Algebraic Extensions, Extension Degree, and Finite Fields

Suppose $[K:F]=1$ and choose a one-element basis $(b)$. By [L2], write $1_K=cb$ with $c\in F$. Since $1_K\mathrm{\ne }0$, one has $c\mathrm{\ne }0$, so $b=c^{-1}{1}_K$. Every $x\in K$ is therefore of the form $db=(d{c}^{-1})1_K$ with $d\in F$, and hence lies in the embedded copy of $F$.

Conversely, if $K=F$, then $(1_F)$ is a basis of $F$ over itself, so [L1] gives $[F:F]=1$.

Step 1.1 gives $K\subseteq F$, while the extension already has $F\subseteq K$, so $K=F$.

Steps 1.1, 1.2, and 2.1 prove both directions.

For $x\in L$, use the $K$-basis to write $x={\sum }_j{b}_j{v}_j$ with $b_j\in K$, then use the $F$-basis to write $b_j={\sum }_i{c}_{ij}{u}_i$. Thus $x={\sum }_{i,j}{c}_{ij}{u}_i{v}_j$, so the products span $L$ over $F$.

Suppose ${\sum }_{i,j}{c}_{ij}{u}_i{v}_j=0$ with $c_{ij}\in F$. Grouping by $v_j$ gives ${\sum }_j({\sum }_i{c}_{ij}{u}_i)v_j=0$. Independence of the $v_j$ makes every inner coefficient zero, and independence of the $u_i$ then makes every $c_{ij}=0$.

The products span and are independent, hence form an $F$-basis.

Choose bases of sizes $m=[K:F]$ and $n=[L:K]$.

By [L1], their $mn$ pairwise products form an $F$-basis of $L$.

Hence $L\mathrm{/}F$ is finite and [L2] gives $[L:F]=mn=[L:K][K:F]$.

Apply [L2] to the spanning set $K$ of the $F$-vector space $K$ to obtain an $F$-basis $B\subseteq K$. This set is independent in the $F$-space $L$, so [L3] makes $B$ finite; hence $K\mathrm{/}F$ is finite.

Any finite $F$-basis of $L$ is also a finite $K$-spanning set of $L$. Applying [L2] over $K$ gives a finite $K$-basis, so $L\mathrm{/}K$ is finite.

The tower law [L1] now applies and writes $[L:F]$ as $[K:F]$ times the natural number $[L:K]$, proving the divisibility.

The $n+1$ vectors $1,a,\dots ,a^n$ lie in the $n$-dimensional $F$-space $K$, so [L2] gives coefficients $c_0,\dots ,c_n\in F$, not all zero, with ${\sum }_{i=0}^n{c}_i{a}^i=0$.

The polynomial $p(t)={\sum }_{i=0}^n{c}_i{t}^i$ is nonzero and satisfies $p(a)=0$, so $a$ is algebraic by [L3].

Since $a$ was arbitrary, the extension is algebraic. The case $n=0$ cannot occur for a field extension because $1_K\mathrm{\ne }0$.

If $a$ is algebraic, [L1] gives a finite power basis of $F(a)$ and the stated degree.

Conversely, if $F(a)\mathrm{/}F$ is finite, [L2] says every element of $F(a)$, in particular $a$, is algebraic over $F$.

These are the two implications of the equivalence.

Put $F_0=F$ and $F_i=F(a_1,\dots ,a_i)$ for $1\le i\le r$. By [L4], the nonzero polynomial over $F$ satisfied by $a_i$ also belongs to $F_{i-1}[t]$, so $a_i$ is algebraic over $F_{i-1}$. Thus [L2] makes $F_i\mathrm{/}{F}_{i-1}$ finite.

Repeated application of [L3] makes $F_r\mathrm{/}F$ finite, with degree equal to the product of the simple-step degrees.

By [L1], $F_r=F(a_1,\dots ,a_r)$. If $r=0$, this is $F\mathrm{/}F$ of degree one, so the boundary case also holds.

Every $c\in F$ is algebraic over $F$, since it is a root of $t-c$, so $F\subseteq A$ and in particular $0,1\in A$.

By [L1], $F(a,b)\mathrm{/}F$ is finite. Thus [L2] makes all of its elements algebraic over $F$, including $a-b$ and $ab$.

If $a\mathrm{\ne }0$, then $a^{-1}\in F(a)\subseteq F(a,b)$ and is algebraic by the same argument.

