The dual number thing clicks when you see that you have numbers of the form a + bi, with i^2 = 0, but i /= 0.

]]>F_3[sqrt(3)] = F_3[X]/(X^2 – 3) = (Z/(3))[X]/(X^2 – 3) = (Z[X]/(3))/(X^2 – 3) = Z[X]/(3, X^2 – 3) = Z[X]/(3, X^2) = … = F_3[X]/(X^2) = F_3[eps]/(eps^2).

Here “A/I” means the factor ring of a ring A modulo the ideal I. “(3)” means the ideal generated by 3.

]]>The modulus part of your version is a bit of a red herring in this setting. First determine the structure of the full group of units, which is a finitely-generated abelian group. In general, it will have some infinite components and some finite components. Then apply your modulus, with the obvious effects on the components.

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