Past exam of the number theory course of the University of Oxford / 2010 (B9, DMAB 2648, DMPB 2648)

Source: Mathematics release, bundle Part B 2007 11.pdf, pages 233–236. The paper was sat on 7 June 2010, 2.30–5.30pm, for Mathematics and Mathematics and Philosophy Part B.

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Any extending field homomorphism $j$ is determined by $j(\alpha )$, since every element of $K(\alpha )$ is represented by a polynomial in $\alpha$ with coefficients in $K$. Applying $j$ to the minimal-polynomial equation shows $i(m_{\alpha })(j(\alpha ))=0$. Thus there are at most $m$ choices.

For each listed root ${\beta }_r$, define $K[x]\to L$ by $\sum a_k{x}^k\mapsto \sum i(a_k){\beta }_r^k$. Its kernel contains $(m_{\alpha })$, so it induces a unital map from the field $K[x]/(m_{\alpha })\cong K(\alpha )$ into $L$. A unital map from a field is injective. This gives an extending homomorphism $j_r$ with $j_r(\alpha )={\beta }_r$. Distinct roots give distinct maps, proving exactly $m$ extensions.

An irreducible polynomial is separable if its roots in a splitting field are distinct, equivalently if it is coprime to its formal derivative. For a general nonzero polynomial, the factorwise separability convention means that every irreducible factor is separable. This is the convention under which the printed splitting field implication holds without an extra hypothesis.

If instead separability of an arbitrary polynomial is defined to require all its roots to be distinct, that implication needs the polynomial to be squarefree. For example, $(x-1{)}^2$ has separable splitting extension $K/K$, but a repeated root. These two conventions agree for irreducible polynomials; repeated factors are the entire distinction here.

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).

First, the number of embeddings extending a prescribed base embedding is at most the extension degree. Prove this for any finite extension by choosing finitely many algebraic generators and applying part (i) successively. At each step the number of extensions is at most the degree of that simple step; multiplication and the tower law for field extensions give the total bound.

If $\tilde{K}/K$ is not separable, choose $\alpha$ with inseparable minimal polynomial of degree $e=[K(\alpha ):K]$. Its image under $i$ has strictly fewer than $e$ distinct roots in any target field: inseparability is preserved under the coefficient embedding. Thus there are $r<e$ embeddings of $K(\alpha )$ extending $i$. Each extends to $\tilde{K}$ in at most $[\tilde{K}:K(\alpha )]$ ways, giving at mostIf instead $i(m_{\alpha })$ does not split in $L$, it again has fewer than $e$ distinct roots in $L$, and exactly the same estimate applies. Hence either obstruction makes the embedding count strictly less than $d$.

Set $K_r=K({\alpha }_1,\dots ,{\alpha }_r)$, beginning with $K_0=K$. Suppose $j:K_{r-1}\to L$ extends $i$. The minimal polynomial of ${\alpha }_r$ over $K_{r-1}$ divides its minimal polynomial over $K$. Its image under $j$ therefore divides $i(m_{{\alpha }_r})$, which splits into distinct linear factors in $L$. It consequently has exactly $[K_r:K_{r-1}]$ distinct roots in $L$. Part (i) supplies exactly this many extensions of $j$ to $K_r$.

Starting with the unique prescribed map on $K$, multiply these branching counts:For the final assertion, let $M$ be a splitting field of a separable polynomial and embed it in an algebraic closure $\Omega$. It is generated by roots whose minimal polynomials are separable and split in $\Omega$. Applying the just-proved count with $L=\Omega$ gives $[M:K]$ embeddings. Were $M/K$ inseparable, part (a) would give strictly fewer. Thus the splitting extension is separable. This works for the factorwise convention as well, since the generators' minimal polynomials divide the separable irreducible factors.

The Galois group $\mathrm{Gal}(L/K)$ is the group of all field automorphisms of $L$ fixing every element of $K$, with composition as its operation. A finite extension is a Galois extension if it is both normal and separable; equivalently, it is a splitting extension of separable polynomials, or its group has order $[L:K]$. Normality means that every irreducible polynomial over $K$ with a root in $L$ splits there.

The Fundamental theorem of Galois theory gives an inclusion-reversing bijection for a finite Galois extension $L/K$ with group $G$:It satisfies $[L:L^H]=|H|$ and $[L^H:K]=[G:H]$. An intermediate extension $M/K$ is Galois precisely when its corresponding subgroup $H$ is normal, in which case restriction induces $\mathrm{Gal}(M/K)\cong G/H$.

The cyclotomic polynomial isBy the allowed irreducibility, this is the minimal polynomial of $\xi$ and its degree is $\phi (n)$. The field $\mathbb{Q}(\xi )$ contains every $n$th root of unity and is the splitting field of $X^n-1$. This polynomial has no repeated roots in characteristic zero, so the extension is Galois.

