Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How many vectors are there in an $n$-dimensional vector space over the field $\mathbb{Z}/(p)$ (where $p$ is prime)? Would the answer be $p^n$?

share|improve this question
1  
Yes. (extra words to meet minimum) –  Brandon Carter Feb 6 '12 at 18:32
    
Thank you very much buddy. –  Hardy Feb 6 '12 at 18:35
    
@BrandonCarter A handy trick I picked from Didier Piau is to insert several $ $ into the comment to meet the minimum. –  Sasha Feb 6 '12 at 18:35
1  
Hint: Use the canonical isomorphism (which is, in particular, a bijection of sets) from your vector space to $(\mathbb{Z}/(p))^n$. –  M Turgeon Feb 6 '12 at 18:36
    
@M Turgeon: there isn't a canonical such isomorphism, as an abstract vector space isn't equipped with a canonical ordered basis (an ordered basis being the same thing as an isomorphism with $(\mathbb{Z}/(p))^n$). But of course there does exist a basis for the vector space, hence such an isomorphism. –  Brad Apr 1 '12 at 20:57
add comment

1 Answer

up vote 1 down vote accepted

Yes,

$$\#(\mathbb{F}_p^{\,n})=(\#\mathbb{F}_p)^n=p^n.$$

The elements can be explicitly constructed as $n$-tuples $(x_1,\dots,x_n)$ with each $x_i\in\{0,1,\dots,p-1\}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.