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A finite straightedge and compass construction repeatedly draws the line through two available distinct points or the circle with an available center and an available distance as radius, and adjoins isolated intersections of two such lines or circles. A finite set is constructible when its points occur after finitely many such steps, starting with the specified points.

If the current coordinate field is $F$, intersections of two lines solve linear equations over $F$. A line and a circle lead to a quadratic equation over $F$. For two distinct circles, subtraction of their equations gives a line, reducing to the previous case. Thus each step adjoins coordinates in an extension of degree one or two. The tower law gives a finite tower of quadratic-or-trivial field extensions containing all output coordinates. If the final coordinate field contains $K_0$, thenbecause its degree divides the power-of-two degree of that tower.

Two implicit qualifications are necessary for this degree statement. The output is obtained in finitely many steps; the aggregate of infinitely many individually constructible points need not generate a finite extension. Also, unless the output retains the original coordinates, $K$ need not contain $K_0$. In that case the degree statement applies to the compositum $K_{0}K$, rather than an undefined $[K:K_0]$. The heptagon application explicitly includes $S_0$, so neither issue obstructs it.

Here $K_0=\mathbb{Q}$, and a vertex adjacent to $(1,0)$ has first coordinate $c=\cos (2\pi /7)$, so $\mathbb{Q}(c)\subseteq K$. Let $\zeta =e^{2\pi i/7}$. The cyclotomic polynomial $1+X+\cdots +X^6$ is irreducible, giving $[\mathbb{Q}(\zeta ):\mathbb{Q}]=6$. Since $\zeta +{\zeta }^{-1}=2c$, it satisfies $X^2-2cX+1$ over $\mathbb{Q}(c)$. The element $\zeta$ is nonreal while $\mathbb{Q}(c)\subset \mathbb{R}$, so this relative degree is exactly two. By the tower law,If the regular regular heptagon were constructible, this degree would divide $[K:\mathbb{Q}]=2^r$, impossible. Therefore the regular heptagon is not constructible from the given unit segment.

An algebraic field extension $E/K$ is separable if every $\alpha \in E$ has a separable minimal polynomial over $K$.

For an inseparable example, take $K={\mathbb{F}}_p(t)$ and $f(X)=X^p-t$. Eisenstein's criterion at the prime $t$ in ${\mathbb{F}}_p[t]$, followed by Gauss lemma for polynomials, proves irreducibility over its fraction field. Its derivative is zero, and in a splitting field it is $(X-t^{1/p}{)}^p$, so it is inseparable.

Now let $M/K$ be a separable splitting extension of a nonzero polynomial $f$. Every irreducible factor $g$ of $f$ has a root $\alpha \in M$, and $g$ is a scalar multiple of the minimal polynomial of $\alpha$. Separability of $M/K$ makes this polynomial separable. Therefore every irreducible factor of $f$ is separable, which proves the requested conclusion in the factorwise convention. The squarefree qualification under the alternative convention is explained in part (a).

A splitting field of a nonconstant polynomial $f$ over a field $K$ is an field extension $L$ in which $f$ factors into linear factors and which is generated over $K$ by those roots. We prove existence and the divisibility simultaneously by induction on $n=\mathrm{deg}f$; constant nonzero polynomials have splitting field $K$.

If $f$ is reducible, write $f=gh$ with positive degrees $r,s<n$. By induction first split $g$, obtaining $E/K$ of degree dividing $r!$, then split $h$ over $E$, obtaining $L/E$ of degree dividing $s!$. The tower law for field extensions gives $[L:K]\mid r!s!\mid (r+s)!$. All adjoined elements are roots of $f$, so $L$ has the required minimality.

If $f$ is irreducible, take $E=K[x]/(f)$ and let $\alpha$ be the image of $x$. Then $[E:K]=n$ and $f=(x-\alpha )h$ in $E[x]$. Apply induction to $h$, of degree $n-1$, to obtain $[L:E]\mid (n-1)!$. ThereforeThis proof also covers inseparable polynomials; it does not incorrectly identify extension degree with automorphism count in that case.

A field extension $K/F$ is finite when $K$, regarded as a vector space over $F$, has finite dimension. Its degree of a field extension isThus a finite basis $u_1,\dots ,u_m$ means that every $k\in K$ is uniquely expressible as ${\sum }_i{c}_i{u}_i$ with coefficients $c_i\in F$.

