Which theorem could be used?

Question

I want to write the $p-$adic expansion of $6!$ in $\mathbb{Q}_3$.

I have to solve the congruence $x \equiv 6! \pmod {3^n}$, right?

I found the following:

$$x_0 \equiv 6! \pmod 3 \Rightarrow x_0 \equiv 0 \pmod 3$$

$$x_1 \equiv 6! \pmod {3^2} \Rightarrow x_1 \equiv 0 \pmod {3^2}$$

$$x_2 \equiv 6! \pmod {3^3} \Rightarrow x_2 \equiv 18 \pmod {3^3}$$

$$x_3 \equiv 6! \pmod {3^4} \Rightarrow x_3 \equiv 72 \pmod {3^4}$$

Using the formula $x_n=\sum_{i=0}^n a_i 3^i$, I found: $a_0=0,a_1=0,a_2=2,a_3=2$.

Is there a theorem I could use, in order to find the coefficients of the infinite series?

A theorem that states, for example, that the solutions we find are periodic?

Answer 1

Every $p$-adic number has a well-defined $p$-adic expansion. When we take an element of $\mathbb{Z}$ and represent it in $\mathbb{Q}_p$, we have very little to do.

You are asking about $6! = 720$ in $\mathbb{Q}_3$. The coefficient of $3^n$ will be $720 \pmod {3^n}$. For $n \geq 6$, we see that $3^n > 720$. So the coefficients for all $3^n$ with $n \geq 6$ will be $720$. It is characteristic of the (normal) integers that they have finite (in this sense) expansions.

You are just left with finding the coefficients for the other $n \leq 5$. Quickly, we see that $$\begin{align} 720 &\equiv 0 &\pmod 3\\ &\equiv 0 &\pmod {3^2}\\ &\equiv 18 &\pmod {3^3}\\ &\equiv 72 &\pmod {3^4}\\ & \equiv 234 &\pmod{3^5}\end{align}$$

and this gives us the entire expansion.

Answer 2

From the comments, it seems you understand now that $720$, just like any other integer, has a finite $3$-adic representation, i.e. one of the form $$720=a_0 3^0+a_1 3^1+\cdots +a_n 3^n,$$ and you want to find the digits $a_0,\cdots, a_n$, where $0\leq a_i\leq 2$.

We will do it step by step. How do we find $a_0$? If we reduce the expression above modulo $3$, we get $720 \equiv a_0\pmod{3}$. Do you understand this step? All positive powers of $3$ go away because we took the remainder modulo $3$.

Now, $720 \equiv 0 \pmod{3}$ so $a_0\equiv 0\pmod{3}$. But there is only one integer $a_0$ between $0$ and $2$ that is zero modulo $3$: the integer $a_0=0$. Thus we found the zeroth digit!

On to the first digit. We reduce the equation $720=a_0 3^0+a_1 3^1+\cdots +a_n 3^n$ modulo $3^2$. What do we get? All powers of $3$ above the first go away and we get $a_0+3a_1 \equiv 0 \pmod{3^2}$ (because $720$ is zero mod $9$). Hmmm, we did not quite get $a_1$, but no worry. We now know $a_0$ from the previous step, so how do we find $a_1$?

Since $a_0+3a_1$ is smaller than $9$ and zero mod nine, it has to equal zero in the integers. This is a crucial property: if a quantity $q$ reduced modulo $M$ is equivalent to $a$ with $0\leq a < M$ and we know that $0 \leq q <M$, then actually $q=a$ as equality of integers. So in our case we get $a_0+3a_1=0$.

Now subtract $a_0$ (a digit we already know) from the equation $a_0+3a_1=0$ and divide by three to get $a_1$. There, we found the first digit which (expectedly) equals zero.

On to the second digit: we reduce the equation $720=a_0 3^0+a_1 3^1+\cdots +a_n 3^n$ modulo $3^3$ to get $$a_0+3a_1 + 9a_2 \equiv 18 \pmod{3^3}.$$ Again, since $a_0+3a_1+9a_2< 27$ we conclude that $$a_0+3a_1 + 9a_2 = 18. $$ But from our previous divsion we had obtained $$a_0+3a_1=0.$$ Subtracting the equations and dividing by $9$, we find $a_2 = 2$. This gives us the second digit.

In general, this is the procedure: reduce your number modulo powers of $3$ consecutively. The first power will immediately give you the zeroth digit. To get the $k$-th digit, reduce $720$ modulo $3^{k+1}$ and modulo $3^k$; consider these remainders as integers, subtract them and divide by $3^k$. The integer you will get is your $k$-th digit.

So for instance you already computed $720 \equiv 234 \pmod{3^5}$ and $720 \equiv 72 \pmod{3^5}$. Subtract $234-72 = 162$ and divide that by $3^4$ to get $a_4=2$.

When do we stop? Your partial expressions $a_0+\cdots +a_k 3^k$ will be smaller than 720 but steadily increasing as you add more and more digits $a_k$, and eventually will become equal to 720. This is where you will stop. (If you continue, the previous process will keep giving you zero digits forever).

I hope this helped, but if you still have unclear points I think you should make a serious review of congruences and their laws before tackling $p$-adic numbers.