Question #179099

Construct a field with 125 elements.


1
Expert's answer
2021-05-07T09:59:19-0400

Because 125=53125=5^3 we need to find an irreducible polynomial of the third degree in the ring Z5[X]\mathbb{Z}_5[X] and perform factorization (because if we perform factorization on irreducible polynomial, we get the field. In this field, the "polynomials" will be at most of the second degree and each coefficient will be from Z5\mathbb{Z}_5. From combinatorial considerations it is clear that the power of the field will be 125)

A suitable polynomial x3+x2+x+4x^3+x^2+x+4

So: Z5[X]/<X3+X2+X+4> F125\mathbb{Z}_5[X]/<X^3+X^2+X+4>\ \simeq \mathbb{F}_{125}



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