The degree of a field extension is the dimension of the larger field as a vector space over the smaller one:The field extension is a finite field extension precisely when this dimension is finite, equivalently when finitely many elements of $K$ span $K$ over $F$. Finiteness here concerns the dimension, not the number of elements of the fields; for example, $\mathbb{C}$ has degree two over $\mathbb{R}$ although both fields are infinite.

Let $[L:F]=n<\infty$. The intermediate field $K$ is an $F$-linear linear subspace of $L$, so the finite-dimensional linear subspace theorem gives $[K:F]\le n$. To handle the other field extension, take an $F$-basis $v_1,\dots ,v_n$ of $L$. Every $\ell \in L$ is a sum $\sum f_i{v}_i$ with $f_i\in F\subseteq K$, so the same list spans $L$ as a $K$-vector space. It need not remain linearly independent, but spanning suffices to give $[L:K]\le n$. Hence both intermediate extensions are finite.

The tower law for field extensions states that, for $F\subseteq K\subseteq L$ with both successive degrees finite,Indeed, if $u_i$ form an $F$-basis of $K$ and $v_j$ form a $K$-basis of $L$, their products $u_i{v}_j$ form an $F$-basis of $L$: expansion gives spanning, and independence follows first over $K$ and then over $F$.

Starting from a unit segment, every straightedge and compass construction introduces coordinates by solving linear or quadratic equations over the coordinates already available. Thus a constructible number belongs to a tower of field extensions whose successive degrees are one or two; the total degree is a power of two. Doubling the volume of a unit cube would require the edge length $\sqrt[3]{2}$. Its minimal polynomial over $\mathbb{Q}$ is $x^3-2$, irreducible by Eisenstein's criterion at the prime number two. Thus $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$. If that number lay in a construction field of degree $2^r$, the tower law would make three divide $2^r$, a contradiction. Therefore doubling the cube by straightedge and compass is impossible.

The degree of a field extension is the dimension of the larger field as a vector space over the smaller one:The field extension is a finite field extension precisely when this dimension is finite, equivalently when finitely many elements of $K$ span $K$ over $F$. Finiteness here concerns the dimension, not the number of elements of the fields; for example, $\mathbb{C}$ has degree two over $\mathbb{R}$ although both fields are infinite.

Let $[L:F]=n<\infty$. The intermediate field $K$ is an $F$-linear linear subspace of $L$, so the finite-dimensional linear subspace theorem gives $[K:F]\le n$. To handle the other field extension, take an $F$-basis $v_1,\dots ,v_n$ of $L$. Every $\ell \in L$ is a sum $\sum f_i{v}_i$ with $f_i\in F\subseteq K$, so the same list spans $L$ as a $K$-vector space. It need not remain linearly independent, but spanning suffices to give $[L:K]\le n$. Hence both intermediate extensions are finite.

The tower law for field extensions states that, for $F\subseteq K\subseteq L$ with both successive degrees finite,Indeed, if $u_i$ form an $F$-basis of $K$ and $v_j$ form a $K$-basis of $L$, their products $u_i{v}_j$ form an $F$-basis of $L$: expansion gives spanning, and independence follows first over $K$ and then over $F$.

Starting from a unit segment, every straightedge and compass construction introduces coordinates by solving linear or quadratic equations over the coordinates already available. Thus a constructible number belongs to a tower of field extensions whose successive degrees are one or two; the total degree is a power of two. Doubling the volume of a unit cube would require the edge length $\sqrt[3]{2}$. Its minimal polynomial over $\mathbb{Q}$ is $x^3-2$, irreducible by Eisenstein's criterion at the prime number two. Thus $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$. If that number lay in a construction field of degree $2^r$, the tower law would make three divide $2^r$, a contradiction. Therefore doubling the cube by straightedge and compass is impossible.

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 cubic has no root in ${\mathbb{F}}_2$, since its values at $0$ and $1$ are both $1$. A cubic over a field is irreducible precisely when it has no root. Consequentlyis a field extension of degree $3$, with eight elements. The Frobenius endomorphism $x\mapsto x^2$ takes a root of a polynomial over ${\mathbb{F}}_2$ to another root, so the roots are $\alpha ,{\alpha }^2,{\alpha }^4$. They are distinct: equality between any two would make the Frobenius endomorphism orbit have length less than $3$, contradicting the minimal polynomial of $\alpha$. They all lie in $K$, and one of them generates $K$. Hence

