Post #2141606

2025-04-06 21:35 UTC

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.

Replies (1)

@antoinechambertloir@mathstodon.xyz 2025-04-06 21:41

This is where a new variant of the criterion can be applied, this time with the prime number 3. Indeed, modulo 3, one has f(T)=T^4+2T^2+1=(T^2+1)^2. So we are almost as in the initial criterion, but the polynomial T is not T^2+1. The first thing that makes this criterion apply is that T^2+1 is irreducible modulo 3. In this case, this is because -1 is not a square mod 3. The criterion also requires of variant of the condition on the derivative — it holds because the polynomial is not zero modulo (T^2+1, 9). Here, one has T^4-10T^2+1=(T^2+1)^2-12T^2 = (T^2+1)^2-12(T^2+1)+12 is equal to 3 modulo (T^2+1, 9). And so we have an Eisenstein-type proof that the polynomial T^4-10T^2+1 is irreducible over the integers. CQFD. #math

Open ##2141607