Let $p ∈ \mathbb N, p ≥ 2$, show that : $$(p − 1)!= −1 [p] \iff p \text{ is prime.}$$ Hint : For one direction, use the fact that $\mathbb F$ is a field and pair each element with its inverse.
I'm not sure where to start with this problem. Any help appreciated. Thanks!
This is known as Wilson's Theorem.
For one direction, notice that if $p$ is prime, then we are looking at a field. Other than $1$ and $-1$, every element has a distinct inverse. $(p-1)!$ is the product of all the invertable elements of the field, so we can reorder and regroup the product as $1\cdot(a_1b_1)\cdots (a_kb_k)(p-1)$ where $a_i$ and $b_i$ are multiplicative inverses. Thus the product is $-1$.
For the other direction, if it's not a prime it's not a field and so there are zero divisors. We ca match up the units as before to see that they multiply to $-1$, but the product of zero divisors is always a zero divisor, so the total product is a zero divisor. Since $-1$ cannot be a zero divisor (it's always a unit), the product isn't $-1$.
Here is one of my favorite proof for the direction ($\Rightarrow$)
Suppose $(p − 1)!= −1 [p]$ then we have : $(p − 1)! +pU =-1 \Leftrightarrow -(p-1)!-pU=1$ with $U \in \mathbb{Z}$
So we have a form like this :
$(p-1)\times (p-2)...\times k\times...2\times 1 -pU=1$
According to Bachet-Bézout theorem we have for each $k\in \{1,...,p-1\}$, $\gcd(k,p)=1$ which means that $p$ is prime. Indeed $k<p$ and $k$ cannot divide $p$ because of Bachet-Bézout relations.
For the direction ($\Leftarrow$) you can see the answer of Stella Biderman who used the properties of field and inverses.
