# Multiple Table for Four Elements

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I already found the multiplication table for a field with 4 elements (and showed that the nonzero elements form a cyclic group). Hence for the second part of the question, please only write down a multiplication table for 9 elements (and show that the nonzero elements form a cyclic group).

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##### Solution Summary

The expert examines a multiplication table for a field with four elements.

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We look for a second-degree monic irreducible polynomial over F_3; that is, a polynomial with no roots

in F_3. A suitable candidate is x^2 + 1. Hence,

F := F_3[x]/(x^2 + 1)

is a field with 9 elements. We use the same letter x to denote the _coset_ containing x. Notice also that X^2 = 2.

The elements of F all have the form a + bX, where a, b are in F_3, i.e. a and b = 0, 1, or 2. Hence

F = {0, 1, 2, x, x + 1, x + 2, 2x, 2x + ...

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