alpha \equiv 1 \hspace{.2cm} mod \hspace{.09cm}p$ must mean that $\alpha_0-1=p.k/pZ=0$ for some $k$ and the rest $\alpha_i \hspace{.3cm}(i \ge 1)$ are just multiples of $p$ like $\alpha_2=p.k_2/p^3Z_p$. Now am i correct, if not please rectify me in easiest language. Please.
I haven't seen any of your other questions, so I'm sorry if this is repetitive.
A $p$-adic integer $a$ isn't just any string of integers $(a_1,a_2,a_3,\cdots)$. The numbers $a_i$ have to be compatible. Each $a_n$ actually determines the numbers before it, i.e. those $a_i$ where $i<n$.
The $a_i$ are in a way better and better approximations of the $p$-adic number $a$ represented by that infinite vector. By analogy with real numbers, it's like representing $\pi$ by the sequence of rationals $(3,3.1,3.14,3.141,\cdots)$.
For the $p$-adics, we have $a_1 \in \mathbb{Z}/p\mathbb{Z}$. This is the most crude approximation to a $p$-adic integer. A slightly better approximation would be a number $a_2$ in $\mathbb{Z}/p^2 \mathbb{Z}$ that refines the approximation of $a_1$, in the sense that $a_2 \equiv a_1 \text{ mod } p$. Similarly $a_3 \in \mathbb{Z}/p^3\mathbb{Z}$ would be yet another refinement, and would have to satisfy $a_3 \equiv a_2 \text{ mod } p^2$ (then automatically $a_3 \equiv a_1 \text{ mod }p$).
So when someone says $\alpha \equiv 1$ mod $p$, that means the digit $a_1=1$. Whatever the other digits are, they have to be compatible with each other, and especially they all have to be $1$ modulo $p$ to be compatible with $a_1$.
So here are some examples of representations. The number $1\in \mathbb{Z}_p$ would corresponds to $(1,1,1,\cdots)$. The number $p$ would be $(0,p,p,\cdots)$. The number $p^2+p+1$ would be $(1,p+1,p^2+p+1,p^2+p+1,\cdots)$.
I find it generally much more convenient to work with the base $p$ representation of $p$-adic numbers. Just like any real number can be written as a (possibly) infinite sum of powers of 10, a $p$-adic number can be written as a possibly infinite sum of powers of $p$, except that in this case it's the positive powers that are allowed to go to infinity, since higher powers of $p$ are smaller in the $p$-adic norm. This means we can represent a $p$-adic number in $\mathbb{Q}_p$ as a sequence of base-$p$ digits that are possibly infinite to the left of the decimal point:
$$ (\cdots b_3 b_2 b_1 b_0.b_{-1} b_{-2} \cdots b_{-n})_p = \sum_{k=-n}^\infty b_k p^k.$$
The integers $\mathbb{Z}_p$ are exactly the ones that don't have digits to the right of the decimal point:
$$ (\cdots b_3 b_2 b_1 b_0)_p = \sum_{k=0}^\infty b_k p^k \in \mathbb{Z}_p.$$
One advantage of this notation is that a normal integer is represented in the ordinary way in base $p$. So for example $51$ as an element of $\mathbb{Z}_5$ in the form above is $(201)_5$, same as its base-$5$ representation.
The relation between the two notations is:
$$a_1 = b_0,\ a_2 = pb_1 +b_0,\ a_3 = p^2 b_2 + p b_1 + b_0 ,\cdots.$$
In other words $a_k$ corresponds to the $k$ most significant digits of the base-$p$ representation.
You can then see that for example $1\in \mathbb{Z}_p$ in vector form will just have $a_k=1$ for all $k$, because the $k$ most significant digits of $1$ in base $p$ will just be 1 again.