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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. | <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. | ||
| + | --- | ||
| + | I feel like I understand it but I am a little confused on the part "Each of the n terms has p possible coefficients, yielding p^n possible elements." can someone explain this a little more? | ||
| + | nshafer | ||
Revision as of 18:46, 16 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. --- I feel like I understand it but I am a little confused on the part "Each of the n terms has p possible coefficients, yielding p^n possible elements." can someone explain this a little more? nshafer