An automorphism sends $\xi$ to another primitive root ${\xi }^a$, where $a$ is a unit modulo $n$. Conversely each such conjugate defines an embedding, and it is an automorphism of $\mathbb{Q}(\xi )$ because $a$ has an inverse modulo $n$. Composition corresponds to multiplication of exponents. Therefore

Let $a=\sqrt[n]{2}$ and $F=\mathbb{Q}(\xi )$. The roots of $X^n-2$ are $a{\xi }^j$, so $L=F(a)$. This splitting extension is separable in characteristic zero, hence Galois over $\mathbb{Q}$. The subextension $F/\mathbb{Q}$ is itself Galois by part (iii), so restriction gives a surjectionwith normal kernel $H=\mathrm{Gal}(L/F)$. Surjectivity follows from the extension theorem for embeddings and normality of $L/\mathbb{Q}$.

For $\sigma \in H$, write $\sigma (a)=a{\xi }^{j(\sigma )}$. This defines an injective homomorphism $H\to ⟨\xi ⟩$: injectivity holds because $L=F(a)$, and multiplicativity holds because $H$ fixes all roots of unity in $F$. A subgroup of a cyclic group is cyclic. The first isomorphism theorem therefore givesThere is no need to assume that $|H|=n$; the intersection with the cyclotomic field can reduce its order.

Let the monic minimal polynomial of $\alpha$ over $L$ have coefficients $a_0,\dots ,a_{r-1}$, and put $F=K(a_0,\dots ,a_{r-1})$. Each coefficient is algebraic over $K$, so adjoining these finitely many coefficients successively gives $[F:K]<\infty$. The same polynomial makes $\alpha$ algebraic over $F$, giving $[F(\alpha ):F]\le r$. The tower law for field extensions yields $[F(\alpha ):K]<\infty$. Its subfield $K(\alpha )$ therefore also has finite degree over $K$. This holds for every $\alpha \in M$, proving $M/K$ is algebraic.

An algebraically closed field is a field in which every nonconstant polynomial has a root, equivalently every such polynomial splits. An algebraic closure of $K$ is an algebraic extension of $K$ which is algebraically closed.

For the displayed subset $\overline{K}\subseteq \mathbb{C}$, if $\alpha ,\beta$ are algebraic over $K$, then $K(\alpha ,\beta )/K$ is finite. Sums, differences, products and nonzero inverses of its elements are algebraic over $K$, so $\overline{K}$ is a subfield. If a nonconstant polynomial has coefficients in $\overline{K}$, those finitely many coefficients generate a finite extension $F/K$. It has a root $z\in \mathbb{C}$ by algebraic closure of $\mathbb{C}$, and $z$ is algebraic over $F$. Part (i) makes it algebraic over $K$, so $z\in \overline{K}$. Hence $\overline{K}$ is algebraically closed and is an algebraic closure of $K$.

Every $z\in \overline{K}$ lies in the finite subextension $K(z)$, and every element of a finite subextension is algebraic over $K$. Therefore $\overline{K}$ is exactly the union of all finite subextensions inside $\mathbb{C}$. The two examples follow.

Take $K=\mathbb{R}$. Every complex number is algebraic over $\mathbb{R}$, since $\mathbb{C}=\mathbb{R}(i)$, so its algebraic closure inside $\mathbb{C}$ is $\mathbb{C}$. This is a proper containment and

Take $K=\mathbb{Q}$. For every positive integer $n$, the real number $\sqrt[n]{2}$ lies in $\overline{\mathbb{Q}}$ and has degree $n$, by Eisenstein's criterion at $2$. If $\overline{\mathbb{Q}}/\mathbb{Q}$ had finite degree, all its subextension degrees would be bounded by that degree, contradicting this unbounded family. Therefore $[\overline{\mathbb{Q}}:\mathbb{Q}]$ is infinite.

Identify $\mathbb{C}$ with the Euclidean plane and start with the points $0,1$. A complex number is a constructible number if a finite sequence of drawing lines through known points and circles with known centres and known distances as radii, and selecting their intersections, produces that point.

Line-line intersections solve linear equations in the current coordinate field. Line-circle and circle-circle intersections require at most one quadratic equation after eliminating a coordinate. Hence all coordinates obtained lie in a finite tower of real quadratic extensions of $\mathbb{Q}$. If $\alpha =x+iy$ is constructible, choose a common such real field $F$ containing $x,y$. Adjoining $i$ if needed still gives an extension $F(i)/\mathbb{Q}$ of power-of-two degree. By the tower law for field extensions,A regular heptagon centred at $0$ with one vertex $1$ would have $\xi =e^{2\pi i/7}$ as another constructible vertex. But its cyclotomic degree is $\phi (7)=6$, which is not a power of two. Thus the regular heptagon is not constructible.

An integral basis is a $\mathbb{Z}$-basis of the ring of integers ${\mathcal{O}}_K$; its elements also form a $\mathbb{Q}$-basis of $K$.

