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| + | Since <math>Z_p</math> with p prime is a field, <math>Z_p[x]/<f(x)></math> is a field (thm 17.5 corollary 1). | ||
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| + | <math>Z_p[x]/<f(x)></math> only has elements that are a linear combination of the n terms 1, x ,x^2, ..., x^(n-2), n^(n-1) because any elements of higher degree can be reduced down to a lower degree using the fact that f(x) = 0. Each of the n terms has p possible coefficients, yielding p^n possible elements. | ||
Revision as of 21:59, 12 November 2008
I don't have a clue... ideas? help?
Since $ Z_p $ with p prime is a field, $ Z_p[x]/<f(x)> $ is a field (thm 17.5 corollary 1).
$ Z_p[x]/<f(x)> $ only has elements that are a linear combination of the n terms 1, x ,x^2, ..., x^(n-2), n^(n-1) because any elements of higher degree can be reduced down to a lower degree using the fact that f(x) = 0. Each of the n terms has p possible coefficients, yielding p^n possible elements.
