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I came across a proof of Wilson's theorem as shown below.

enter image description here

I understand why the degree of $f(X)$ has to be strictly less than $p - 1$, and why there are $p - 1$ solutions to $f(a) \equiv 0$ (mod $p)$ in {$1,2,...p-1$}. However, it is not clear to me why, from these two statements, it can be concluded that all coefficients of $f(X)$ are divisible from $p$ [the highlighted statement]. I feel like it has something to do with the fact that $f(X)$ has (at most) $p-2$ terms i.e. there are $p - 2$ coefficients, and that reducing the coefficients mod $p$ will result in at least two coefficients being equal to each other. From there, it is clear to me how to conclude that $f(0) \equiv 0$ (mod $p$), and how this can be used to prove Wilson's theorem.

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If $f$ is not the zero polynomial when reduced modulo $p$, then its reduction to $\mathbf{F}_p[X]$ must have strictly less than $p-1$ roots in $\mathbf{F}_p$ (this is because its degree is already strictly less than $p-1$; any nonzero polynomial of degree $d$ over a field $F$ has at most $d$ roots in $F$). This contradicts the fact that $1, \ldots, p-1$ are all roots of the reduction of $f$ to $\mathbf{F}_p[X]$. As a result, $f$ must be identically zero when reduced modulo $p$, i.e. all of its coefficients are divisible by $p$.

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  • $\begingroup$ Is reducing to $F_p [X]$ just taking the coefficients and reducing them mod $p$? $\endgroup$
    – 1123581321
    Jun 12, 2019 at 6:54
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    $\begingroup$ @1123581321 That is right! The "projection" $\mathbf{Z}[X] \to \mathbf{F}_p[X]$ is just given by reducing all the coefficients mod $p$. $\endgroup$
    – babu_babu
    Jun 12, 2019 at 23:10

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