In Michael Artin states in his Algebra book chapter 13, paragraph 6, the following. Let $L$ be a finite field. Then $L$ contains a prime field $\mathbb{F}_{p}$. Now let us denote $\mathbb{F}_{p}$ by K. If the degree of the field extension $[L:K]=r$, then $L$ as a vector space over $K$ is isomorphic to $K^r$.

My three questions are:

1) Why are $L$ and $K^r$ isomorphic and what is the explicit isomorphism?

2) Over what space is $K^r$ a vector space?

3) Would this then imply that $[\mathbb{F}_{p^n}:\mathbb{F}_{p}]=n$ ?

Thanks a lot in advance!


One common definition of $[L:K]$ is $\dim_K(L)$, the dimension of $L$ as a $K$-vector space. If that's the definition you're using, then the answer to your first question is

1) $[L : K] = r$ means by definition that $L \cong K^r$, since every vector space of dimension $r$ over $K$ is isomorphic to $K^r$ (in a highly non-unique way!).

If $L$ is a principal extension, meaning that $L = K(\alpha)$ for some single element $\alpha \in L$, then this is the same as the degree of the minimal polynomial of $\alpha$ over $K$.

In our case, since finite fields are separable extensions of the prime field $K = \mathbb{F}_p$, the primitive element theorem tells us that $L \cong K(\alpha)$ for some $\alpha$. Let $$ m(x) = x^r + a_1x^{r-1} + \cdots + a_r $$ be the minimal polynomial of $\alpha$ over $K$. Then I claim that a canonical basis for $L$ over $K$ is given by the elements $1, \alpha, \alpha^2, \dots, \alpha^{r-1}$. Indeed, we know that these powers of $\alpha$ are linearly independent over $K$, since an equation of linear dependence (supposing the coefficient on $\alpha^{r-1}$ is nonzero, and so can be taken to be 1) $$ \alpha^{r-1} + b_2 \alpha^{r-2} + \cdots + b_{r-1} = 0 \quad \text{ with $b_i \in K$}$$ gives a polynomial $P(x)$ (by replacing $\alpha$ with $x$) of smaller degree than $m(x)$ such that $P(\alpha) = 0$, contradicting that this was the minimal polynomial. On the other hand, since $m(\alpha) = 0$ we can re-write any polynomial in $\alpha$ of degree $\ge r$ in terms of $1, \dots, \alpha^{r-1}$ using the relation $$ \alpha^r = -(a_1 \alpha^{r-1} + \dots + a_r), $$ coming from $m(\alpha) = 0$. We conclude that $1, \alpha, \dots, \alpha^{r-1}$ is a basis for $L$ over $K$ of size $r$.

2) $K^r$ is a vector space over $K$.

3) Yes.

Let me know if what I wrote is still confusing. I'm afraid I don't have a copy of Artin's Algebra nearby, so I can't check what the definition of $[L : K]$ he's using is.

  • $\begingroup$ About 1). What would be the explicit isomorphism between $L$ and $K^r$? Maybe this is really really trivial, sorry. $\endgroup$ – vaoy Jun 10 '17 at 15:19
  • $\begingroup$ So if we're using the basis $1, \alpha, \dots, \alpha^{r-1}$ that I discussed, and we're viewing $K^r$ as $r$-tuples $(b_1, \dots, b_r)$ of elements in $K$, then we can define the explicit isomorphism by $$ b_{1}\alpha^{r-1} + b_{2}\alpha^{r-2} + \cdots + b_r \mapsto (b_1, b_2, \dots, b_r).$$ The fact that this is an isomorphism is exactly what checking linear independence does. On the other hand, if we're just going with the definition $[L : K] = r$ when $\dim_K(L) = r$, then we pick any $r$ linearly independent elements and send them to $(1, 0, \dots, 0), (0, 1, 0, ...)$ $\endgroup$ – Eric Canton Jun 10 '17 at 15:21
  • $\begingroup$ I know I am a Little bit late to this. But how do I see why 3) is true using 1) and 2)? $\endgroup$ – vaoy Jun 16 '17 at 16:36

Good exercice. I hope I didn't introduce some unnecessary steps.

  • Iteratively construct $\{\alpha_i\}_{i=1}^n$ such that $\{\sum_{i=1}^m c_i\alpha_i, c_i \in\mathbb{F}_p\}$ is a set with $p^m$ elements (start with $m=1$, then $m=2, \ldots$). You automatically get that $$L = \{\sum_{i=1}^n c_i \alpha_i , c_i \in \mathbb{F}_p \}$$ is a $\mathbb{F}_p$ vector space with $p^n$ elements, which means by definition $[L:\mathbb{F}_p] = n$.

  • Using that in characteristic $p$ : $$(\alpha+\beta)^{p^n} = \sum_{j=0}^{p^n} {p^n \choose i} \alpha^i \beta^{p^n-i} = \alpha^{p^n}+ \beta^{p^n}$$ we get that the roots of $X^{p^n}-X \in \mathbb{F}_p[X]$ form an integral domain and hence a field $K$ with $p^n$ elements.

  • In particular for each $d | n$, $K$ contains the roots of $X^{p^d}-X$, ie. a subfield with $p^d$ elements. But this means the other elements $K$ not in those subfields have multiplicative order $p^n-1$. In particular we have $\alpha \in K$ of order $p^n-1$ which means that $$K = \mathbb{F}_p(\alpha)$$

    As any algebraic element, $\alpha$ is defined by its minimal polynomial $f(X) \in \mathbb{F}_p[X]$, and we have the field isomorphisms $$\mathbb{F}_p(\alpha) \cong \mathbb{F}_p[X](f(X)) \cong \mathbb{F}_p(\alpha')$$ for any other root $\alpha'$ of $f$.

  • $L$ is also a field with $p^n$ elements, its multiplicative group is a group with $p^n-1$ elements, ie. $X^{p^n}-X$ splits completely in $L$ and hence $ L$ contains a root $\alpha'$ of $f$. Qed. $$L = \mathbb{F}_p(\alpha') \cong K$$

    That is to say all the finite fields with $p^n$ elements are isomorphic, and we denote it $\mathbb{F}_{p^n}$.


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