@antoinechambertloir@mathstodon.xyz
Post #2141605
2025-04-06 21:31 UTC
Indeed, for a monic polynomial f in Z[T], a variant of the criterion states:
take an integer a and a prime number p, and assume that :
* f(T) = (T-a)^d mod p
* f'(a) (derivative) is not divisible by p^2.
Then f is irreducible.
For the above polynomial, f(T) = T^4+T^3+T^2+T+1, still taking p=5,
one has f(T) = (T-1)^4 mod 5 and f'(1)=4·5/2=10.
The reason, that avoids any strange computation, is that f(T)=(T^5-1)/(T-1),
which modulo 5 is (T-1)^4 (by the divisibility of the binomial coefficients).
Replies (1)
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@antoinechambertloir@mathstodon.xyz 2025-04-06 21:35
To go back to the initial example, T^4-10T^2+1, there are indeed no prime numbers with which the Eisenstein criterion can be applied. This is obvious in the standard form, because the constant coefficient is 1. But the variant doesn't help neither. The only prime it could is 2, but its derivative at 1 is equal to -16, and is divisible by 4.