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A field extension $L/K$ consists of a field $L$ containing a specified subfield $K$. Its degree $[L:K]$ is the dimension of $L$ as a vector space over $K$, possibly infinite. The inclusion specifies which scalars are available; enlarging them can supply missing roots of a polynomial.^[ref]

For example, $\mathbb{C}/\mathbb{R}$ has basis $1,i$ and degree two, while $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ has basis $1,\sqrt{2}$. The field extension $\mathbb{Q}(t)/\mathbb{Q}$ with $t$ an indeterminate is not algebraic: no nonzero rational-coefficient polynomial vanishes at $t$. The tower law multiplies degrees along finite towers $K\subseteq L\subseteq M$.

A Gaussian period is a sum of roots of unity indexed by a multiplicative coset in the group of nonzero residues modulo a prime number. Such sums describe subfields of a cyclotomic field and occur as values of induced characters of certain semidirect products.

For $m\ge 3$, put $\eta ={\zeta }_m+{\zeta }_m^{-1}$. Since $\eta$ is real and ${\zeta }_m$ is nonreal but satisfies $T^2-\eta T+1=0$, the cyclotomic field $\mathbb{Q}({\zeta }_m)$ has degree two over $\mathbb{Q}(\eta )$. Complex conjugation fixes $\eta$ and is nontrivial. The Fundamental theorem of Galois theory therefore identifies its fixed field with $\mathbb{Q}(\eta )$. That fixed field is exactly $\mathbb{Q}({\zeta }_m)\cap \mathbb{R}$, proving maximality among real subfields.

The splitting field is the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q}$ and all the roots of a polynomial under consideration. Equivalently it is generated over $\mathbb{Q}$ by those roots, and the polynomial splits there into linear factors.

Put ${\zeta }_6=e^{\pi i/3}$. All sixth roots of unity are powers of ${\zeta }_6$, so the required splitting field isIndeed ${\zeta }_6=(1+\sqrt{-3})/2$, and conversely $\sqrt{-3}=2{\zeta }_6-1$. Its field extension degree is two.

The notation $K=F(\delta )$ means that $K$ is the smallest subfield containing both $F$ and $\delta$, namely the simple field extension generated by $\delta$. Its elements can be expressed as rational functions in $\delta$ with coefficients in $F$ and nonzero denominator.

Write $d={\delta }^2\in F$. The element $\delta$ is a root of $X^2-d$, so its minimal polynomial over $F$ has degree at most two. Since $\delta \notin F$, that degree cannot be one. The degree of the simple field extension equals the degree of the minimal polynomial, givingIn particular $1,\delta$ form an $F$-basis, so every element is uniquely $a+b\delta$. This argument also works in characteristic two; separability is not needed for this part.

Choose $\alpha \in K\setminus F$. Since $[K:F]=2$, the linearly independent elements $1,\alpha$ form a basis of $K$ over $F$. Thus there are $a,b\in F$ with ${\alpha }^2=a+b\alpha$. The assumption on the characteristic of a field allows division by two. SetCompleting the square givesAlso $\delta \notin F$, since otherwise $\alpha =\delta +b/2$ would lie in $F$. Because $\alpha \in F(\delta )$ and $\delta \in F(\alpha )$, these generated subfields agree. The basis representation shows $F(\alpha )=K$, henceThe square-root generator is obtained by translating an arbitrary generator to remove the linear term of its quadratic relation.

The characteristic of a field is the least positive integer $p$ with $p{1}_F=0$, or zero if none exists. Finiteness forces a positive characteristic, since the sequence of integer multiples of one must repeat. If this least integer factored into two smaller positive factors, their nonzero field multiples would multiply to zero, contradicting the absence of zero divisors. Hence it is prime. The field is a finite-dimensional vector space over its prime subfield ${\mathbb{F}}_p$, so choosing a basis gives $|F|=p^a$.

A nonzero polynomial of degree $n$ over a field has at most $n$ roots: each root supplies a linear factor by polynomial division, and induction applies to the quotient. In particular $X^n-1$ has at most $n$ roots. The elements of $F^{\times }$ whose orders divide $n$ are precisely those roots. The permitted finite-group criterion therefore proves $F^{\times }$ is cyclic.

For the final assertion take an odd Fermat prime $p=2^m+1$, $m\ge 1$. Primality forces $m$ to be a power of two: an odd factor $s>1$ in $m=2^{r}s$ would factor $(2^{2^r}{)}^s+1$ nontrivially. Since $2^m\equiv -1(\mathrm{mod}p)$, the order of two divides $2m$ but not $m$. As $2m$ is a power of two, its only divisor not dividing $m$ is $2m$ itself. ThusIt generates the full group exactly when $2m=2^m$, which holds for $m=1,2$ and fails for $m\ge 3$. Therefore the possibilities are $p=3,5$, equivalently $p\le 5$ among odd Fermat primes. The oddness is implicit in referring to two as a unit; $p=2$ is not an admissible case for that claim.

The characteristic of a field is the least positive integer $p$ with $p{1}_F=0$, or zero if none exists. Finiteness forces a positive characteristic, since the sequence of integer multiples of one must repeat. If this least integer factored into two smaller positive factors, their nonzero field multiples would multiply to zero, contradicting the absence of zero divisors. Hence it is prime. The field is a finite-dimensional vector space over its prime subfield ${\mathbb{F}}_p$, so choosing a basis gives $|F|=p^a$.

