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Wilson's theorem states that a natural number $n > 1$ is a prime number if and only if the product of all the positive integers less than $n$ is one less than a multiple of $n$.

Is there a version of the theorem in $\mathbb Z/p\mathbb Z[t]$ where $p$ is a prime?

Are there any other non-trivial versions?

My motivation is the following.

There is no efficient algorithm for primes without Cramer's conjecture and so perhaps working with irreducible polynomials avoids conjectures and so is finding irreducibles in $\mathbb Z/p\mathbb Z[t]$ any easier?

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    $\begingroup$ From an algebraic point of view, Wilson's Theorem is the observation that the product of all the elements of the units group $\mathbb{Z}/p\mathbb{Z}^\times \cong C_{p-1}$ is the unique involution, namely $-1$, while taking the product of nonzero elements of $\mathbb{Z}/n\mathbb{Z}$ gives $0$, because the ring is not an integral domain. You could run this same argument in $\mathbb{Z}/p\mathbb{Z}[t]$ modulo an ideal. $\endgroup$
    – Mark Wildon
    Commented Jun 6, 2021 at 10:32
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    $\begingroup$ What is "algorithm for primes"? $\endgroup$
    – Max Alekseyev
    Commented Jun 6, 2021 at 11:12
  • $\begingroup$ Guess $x$ and check by $AKS$ (en.wikipedia.org/wiki/AKS_primality_test) and if not prime check $x+1$ and so on until Cramer's conjecture guarantees. The motivation is to utilize a single prime $p$ (say $2$). $\endgroup$
    – Turbo
    Commented Jun 6, 2021 at 11:15
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    $\begingroup$ As Mark says, there is a Wilson's theorem for polynomials. I'm not sure what you want of it regarding avoiding Cramer type conjectures, the usual Wilson's theorem doesn't have anything to do with them. $\endgroup$
    – Wojowu
    Commented Jun 6, 2021 at 13:01
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    $\begingroup$ As for the last question (which I have no idea what it has to do with Wilson's theorem) - $\mathbb Z/p\mathbb Z[t]$ has irreducible polynomials of any positive degree $n$, which is the closest analogue to Cramer I can think of (it is known unconditionally though). For some algorithms for that, see here $\endgroup$
    – Wojowu
    Commented Jun 6, 2021 at 13:07

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