The polynomial $X^3+X+1$ has no root in ${\mathbb{F}}_2$, so it is an irreducible polynomial. The quotient ring of ${\mathbb{F}}_2[X]$ by the ideal it generates is therefore a field with eight elements. The seven nonzero elements form a group of order seven, so each satisfies $x^7=1$ by Lagrange's theorem. Thus this field splits $X^7-1$. A root different from one has order seven; it cannot lie in ${\mathbb{F}}_2$ and generates this degree-three field extension, since three is prime and the tower law applies. Therefore

An element $\alpha$ is an algebraic element over $F$ if some nonzero polynomial in $F[X]$ vanishes at $\alpha$. A field extension is an algebraic extension if each of its elements is an algebraic element over $F$. If $[K:F]=d<\infty$, the $d+1$ vectors $1,\alpha ,\dots ,{\alpha }^d$ are linearly dependent, producing such a polynomial. Thus every finite field extension is an algebraic extension.

The converse fails for the field of algebraic numbers $\overline{\mathbb{Q}}$. It is an algebraic extension of $\mathbb{Q}$ by definition, but it contains primitive $p$th roots of unity of degree $p-1$ for arbitrarily large primes $p$, by the permitted irreducibility of the cyclotomic polynomial. The tower law would make every such degree at most $[\overline{\mathbb{Q}}:\mathbb{Q}]$ if the latter were finite, a contradiction.

Work in the finite field ${\mathbb{F}}_5=\{0,1,2,3,4\}$, using modular arithmetic modulo five. Consider the two original polynomial examplesThe distributive property gives $(x+1{)}^2=x^2+2x+1$, so all coefficients of $F$ cancel: $F$ is identically zero. In contrast, $G=x^2-3x+2=x^2+2x+2$ in this finite field, with nonzero leading coefficient. Its roots of a polynomial are exactly one and two: a product in a field is zero only if a factor is zero.

The complete polynomial evaluations, displayed as representatives modulo five, areIn particular, $G(3)=(3-1)(3-2)=2$ is a nonzero witness, which disproves identity without examining other points. At one or two, a zero value fails to reveal that $G$ is nonzero. A random variable $a$ with a uniform distribution on a finite set ${\mathbb{F}}_5$ hits those two polynomial roots with probability $2/5$.

A crucial limit appears with $H(x)=x^5-x$. Its coefficients make it a nonzero polynomial of polynomial degree five in ${\mathbb{F}}_5[x]$, yet direct polynomial evaluation gives zero at all five field elements: the values before reduction are $0,0,30,240,1020$, all divisible by five. Thus $H$ induces the same polynomial function as the zero polynomial. Any number of trials restricted to ${\mathbb{F}}_5$ will falsely accept this identity. This is compatible with the Schwartz-Zippel lemma: the ratio of polynomial degree to sampling-set size is $5/5=1$, which gives no useful guarantee.

When a polynomial's representation permits polynomial evaluation over a larger field extension, a larger sampling set can restore the useful degree ratio. Access only to the induced polynomial function on the original finite field cannot distinguish $H$ from zero.^[ref]

If $K(\alpha )/K$ is a simple algebraic field extension of degree $n$, the displayed powers form a basis. The minimal polynomial of $\alpha$ reduces every higher power to a linear combination of them, while a nontrivial linear relation among the first $n$ powers would give a polynomial of degree below the minimal polynomial vanishing at $\alpha$. If $\alpha$ is an algebraic integer over the rational numbers, this is a basis of the order $\mathbb{Z}[\alpha ]$ as a free abelian group; it need not be an integral basis of the whole ring of integers.

A primitive element of ${\mathbb{F}}_q$ generates its multiplicative group. This implies ${\mathbb{F}}_q={\mathbb{F}}_p(\alpha )$, but an element generating the field extension need not generate the multiplicative group. For instance $u\in {\mathbb{F}}_3[u]/(u^2+1)$ generates the quadratic extension but has order four, whereas $1+u$ has order eight and is multiplicatively primitive.

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.

If $L,L^{\prime }$ are field extensions of $K$, choose a $K$-basis of $L$ containing $1$. After tensoring with $L^{\prime }$, that basis expresses $L{\otimes }_K{L}^{\prime }$ as a nonzero direct sum of copies of $L^{\prime }$, and $1\otimes 1$ is nonzero. A prime ideal of this nonzero ring supplies a point used in the point-lifting property of a scheme fibre product.

A transcendence basis of a field extension $L/k$ is an algebraically independent set $T$ such that $L/k(T)$ is algebraic. Its cardinality is the transcendence degree. For finitely generated extensions it is finite, and the remaining algebraic extension is finite.