Two Questions About the Field of p-adic Numbers These questions come from exercise 5 of Gouvea's "$p$-adic Numbers: An Introduction."

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*How do we show that $\mathbb Q_p$ is a field if we only know of its elements as Laurent series in powers of $p$ with finite tails (no norms, no Cauchy sequences, no completeness, etc.)? The only thing I have not been able to show rigorously is that every nonzero element has a multiplicative inverse. The naive answer is something like "because 1 divides by such a Laurent series can be reduced to another such Laurent series," but I cannot prove this.


*How do we show that the injection $\mathbb Q \to \mathbb Q_p$ given by sending a rational to its unique Laurent series expansion is a ring homomorphism, again, only knowing the elements on $\mathbb Q_p$ as Laurent series? I am completely stuck on this.
 A: Let $a = \sum_{i=0}^\infty a_i p^i$ with all $a_i \in \{0,...,p-1\}$ and $a_0 \neq 0$. I claim there is (and you can iteratively find) a (unique) sequence $(b_i)_{i \ge 0}$ with all $b_i \in \{0, ..., p-1\}$ such that for all $n \ge 0$,
$$(\sum_{i=0}^n a_i p^i) \cdot (\sum_{i=0}^n b_i p^i) \equiv 1 \text{ mod } p^{n+1},$$
as follows:
Since $a_0 \neq 0$, there is a unique $b_0 \in \{0,...,p-1\}$ such that $a_0b_0 \equiv 1$ mod $p$.
We now need $a_0b_0 +p (a_0b_1 +a_1b_0) \equiv 1$ mod $p^2$. Because we constructed $b_0$ so that $a_0b_0-1 = c_1 p$ for some $c_1 \in \mathbb Z$, this is equivalent to setting $b_1 \equiv (-c_1-a_1b_0)\cdot {a_0}^{-1}$ mod $p$ which determines $b_1 \in \{0,...,p-1\}$ uniquely.
Next we need $a_0b_0 +p (a_0b_1 +a_1b_0) +p^2 (a_0b_2+a_1b_1+a_2b_0) \equiv 1$ mod $p^3$. Because we constructed $b_0, b_1$ so that $a_0b_0 +p (a_0b_1 +a_1b_0) -1 \equiv c_2p^2$ for some $c_2 \in \mathbb Z$, this is equivalent to setting $b_2 \equiv (-c_2-a_2b_0-a_1b_1)\cdot {a_0}^{-1}$ mod $p$ which determines $b_2 \in \{0,...,p-1\}$ uniquely.
In the general inductive step, we have $a_0b_0 + ... + p^n(a_0b_n+a_1b_{n-1} + ...+ a_n b_0) -1 = c_{n+1}p^{n+1}$ for some $c_{n+1} \in \mathbb Z$ by induction hypothesis, so we are forced to choose $b_{n+1}$ as the unique element of $\{0,...,p-1\}$ such that $b_{n+1} \equiv (-c_{n+1}-a_{n+1}b_{0}-a_n b_1-...- a_1 b_n)\cdot {a_0}^{-1}$ mod $p$.
This suffices to prove 1 (the general case is reduced to this via multiplying through with an appropriate power of $p$).

For 2, let's denote the formulae for addition and multiplication of $p$-adic expansions by $x \oplus y$ and $x \odot y$, respectively. We want to show that for all $x,y \in \mathbb Q$, they agree with the expansion of the "usual" sum $x+y$ and product $x \cdot y$.
As above, by multiplying with appropriate $p$-powers assume w.l.og. both are in $\mathbb Z_p$ i.e. they have non-negative $p$-adic valuation i.e. they are of the form $n/d$ with $n,d \in \mathbb Z$, $gcd(n,d)=1$, $p$ does not divide $d$, i.e. their $p$-adic expansions contains no negative $p$_powers. Let $x= \sum_{i=0}^\infty a_i p^i, y= \sum_{i=0}^\infty b_i p^i$ be their $p$-adic expansions.
Remember what that means: It means that for any $n$, $a_0, ..., a_n$ are the (unique!) elements $\in \{0, ..., p-1\}$ such that the rational number $x- \sum_{i=0}^n a_i p^i$ has $p$-adic valuation $\ge n+1$, which we denote by $x \equiv \sum_{i=0}^n a_i p^i$ mod $p^{n+1}$. Let's call that $n$-th truncated sum
$$x_n := \sum _{i=0}^n a_i p^i.$$
Note this is a non-negative integer. You say you believe that the formula for sum and product of such finite expansions agree with standard sum and product, i.e.
$$x_n \oplus y_n = x_n + y_n \quad \text{ and } \quad x_n \odot y_n = x_n \cdot y_n.$$
But now inspect the sum and product procedure for general expansions, to see that modulo $p^{n+1}$, the $n$-th truncated part of the sum (or product) of expansions agrees with the $n$-th truncated part of the sum (or product) of the respective $n$-th truncated parts,
$$(x\oplus y)_n \equiv (x_n \oplus y_n)_n \text { mod } p^{n+1} \text{ and } (x\odot y)_n \equiv (x_n \odot y_n)_n \text { mod } p^{n+1}$$
or, combining with what you accept,
$$(x\oplus y)_n \equiv (x_n + y_n)_n \text { mod } p^{n+1} \text{ and } (x\odot y)_n \equiv (x_n \cdot y_n)_n \text { mod } p^{n+1}.$$
But from $x \equiv x_n$ mod $p^{n+1}$ it also follows that $(x_n + y_n)_n \equiv (x+y)_n \text { mod } p^{n+1}$ and $(x_n \cdot y_n)_n \equiv (x \cdot y)_n \text { mod } p^{n+1}$, so that
$$(x\oplus y)_n \equiv (x + y )_n \text { mod } p^{n+1} \text{ and } (x\odot y)_n \equiv (x \cdot y)_n \text { mod } p^{n+1}.$$
Or in words: The ("new") sum / product of the $p$-adic expansions agrees with the  $p$-adic expansion of the ("usual") sum / product up to the $p^n$-term. Since this is true for all $n$, they agree in toto.
Note: This is just a very down-to-earth way to spell out the more algebraic definition $\mathbb Z_p = \projlim \mathbb Z/ p^n$ (which gives the ring structure for free), and a very common technique to check something in $\mathbb Z_p$ by checking that it is true modulo $p^n$ for all $n$. One should eventually familiarize oneself with this projective limit definition anyway.