Therefore $A$ satisfies the subfield criterion [L3] and is a subfield containing $F$.

Since $a$ is algebraic over $K$, choose a nonzero polynomial $c_0+c_{1}t+\cdots +c_d{t}^d\in K[t]$ with value zero at $a$.

Every coefficient $c_i$ is algebraic over $F$. Hence $M=F(c_0,\dots ,c_d)$ is finite over $F$ by [L1].

The same polynomial lies in $M[t]$, so $a$ is algebraic over $M$ and [L2] makes $M(a)\mathrm{/}M$ finite. The tower law [L3] makes $M(a)\mathrm{/}F$ finite.

By [L4], $a$ is algebraic over $F$. Since $a$ was arbitrary, $L\mathrm{/}F$ is algebraic.

By [L1], every element of $A$ is algebraic over $F$, so $A\mathrm{/}F$ is algebraic.

The element $b$ is algebraic over $A$ by hypothesis, so [L3] makes $A(b)\mathrm{/}A$ finite and [L4] makes it algebraic. Applying [L2] to $F\subseteq A\subseteq A(b)$ shows that $b$ is algebraic over $F$.

By [L1], an element of $K$ algebraic over $F$ belongs to $A$, hence $b\in A$.

Choose $F$-bases $(u_1,\dots ,u_m)$ of $E$ and $(v_1,\dots ,v_n)$ of $E^{\mathrm{\prime }}$. The $F$-span $A$ of the $mn$ products $u_i{v}_j$ contains $E$ and $E^{\mathrm{\prime }}$ and is closed under addition and multiplication, because products are reduced separately in the two bases.

By [L4], every $u_i$ and $v_j$ is algebraic over $F$, so [L5] makes $E{E}^{\mathrm{\prime }}=F(u_1,\dots ,u_m,v_1,\dots ,v_n)$ finite over $F$. Every $r\in A\subseteq E{E}^{\mathrm{\prime }}$ is therefore algebraic over $F$.

If $0\mathrm{\ne }r\in A$, take a nonzero annihilating polynomial, factor out its largest power of $t$, and cancel the corresponding nonzero power of $r$ in the ambient field. This gives $c_0+c_{1}r+\cdots +c_d{r}^d=0$ with $c_0\mathrm{\ne }0$. Then $r^{-1}=-c_0^{-1}(c_1+c_{2}r+\cdots +c_d{r}^{d-1})\in A$, so $A$ is a field.

Since $A$ is a field containing $E\cup E^{\mathrm{\prime }}$, [L1] gives $E{E}^{\mathrm{\prime }}\subseteq A$; the reverse inclusion is clear from the product span, so $A=E{E}^{\mathrm{\prime }}$.

The $mn$ products span $E{E}^{\mathrm{\prime }}$. By [L3] they contain a basis of at most $mn$ elements, so [L2] yields $[E{E}^{\mathrm{\prime }}:F]\le mn=[E:F][E^{\mathrm{\prime }}:F]$.

If no positive multiple of $1_F$ is zero, [L1] gives characteristic $0$. Otherwise write $n=\mathrm{char}F$. Since $1_F\mathrm{\ne }0$, one has $n>1$.

Suppose, for contradiction, that this positive $n$ is not prime. By [L2], write $n=ab$ with $1<a<n$ and $1<b<n$.

Minimality of $n$ gives $a\cdot 1_F\mathrm{\ne }0$ and $b\cdot 1_F\mathrm{\ne }0$, while their product is $(ab)\cdot 1_F=n\cdot 1_F=0$. This contradicts [L3].

Therefore the positive characteristic is prime, completing both cases.

Suppose $\mathrm{char}F=p>0$. The map $\mathbb{Z}\mathrm{/}p\to F$ given by $[n]\mapsto n\cdot 1_F$ is well defined, is a field homomorphism, and is injective; its image is a subfield contained in every subfield of $F$.

Suppose $\mathrm{char}F=0$. The map $\mathbb{Z}\to F$, $n\mapsto n\cdot 1_F$, is injective. Sending a rational class $a\mathrm{/}b$ with $b\mathrm{\ne }0$ to $(a\cdot 1_F)(b\cdot 1_F{)}^{-1}$ is well defined and gives an injective field homomorphism $\mathbb{Q}\to F$.

By [L1], that image equals $F_0$, proving the first classification.

