Past exam of the number theory course of the University of Oxford / 2007 / A

By Vieta's formulas, the elementary symmetric polynomials in the roots giveFor each root, ${\alpha }_j^3+c_1{\alpha }_j^2+c_2{\alpha }_j+c_3=0$. Multiplying by ${\alpha }_j^k$ and summing gives the power sum recurrenceIts initial values are $s_0=3$, $s_1=-c_1$ and $s_2=c_1^2-2c_2$. They are polynomials with integer coefficients in $c_1,c_2,c_3$. The recurrence uses only addition and multiplication by those coefficients, so induction proves the same for every $s_k$. This is the cubic case of Newton's identities.

For $c_1=0$, put ${\alpha }_1=u$, ${\alpha }_2=v$, ${\alpha }_3=-u-v$. Then $c_2=-(u^2+uv+v^2)$ and $c_3=uv(u+v)$. Expanding the product of root differences givesThis is the discriminant of a depressed cubic in depressed form, and the displayed identity also proves its expression in integer-coefficient symmetric polynomials.

Scaling every root scales its elementary symmetric polynomials by the corresponding powers of $d$. Therefore the monic polynomial isThis remains valid for $d=0$, when all three roots are zero; deriving it by symmetric coefficients avoids dividing by $d$.

Reciprocal roots require $c_3\ne 0$, equivalently every original root is nonzero. Their monic polynomial is the normalized reciprocal polynomialIf $c_3=0$, at least one reciprocal in the source is undefined, so that part has no polynomial interpretation without the nonzero-root qualification.

The sum of squared roots is $c_1^2-2c_2$. The sum of their pairwise products isTheir product is $c_3^2$. Thus the required monic polynomial is

An algebraic element is a separable algebraic element over $K$ if its minimal polynomial of an algebraic element has distinct roots in a splitting field. A field extension is a separable extension if all its elements are separable algebraic elements.

The formal derivative is additive and satisfies $D(X^m)=m{X}^{m-1}$. For monomials, $D(X^r{X}^s)=(r+s)X^{r+s-1}=X^{r}D(X^s)+X^{s}D(X^r)$. Distributing over the finitely many coefficients proves the product rule $D(fg)=fDg+gDf$ over any field. If $f=(X-\alpha {)}^{2}h$ in an extension, the rule gives $Df=2(X-\alpha )h+(X-\alpha {)}^{2}Dh$, so $Df(\alpha )=0$, including in characteristic two. If $f,Df$ were relatively prime in $K[X]$, the Bézout's identity for polynomials would give $uf+vDf=1$, whose evaluation at $\alpha$ is impossible. Hence they are not relatively prime.

In characteristic zero, the formal derivative of a positive-degree irreducible polynomial $m$ is nonzero and has smaller degree. Thus $\gcd (m,Dm)=1$, since an irreducible polynomial has no proper nonconstant factor. By the previous argument, $m$ has no repeated root. Every element of a finite field extension is algebraic, so the extension is a separable extension.

For $K={\mathbb{F}}_p(t)$, let ${\alpha }^p=t$. The characteristic-$p$ identity gives $X^p-t=(X-\alpha {)}^p$, so $K(\alpha )$ is already a splitting field. It is a proper extension by the allowed no-root fact. More precisely, the minimal polynomial of an algebraic element $\alpha$ divides $X^p-t$ and, over $K(\alpha )$, equals $(X-\alpha {)}^r$ for some $1\le r\le p$. If $r<p$, its next-to-leading coefficient is $-r\alpha \in K$, forcing $\alpha \in K$ because $r\ne 0$ in ${\mathbb{F}}_p$, a contradiction. Hence its minimal polynomial of an algebraic element is $X^p-t$, with its one root repeated $p$ times. The extension is therefore inseparable, of degree $p$.

By contrast, every finite field extension $L/{\mathbb{F}}_p$ is a separable extension. If $[L:{\mathbb{F}}_p]=n$, then $|L|=p^n$. Lagrange's theorem in its nonzero multiplicative group gives $a^{p^n}=a$ for all $a\in L$. Thus every element is a root of $X^{p^n}-X$, whose formal derivative is $-1$; it has no repeated roots. Every minimal polynomial of an algebraic element divides this separable polynomial, proving the assertion.

Put $\theta =\sqrt{p+\sqrt{p}}$. Then $({\theta }^2-p{)}^2=p$, so $\theta$ is a root ofThis polynomial satisfies Eisenstein's criterion at $p$: every nonleading coefficient is divisible by $p$, while the constant coefficient $p(p-1)$ is not divisible by $p^2$. ThereforeNow suppose $p=n^2+1$, taking $n>0$ without loss of generality. Let $\beta =\sqrt{p-\sqrt{p}}$. Since $\theta \beta =n\sqrt{p}$ and $\sqrt{p}={\theta }^2-p$,Consequently all four distinct roots $\pm \theta ,\pm \beta$ lie in this field. This gives a cyclic quartic field from a nested quadratic radical. It is a splitting field in characteristic zero, hence a normal extension and a Galois extension.

There is a Galois automorphism $\sigma$ sending $\theta$ to $\beta$: a root image defines an embedding, and normality makes it an automorphism of the splitting field. It sends $\sqrt{p}={\theta }^2-p$ to $-\sqrt{p}$, soThus $\theta \mapsto \beta \mapsto -\theta \mapsto -\beta \mapsto \theta$ is a four-cycle. The Galois group has order four and contains an element of order four, givingFor the requested lower degrees, $(q,r)=(3,4)$ gives $\sqrt{q+\sqrt{r}}=\sqrt{5}$ and degree two; $(q,r)=(3,1)$ gives $2$ and degree one. Both pairs use positive integers.

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.