# Finite field extension isomorphism (Algebra by Michael Artin)

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 $[\mathbb{F}_{p^n}:\mathbb{F}_{p}]=n$ ?

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.

• About 1). What would be the explicit isomorphism between $L$ and $K^r$? Maybe this is really really trivial, sorry. – vaoy Jun 10 '17 at 15:19
• 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 $(1, 0, \dots, 0), (0, 1, 0, ...)$ – Eric Canton Jun 10 '17 at 15:21
• I know I am a Little bit late to this. But how do I see why 3) is true using 1) and 2)? – 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 $$\mathbb{F}_{p^n}$$.