Its image is a subfield and every subfield of $F$ contains all integer multiples of $1_F$ and their nonzero quotients. Hence the image is contained in every subfield and equals $F_0$ by [L1].

Steps 2.1 and 2.2 exhaust the alternatives in [L2].

For $n=0$, both sides are $1$: the left by [L3], and the right is the sole term $\left(\genfrac{}{}{0px}{}{0}{0}\right)x^0{y}^0=1$.

Assume the formula holds for $n$.

Multiply the inductive formula by $x+y$, distribute using [L1], and reindex to obtain the coefficient $\left(\genfrac{}{}{0px}{}{n}{k-1}\right)+\left(\genfrac{}{}{0px}{}{n}{k}\right)$ on $x^k{y}^{n+1-k}$ for each interior $k$.

By [L4] these interior coefficients are $\left(\genfrac{}{}{0px}{}{n+1}{k}\right)$, and [L2] supplies the two endpoint coefficients $1$. Thus the formula holds for $n+1$.

Induction proves the identity for every natural $n$.

Applying [L2] to $(p,k)$ and $(p-1,k-1)$ and cancelling the common nonzero factorial factors yields the integer identity $k\left(\genfrac{}{}{0px}{}{p}{k}\right)=p\left(\genfrac{}{}{0px}{}{p-1}{k-1}\right)$.

Thus $p$ divides $k\left(\genfrac{}{}{0px}{}{p}{k}\right)$. Since $0<k<p$, [L3] gives $p\nmid k$.

Euclid's lemma [L4] therefore forces $p\mid \left(\genfrac{}{}{0px}{}{p}{k}\right)$. The excluded endpoints have coefficient $1$ and are not part of the claim.

By [L1] and [L2], all intermediate terms in $(x+y{)}^p$ have coefficients divisible by $p$ and hence vanish in $F$, so $(x+y{)}^p=x^p+y^p$.

Commutativity gives $(xy{)}^p=x^p{y}^p$, and $1^p=1$, so Frobenius is an endomorphism by [L4].

If $x^p=y^p$, then step 1.1 gives $(x-y{)}^p=0$. A field has no nonzero nilpotents, so $x-y=0$ and the map is injective.

If $F$ is finite, [L5] turns this injection into a bijection, hence an automorphism.

Iterating and using $(x^{p^r}{)}^p=x^{p^{r+1}}$ gives ${\mathrm{Fr}}_F^n(x)=x^{p^n}$, including $n=0$ as the identity.

The characteristic cannot be zero, because the distinct integer multiples of $1_F$ would give infinitely many elements. Hence it is a unique prime $p$, and [L2] identifies the prime subfield with ${\mathbb{F}}_p$.

The finite set $F$ spans itself over ${\mathbb{F}}_p$, so [L4] supplies a finite basis. Its size $n$ is positive because $F$ is not the zero vector space.

By [L5], taking coordinates is a bijection from $F$ to the functions from an $n$-element basis index set to ${\mathbb{F}}_p$. Thus [L6] gives $\mathrm{\mid }F\mathrm{\mid }=p^n$.

Steps 1.1 and 1.2 exhibit the pair $(p,n)$ with $p=\mathrm{char}F$ and $n=[F:{\mathbb{F}}_p]$. For uniqueness, suppose also $\mathrm{\mid }F\mathrm{\mid }={\mathrm{\ell }}^m$ with $\mathrm{\ell }$ prime and $m\ge 1$. Then $p$ divides ${\mathrm{\ell }}^m$, so [L7] gives $p\mid \mathrm{\ell }$, and primality of $\mathrm{\ell }$ forces $p=\mathrm{\ell }$. Now $p^n=p^m$ with $p\ge 2$ forces $n=m$, since $n<m$ would give $1=p^{m-n}\ge p\ge 2$ and symmetrically for $m<n$.

By [L2], $F^{\times }$ is the unit group of the integral domain $F$. It is finite by [L1].

Apply [L3] to the finite subgroup $F^{\times }$ of itself to conclude that it is cyclic.

The elements $0$ and $1$ lie in $R_q$. Since the map $a\mapsto a^q$ is a field endomorphism by [L1], if $a,b\in R_q$ then $(a-b{)}^q=a^q-b^q=a-b$ and $(ab{)}^q=a^q{b}^q=ab$.

By [L3], the derivative of $t^q-t$ is $q{t}^{q-1}-1=-1$, because $q=p^n$ is zero in characteristic $p$. It vanishes nowhere.