A nonzero polynomial of degree $n$ over a field has at most $n$ roots: each root supplies a linear factor by polynomial division, and induction applies to the quotient. In particular $X^n-1$ has at most $n$ roots. The elements of $F^{\times }$ whose orders divide $n$ are precisely those roots. The permitted finite-group criterion therefore proves $F^{\times }$ is cyclic.

For the final assertion take an odd Fermat prime $p=2^m+1$, $m\ge 1$. Primality forces $m$ to be a power of two: an odd factor $s>1$ in $m=2^{r}s$ would factor $(2^{2^r}{)}^s+1$ nontrivially. Since $2^m\equiv -1(\mathrm{mod}p)$, the order of two divides $2m$ but not $m$. As $2m$ is a power of two, its only divisor not dividing $m$ is $2m$ itself. ThusIt generates the full group exactly when $2m=2^m$, which holds for $m=1,2$ and fails for $m\ge 3$. Therefore the possibilities are $p=3,5$, equivalently $p\le 5$ among odd Fermat primes. The oddness is implicit in referring to two as a unit; $p=2$ is not an admissible case for that claim.

The characteristic of a field is the least positive integer $r$ with $r{1}_K=0$, or zero if none exists. In a finite field a positive such integer exists because the sequence $0,1_K,2\cdot 1_K,\dots$ must repeat.

The prime subfield is the intersection of all subfields of $K$, equivalently the subfield generated by $1_K$. In characteristic $p$ it is the copy of ${\mathbb{F}}_p$ formed by the multiples of $1_K$. Every field automorphism preserves $1_K$ and addition, so it fixes every element of this subfield.

Finally, once the order-$n$ Frobenius automorphism below is established, all its $n$ powers are distinct. Any automorphism fixes $F$ and is determined by its value at the generator $\lambda$. That value must be a root of the degree-$n$ minimal polynomial of $\lambda$, so there are at most $n$ automorphisms. Hence

A splitting field of $f$ over $F$ is a field extension $K/F$ in which $f$ is a product of linear factors and which is generated over $F$ by its roots. The generation condition rules out adjoining unrelated elements.

For existence, induct on the degree of a polynomial. Choose an irreducible polynomial divisor $p$ of $f$. The quotient ring $F[X]/(p)$ is a field because $(p)$ is a maximal ideal; the class $\alpha$ of $X$ is a root of $p$, hence of $f$. Over this field extension, divide out $X-\alpha$ and apply induction to the remaining polynomial. This gives an extension in which $f$ splits. Inside it, the subfield generated by all roots is a splitting field. The induction starts with a linear polynomial, whose splitting field is already $F$. Thus every positive-degree polynomial has a splitting field; being monic is unnecessary for existence.

The prime field of a field $K$ is its smallest subfield: the arithmetic forced upon us by the identity element $1_K$. Every subfield must contain $1_K$, its repeated sums, their additive inverses, and the multiplicative inverses of nonzero results. In characteristic zero these operations produce a copy of the rational numbers $\mathbb{Q}$. In positive characteristic of a field $p$, they produce a copy of ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$, with addition and multiplication given by modular arithmetic.

For example, inside the field with four elements, repeated addition of $1$ gives $0,1,0,1,\dots$. Thus its prime field is $\{0,1\}$, and the other two elements cannot be reached by repeatedly adding $1$. This is why a finite field need not consist entirely of ordinary integer residues: a larger finite field extends the arithmetic of its prime field with additional elements.

The positive characteristic of a field must be a prime number. If its least vanishing multiple of $1_K$ were $rs$ with $1<r,s<rs$, then $(r{1}_K)(s{1}_K)=0$ would exhibit zero divisors. The absence of zero divisors rules this out. The prime field supplies the scalars when viewing $K$ as a vector space; its basis then explains the possible sizes of a finite field.

A simple field extension is a field extension generated by one element: $F(\alpha )$ is the smallest subfield of an ambient field containing $F$ and $\alpha$. Its elements are rational expressions in $\alpha$ with coefficients in $F$ and nonzero denominators. If $\alpha$ is algebraic, its minimal polynomial $m_{\alpha }$ identifies $F(\alpha )\cong F[t]/(m_{\alpha })$, so $[F(\alpha ):F]=\mathrm{deg}{m}_{\alpha }$. The classes of $1,t,\dots ,t^{d-1}$ give a basis when $d=\mathrm{deg}{m}_{\alpha }$.

In particular, ${\alpha }^2\in F$ but $\alpha \notin F$ implies $[F(\alpha ):F]=2$. Conversely, in characteristic different from two, any degree-two extension is of this form: for $\beta \notin F$, express ${\beta }^2=a+b\beta$ in the basis $1,\beta$, then $\alpha =\beta -b/2$ has ${\alpha }^2=a+b^2/4\in F$ and generates the same extension. The restriction on the characteristic makes division by two possible.

The Cyclotomic character of a finite cyclotomic extension and the Chinese remainder theorem give the displayed Galois group. It has three order-two subgroups and three order-four subgroups; the latter are kernels of its three nonzero group homomorphisms to $C_2$. By the Fundamental theorem of Galois theory, its proper intermediate fields strictly above $\mathbb{Q}$ are the three quadratic fieldsand the three quartic fieldsThe first quartic has cyclic Galois group of order four and unique quadratic subfield $\mathbb{Q}(\sqrt{5})$; the second is a biquadratic extension, so they differ. The last is real whereas the others are not, by the maximal real subfield of a cyclotomic field formula. Also $\sqrt{-3}+\sqrt{5}$ generates the second quartic: if $s=\sqrt{-3}+\sqrt{5}$, then $\sqrt{-3}=(s-8/s)/2$ and $\sqrt{5}=(s+8/s)/2$.