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I am stuck reading the proof that the size of every finite field is a power of a prime number, and want to get unstuck.

The lecturer's notes are here. On the second page of his notes, he gives the proof.

I'm OK with this part:

Assume $F$ is a finite field. Consider the canonical map $h:\mathbb{Z} \rightarrow F$, where $n \mapsto n_{F}$ (where $n_{F}$ is $1_{F} + ... + 1_{F}$ n times.) This map $h$ can't be injective so $ker(h) \neq \{0\}$, so it must be $p\mathbb{Z}$ for some prime $p$. (This has previously been established).

So the first isomorphism theorem gives us a homomorphism

$f:\mathbb{Z}/p\mathbb{Z} \hookrightarrow F$

Next, he says "This gives $F$ the structure of a vector space over the field $\mathbb{Z}/p\mathbb{Z}$" and concludes from this that, since $F$ is finite, the size of $F$ must be a power of $p$.

This assertion about the structure of a vector space seems to come out of nowhere, and I can't see what it's based on. What is the principle that justifies this that I'm missing? If you can inject field A into field B, does that mean that field B has the structure of a vector space over field A? Or what? Many thanks.

(Edited to fix mistake. Thank you Michael Albanese.)

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I think you mean $\ker(h)\neq\{0\}$. –  Michael Albanese Jan 9 '13 at 6:19
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To your question about injecting a field $A$ into a field $B$, recall that any field $B$ is a vector space over itself. But any vector space $V$ over a field $F$ is also a vector space over any subfield $k\subset F$, because you already know how elements of $k$ act with $V$ by viewing them as elements of $F$. So $B$ can indeed be viewed as a vector space over $A$ by identifying $A$ with its isomorphic copy contained in $B$. –  Ben West Jan 9 '13 at 6:25
    
Yes, if $A,B$ are fields and $A\subseteq B$ then $B$ can be viewed as a vector space over $A$. For example $\mathbb C$ is a vector space over $\mathbb R$ or $\mathbb Q$, $\mathbb R$ is a vector space over $\mathbb Q$ and in your example $\mathbb F$ is a vector space over $f(\mathbb Z/ p\mathbb Z)\simeq \mathbb Z/ p\mathbb Z.$ –  P.. Jan 9 '13 at 6:31
    
Ah. Yes. Great. Thanks, everyone. –  infinitecardinal Jan 9 '13 at 6:59

2 Answers 2

up vote 2 down vote accepted

Suppose $F\subseteq K$ and $F$ is a subfield of $K$. Then $K$ can be given a vector space structure over the field $F$ by the following : vector space addition is same as the addition in the field $K$ and scalar multiplication $ax$ where $a\in F$ and $x\in K$ is defined as the product $ax$ considering both $a$ and $x$ as elements of $K$. Now try to verify that $K$ is a vector space over $F$.

So in your case we have $F$ is a vector space over the field $f(\mathbb{Z}/p\mathbb{Z})$ and suppose the dimension is $n$ (the dimension has to be finite because $F$ is finite). Then $|F|=|f(\mathbb{Z}/p \mathbb{Z})|^n=|\mathbb{Z}/p\mathbb{Z}|^n$, because $f$ is injective and hence $|F|=p^n$.

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Recall from linear algebra that any vector space $V$ over a field $E$ is also a vector space over any subfield $k\subseteq E$. This follows since we already know how scalars in $k$ act on vectors in $V$ by viewing them as scalars in $E$.

In this case, $F$ may be viewed as an $F$-vector space over itself. The first isomorphism theorem gives us an imbedding of $\mathbb{Z}/p\mathbb{Z}$ into $F$, so $F$ contains a subfield isomorphic to $\mathbb{Z}/p\mathbb{Z}$, namely $f(\mathbb{Z}/p\mathbb{Z})$. So $F$ is a finite dimensional vector space of some dimension $n$ over a field of order $p$. It follows then that $|F|=p^n$, so the order of $F$ must be a power of $p$.

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