Suppose $V$ is an $n$-dimensional vector space over the finite field $\mathbb{F}_p$ for some prime $p$. How do I show that $V$ has $p^n$ elements? I was thinking that considering $n$ basis elements one can show that each element of the basis spans its own subspace which doesn't have any other basis elements, but how do I show that there are exactly $p^n$ elements overall?
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Let $(e_1,\dots,e_n)$ be a base of $V$ as a $\mathbb{F}_p$-vector space. By definition of a basis, the application $$ \begin{array}{rcl}(\mathbb{F}_p)^n & \longrightarrow & V\\(x_1,\dots,x_n) & \longmapsto & x_1e_1 + \dots + x_n e_n\end{array} $$ yields an isomorphism $V \simeq (\mathbb{F}_p)^n$. In particular, $\text{card}V = \text{card}(\mathbb{F}_p)^n = p^n$. |
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Since $V$ is $n$-dimensional over $\mathbb{F}_p$, any given element $v\in V$ can be written as $$v=\lambda _1 v_1 + \cdots +\lambda _nv_n$$ for some $\lambda _1,\dots ,\lambda _n \in \mathbb{F}_p$, where $\{v_1 ,\dots ,v_n\}$ is a basis of $V$ over $\mathbb{F}_p$. For each $v_i$ you have $p$ possibilities for $\lambda _i$, it follows that, since $\{v_1 ,\dots ,v_n\}$ is a basis, you have $p^n$ such $v$'s. |
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Every $n$-dimensional vector space is isomorphic to $\mathbb{F}_p^n$ by choosing a basis. In particular it suffices to prove the statement for $\mathbb{F}_p^n$, where it follows by counting. |
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It is, as Julian says, a simple counting problem. Consider an "empty n-tuple" to be filled with elements from a field, F, with p elements. You have p choices for each of n spaces to fill. Therefore p^n elements. |
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