Algebraic Extensions, Extension Degree, and Finite Fields — Examples

By [L3], $s^2=2$. The polynomial $t^2-2$ is Eisenstein at $2$, so [L2] makes it irreducible over $\mathbb{Q}$.

Apply [L1] to obtain the basis $(1,s)$ and degree $2$.

Multiplying $(a+bs)(c+ds)$ and reducing $s^2$ to $2$ gives the displayed coordinate formula.

By [L3], $a^3=2$. The polynomial $t^3-2$ is Eisenstein at $2$, so it is irreducible by [L2].

Its minimal polynomial therefore has degree $3$, and [L1] gives degree $3$ and basis $(1,a,a^2)$.

The polynomial $t^2-2$ is irreducible over $\mathbb{Q}$. Also $v\mathrm{\notin }\mathbb{Q}(u)$: if $v=r+su$ with rationals $r,s$, squaring and using uniqueness of the coordinates $(1,u)$ gives $2rs=0$ and $r^2+2s^2=3$; either case would make $3$ or $3\mathrm{/}2$ a rational square, contradicted by comparing the parity of prime exponents in numerator and denominator.

Since $(u+v)(v-u)=v^2-u^2=1$, one has $a^{-1}=v-u$. Therefore $v=(a+a^{-1})\mathrm{/}2$ and $u=(a-a^{-1})\mathrm{/}2$ both lie in $\mathbb{Q}(a)$.

Hence both steps in $\mathbb{Q}\subset \mathbb{Q}(u)\subset \mathbb{Q}(u,v)$ have degree $2$. By [L1] the total degree is $4$, and [L2] gives the product basis $(1,u,v,uv)$.

Thus $\mathbb{Q}(u,v)\subseteq \mathbb{Q}(a)$, while the reverse inclusion follows from $a=u+v$ and [L3]. The fields are equal.

The nonzero real polynomial $(t-a{)}^2+b^2$ vanishes at $z$, because $(z-a{)}^2+b^2=(bi{)}^2+b^2=0$.

Thus every $z\in \mathbb{C}$ is algebraic over $\mathbb{R}$, so [L1] identifies the relative algebraic closure with all of $\mathbb{C}$.

The polynomial $t^2+t+1$ has value $1$ at both $0$ and $1$, so it has no root in ${\mathbb{F}}_2$ and is irreducible. By [L1] and [L2], $A$ is a field.

Every residue has the unique form $c+da$ with $c,d\in {\mathbb{F}}_2$, and the relation is $a^2+a+1=0$, hence $a^2=a+1$.

Applying characteristic-two addition and the reduction in step 2.1 gives every entry in the two displayed tables. In particular $a(a+1)=1$ and $(a+1{)}^2=a$, so the three nonzero elements form the cyclic group $1,a,a+1$ predicted by [L3].

A reducible cubic over a field has a linear factor. The polynomial $t^3+t+1$ has value $1$ at both elements of ${\mathbb{F}}_2$, so it is irreducible; [L1] and [L3] make $A$ a field.

By [L2], the eight coefficient triples in the basis $(1,a,a^2)$ are the eight elements of $A$, and $a^3=a+1$.

Successive multiplication by $a$ and reduction by $a^3=a+1$ gives the displayed powers. Together with $a^0=1$, $a^1=a$ and $a^2$, the powers $a^0,\dots ,a^6$ are the seven distinct nonzero basis combinations, and the next product returns $a^7=1$.

The squares in ${\mathbb{F}}_3$ are $0$ and $1$, so $t^2+1$ has no root and is irreducible. Thus [L1] and [L2] make $A$ a field with the nine elements $r+sa$, where $r,s\in {\mathbb{F}}_3$, and $a^2=2$.

One has $a^2=2$ and $a^4=1$, with $a^2\mathrm{\ne }1$, so $a$ has order $4$. Put $b=1+a$. Then $b^2=2a$, $b^4=2$, and $b^8=1$.

