$5$-adic expansion

Question

Given $x = 2+1\cdot5+3\cdot5^2 +2\cdot5^3 +\ldots ∈ \mathbb{Z_5}$, find $\frac{1}{x}$, expressing it similarly as a $5$-adic expansion. (First 4 digits only).

I'm new to p-adic numbers and was struggling to do this question. Any hints?

Answer 1

If you're only interested in the $4$ first "digits", then it means that you are working in $\mathbb{Z}/5^4\mathbb{Z}$. So what you want is the inverse of $332$ in $\mathbb{Z}/625\mathbb{Z}$ (if my mental calculation is correct).

In general, remember that truncating a $p$-adic expansion is just reducing from $\mathbb{Z}_p$ to $\mathbb{Z}/p^n\mathbb{Z}$.

Answer 2

It’s really easiest to do this by writing in standard $5$-ary notation, as you did in elementary school, and as I explained in depth in a recent post. Your number then is written as $2312;$ — the rightmost digit is the units digit, the next to the left is the $5$’s digit, etc.

Division is easily described on a blackboard, but the published word is comparatively unsatisfactory; I’ll try nonetheless. You write the dividend $1$ on the left, but include sufficiently many zero-digits, so that’ll be $0001;$, and on the right you write your divisor, thus: $$ \begin{matrix} 0&0&0&1;&\Big|&2&3&1&2; \end{matrix} $$ Now you look at your rightmost digits, and ask what number mod $5$ gives $1$ when multiplied by $2$, in other words, you want the mod-$5$ reciprocal of $2$. That’s $3$, of course, so that’s the first digit of your quotient. You write that above, and multiply that $3$, now as an ordinary integer, times your divisor $2312;$, and put the product $2441$ below the $0001;\,$. (To do the multiplication, you may find that it helps to prepare yourself with a $5$-by-$5$ multiplication table, so that you know that $3\times3=14$, for instance. So far, you have this: $$ \begin{matrix} &&&3;\\ 0&0&0&1;&\Big|&2&3&1&2;\\ 2&4&4&1; \end{matrix} $$ Next step is to subtract, right? This is long division, don’t forget. When you subtract $2441;$ from $0001;$, you get $2010;$, but I’ll omit that rightmost zero, so that the next few steps look like this: $$ \begin{matrix} &&3&3;\\ 0&0&0&1;&\Big|&2&3&1&2;\\ 2&4&4&1;\\ 2&0&1\\ 4&4&1 \end{matrix} $$ and I’ll let you finish it out. You should get the quotient $4333;$