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How many vectors there are in: $(\Bbb Z_p)_n[x] \ \{n\in\Bbb N, \ p \ is \ a \ prime\}$

I really doubt that we need to use Gauss' formula since we haven't covered that, so there must be a simpler way to count all the vectors in mod(p) up to the nth polynomial but I'm clueless.

Any hints please ?

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GinKin
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  • The user who deleted his answer said that the answer is $P^{n+1}$ without almost any explanation. Why ? How do you count polynomials up to a certain n ? What did he mean by positions ? An example could help. – GinKin Nov 30 '13 at 17:14
  • Is $;\left(\Bbb Z_p\right)_n[x]=$ the set of all polynomials of degree up to $;n;$ with coefficients in the field $;\Bbb F_p:=\Bbb Z/p\Bbb Z=:\Bbb Z_p;$ ? – DonAntonio Nov 30 '13 at 17:26
  • @DonAntonio I'm not really sure what does $\Bbb Z/p\Bbb Z$ means. I wrote all the info I am given. As I understood it, the coeffcients are in modulo P and the field is $\Bbb Z$. – GinKin Nov 30 '13 at 17:32
  • Well, $;\Bbb Z;$ definitely is not a field, and I was almost sure you must know about the finite, prime field $;\Bbb F_p;$ as I wrote in my past comment, otherwise it's not clear to me how you can approach this question... – DonAntonio Nov 30 '13 at 17:34
  • @DonAntonio Well let's ignore what I said about $\Bbb Z$ being a field. I don't think ever heard of the prime field, nor see this $\Bbb F_p:=\Bbb Z/p\Bbb Z$. Maybe this could shed some light, in a related question we were asked to find how many vectors there are in the space: $(\Bbb Z_p)^n$, (the answer is $p^n$). – GinKin Nov 30 '13 at 17:43
  • This is already too long, look at my answer... – DonAntonio Nov 30 '13 at 17:47

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Ok, so let us think only of vectors with $\;n\;$ entries each of which is taken from the set $\;\{0,1,2,...,p-1\}\;$: clearly there are $\;p^n\;$ such vectors (for each entry there are $\;p\;$ choices...).

But you originally asked about $\;(\Bbb Z_p)_n[x]\;$ which, imo, is the set of all polynomials of degree up to $\;n\;$ (and sometimes: of degree less than $\;n\;$ ) with coefficients from $\;\{0,1,2,...,p-1\}\;$ , and again: in the first case there are $\;p^{n+1}\;$ , in the second $\;p^n\;$.

Why? Because each such polynomial can be seen as "a vector $\;(a_0, a_1,...,a_{n-1})\;$ with coefficients in $\;\{0,1,...,p-1\}\;$ (or up to $\;a_n\;$ , in the first case).

DonAntonio
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