Since $b^2\mathrm{\ne }1$ and $b^4\mathrm{\ne }1$, its order is $8$. Hence it exhausts the eight nonzero elements and is a generator, as [L3] guarantees some element must be.

The modulus has no root in ${\mathbb{F}}_2$, so [L2] and [L3] give the field with relation $a^2=a+1$.

Squaring gives $0^2=0$, $1^2=1$, $a^2=a+1$, and $(a+1{)}^2=a^2+1=a$.

Thus Frobenius swaps the two non-prime-field elements and fixes the prime field. Applying the swap twice is the identity, agreeing with $x^4=x$ from [L1].

The positive divisors of $6$ are $1,2,3,6$.

Apply [L1] with $p=2$ and these four divisors to obtain the orders $2^1,2^2,2^3,2^6$, namely $2,4,8,64$, and the displayed root-set descriptions.

The cubic has no root in ${\mathbb{F}}_2$, so [L1] and [L3] make the quotient a field, and [L2] gives its eight elements. The defining relation is $a^3=a+1$.

Reduction gives $a^0=1$, $a$, $a^2$, $a^3=a+1$, $a^4=a^2+a$, $a^5=a^2+a+1$, $a^6=a^2+1$, and $a^7=1$.

The first seven displayed powers are distinct and are all nonzero elements, so $a$ has order $7$ and generates $A^{\times }$, in agreement with [L4].

By [L4], the elements $a_m$ exist, and $a_{m+1}^2=a_m$, so $K_m\subseteq K_{m+1}$ and the union $K$ is a field.

By [L5] and [L6], $[K_m:\mathbb{Q}]=2^m$ for every $m$.

Every element of $K$ lies in some $K_m$. By [L5] and [L6], the extension $K_m\mathrm{/}\mathbb{Q}$ has finite degree $2^m$, so [L1] makes each of its elements algebraic over $\mathbb{Q}$. Thus $K\mathrm{/}\mathbb{Q}$ is algebraic.

Suppose, for contradiction, that $K\mathrm{/}\mathbb{Q}$ has finite degree $N$. Then [L3] makes $2^m=[K_m:\mathbb{Q}]$ divide $N$ for every $m$. Choosing $m$ with $2^m>N$ is impossible.

Thus $K\mathrm{/}\mathbb{Q}$ is algebraic by step 2.1 but not finite, refuting the statement.

Suppose, for contradiction, that $F$ exists. By [L1], $6=p^n$ for some prime $p$ and $n\ge 1$.

The prime $p$ divides $6=2\cdot 3$, so [L3] gives $p\mid 2$ or $p\mid 3$, and [L2] forces $p=2$ or $p=3$. But no positive power of $2$ is $6$, and no positive power of $3$ is $6$: at exponent one the values are $2$ and $3$, while at exponent at least two they are divisible by $4$ or $9$.

This contradiction proves that no six-element field exists.

By [L1], ${\mathbb{F}}_4$ is a field. In $\mathbb{Z}\mathrm{/}4\mathbb{Z}$, the class $[2]$ is nonzero but $[2{]}^2=[4]=[0]$.

Thus $\mathbb{Z}\mathrm{/}4\mathbb{Z}$ has a nonzero nilpotent and is not a field by [L3], so it cannot be isomorphic to ${\mathbb{F}}_4$.

This single case refutes the universal identification. Equal cardinality does not determine a ring structure.

Choose $L={\mathbb{F}}_{64}$ using [L3], let $F={\mathbb{F}}_2$, and let $K={\mathbb{F}}_4\subseteq L$ from [L2]. Then $[K:F]=2$ and $[L:F]=6$.

By the true tower law [L1], $6=[L:K]\cdot 2$, so $[L:K]=3$.

The false additive formula would give $[L:F]=3+2=5$, contradicting the actual value $6$. Thus degrees multiply, rather than add, in a tower.