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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 $p^n$?

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    @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.2012-04-01

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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 $x_i\in\{0,1,\dots,p-1\}$.