Calculate the power series

Question

I want to find the power series of $\frac{1}{3!}$ in the field $\mathbb{Q}_3$.

In order to do this, do I have to solve the congruence $3!x \equiv 1 \pmod{3^n} \Rightarrow 6x \equiv 1 \pmod 3$?

If so, for $n=1$, we get $0 \equiv 1 \pmod 3$, that is a contradiction.

What does this mean?

Answer 1

It's not a power series but a Laurent series. You can express elements of $\mathbb{Q}_p$ as series

$$\sum_{k=m}^\infty a_k p^k,$$

where $m\in\mathbb{Z}$ and each $a_k$ is an integer between $0$ and $p-1$ inclusive.

For $\frac{1}{3!}$ in $\mathbb{Q}_3$, the shortcut way to find the series is to write

$$\frac{1}{3!} = \frac{1}{3}\cdot \frac{1}{2} = \frac{1}{3}\left(1 + \frac{1}{-2}\right) = \frac{1}{3}\left(1+\frac{1}{1-3}\right).$$

That form should give you an idea on how to continue.

Answer 2

Can I cut through the fog with a somewhat different approach? You can write elements of $\mathbb Z_p$ as infinite $p$-ary expansions going to the left, a typical one would look like $$ \cdots a_4a_3a_2a_1a_0; $$ which means $a_0+pa_1+p^2a_2+\cdots=\sum_{i\ge0}a^ip^i$, and as appropriate for a $p$-ary expansion, all the $a_i$ are integers between $0$ and $p-1$ inclusive. You can add, subtract, and multiply any two of these things using exactly the methods you learned in elementary school, the carries proceeding to the left.

In this notation, $-1$ has the expansion where all the entries are $p-1$. In particular, in $\mathbb Z_3$, $-1=\cdots22222;\,$. You can check that this is right by applying the formula for geometric series, $a/(1-r)$, where in this case $a=2$ and $r=3$, it evaluates to $-1$.

In particular, $-1/2=\cdots11111;\,$, and you get $1/2$ by adding $1$ to this expansion to get $1/2=\cdots111112;\,$.

Your question was to expand $1/6$, which is $3^{-1}\times1/2$. But you know how to divide by the radix $3$ in $3$-ary expansion: move the radix point one to the left, just as you divide by $10$ in decimal by moving the decimal point one to the left. So your expansion is $1/6=\cdots11111;2\,$. You can check that this is right by noticing that your expansion claims that $1/6=2/3-1/2$, which is just right.