To prove existence, first multiply the elements of a rational basis by suitable nonzero integers to obtain an integral rational basis $w_1,\dots ,w_n$. Put $M={\sum }_j\mathbb{Z}{w}_j$. The trace pairing is nondegenerate in characteristic zero, and the matrix $T=(\mathrm{Tr}(w_i{w}_j))$ has integer entries and nonzero determinant $\Delta$. If $\alpha ={\sum }_j{c}_j{w}_j$ is integral, all $\mathrm{Tr}(\alpha w_i)$ are integers, so $Tc\in {\mathbb{Z}}^n$. The adjugate formula gives $c\in {\Delta }^{-1}{\mathbb{Z}}^n$. ConsequentlyA subgroup of a free abelian group is free, and these inclusions force rank $n$. A $\mathbb{Z}$-basis of ${\mathcal{O}}_K$ is therefore an integral basis.

Now put $s={\theta }^2=\sqrt{d}$ and $F=\mathbb{Q}(s)$. The polynomial $X^4-d$ is irreducible by Eisenstein at any prime dividing the squarefree $d$, except for $d=-1$, where $(X+1{)}^4+1$ is Eisenstein at $2$. Thus $[K:\mathbb{Q}]=4$ and $[K:F]=2$. Every $\gamma \in K$ has the unique form $\alpha +\beta \theta$ with $\alpha ,\beta \in F$. The nontrivial $F$-automorphism sends $\theta$ to $-\theta$, so an integral $\gamma$ has integral conjugate $\alpha -\beta \theta$. Relative trace and norm of integral elements are integral in $F$. Applying them to $\gamma$ and taking the trace of $\theta \gamma$ givesThe power basis has discriminantIndeed, the resultant of $X^4-d$ and $4X^3$ is $4^4\prod {\theta }_j^3=-256d^3$; the discriminant sign factor is $(-1{)}^6=1$. By the discriminant-index formula for an integral lattice, a prime dividing $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$ must divide $2d$.

No odd prime $p\mid d$ can divide this index. The Eisenstein equation gives ${\theta }^4/p\in {\mathcal{O}}_K$, while $r{\theta }^i/p$ has nonintegral field norm for $p\nmid r$ and $0\le i<4$: its $p$-adic norm valuation is $i-4<0$. If $p^{-1}{\sum }_{i=0}^3{r}_i{\theta }^i$ were integral, multiplication by ${\theta }^3$ first forces $p\mid r_0$; subtracting $r_0/p$ and multiplying by ${\theta }^2$ forces $p\mid r_1$. Continuing forces all coefficients divisible by $p$. This excludes an index element of order $p$.

To exclude $2$, any index element of order two would be represented by an integralwith integers $a,b,c,e$ not all even. Here $\alpha =(a+cs)/2$ and $\beta =(b+es)/2$. The integrality of ${\alpha }^2-{\beta }^{2}s$ forcesSince $d\equiv 3(\mathrm{mod}4)$, the first congruence implies $a,c$ have the same parity. If both were odd, the second would require $b^2+3e^2\equiv 2(\mathrm{mod}4)$, impossible. Hence $a,c$ are even. The second congruence then forces $b,e$ to have the same parity, while the first forces $be$ even. Thus $b,e$ are even too, a contradiction. Therefore the index is one andFor the last request, let $t=\sqrt[4]{5}$ and take $\delta =(t+t^3)/2$. It satisfies ${\delta }^2=(5+3\sqrt{5})/2$, and henceThis monic integer polynomial proves integrality. Its coordinates in the rational power basis have halves, so $\delta \notin \mathbb{Z}[t]$. Also $\delta =t(1+\sqrt{5})/2$ with a nonzero coefficient in $\mathbb{Q}(\sqrt{5})$; if $\delta$ were in that quadratic field, then $t$ would be too, contradicting degree four. Thus it meets both exclusions.

The product of ideals $AB$ consists of finite sums ${\sum }_j{a}_j{b}_j$ with $a_j\in A$, $b_j\in B$. Ideal divisibility $P\mid A$ means $A=PC$ for an integral ideal $C$; in this ring it is equivalent to $A\subseteq P$. The ideal norm is $N(A)=|{\mathcal{O}}_K/A|$, equivalently its additive lattice index.

If the prime ideal $P$ divides $AB$ but does not divide $A$, choose $a\in A\setminus P$. For every $b\in B$, $ab\in AB\subseteq P$, so primality gives $b\in P$. Thus $B\subseteq P$, proving $P\mid B$.

Every nonzero proper ideal $A$ is contained in a maximal ideal $P$, which is prime. Then $P\mid A$, so $A=PB$. Norm multiplicativity gives $N(B)=N(A)/N(P)<N(A)$ because $N(P)\ge 2$. Induction on the positive integer $N(A)$ gives a prime-ideal factorization, ending at norm one. For uniqueness, a prime in one factorization divides some prime in the other. Every nonzero prime ideal is maximal, since its quotient is a finite integral domain and hence a field. The two primes therefore coincide. Cancel that common factor using the permitted cancellation law, and continue. This proves uniqueness up to order.

The finite additive group ${\mathcal{O}}_K/A$ has order $N(A)$, so that integer annihilates the class of $1$. ThereforeIn particular an ideal of norm $m$ contains $(m)$. Such ideals correspond to certain additive subgroups of the finite group ${\mathcal{O}}_K/(m)$, so there are finitely many. Finiteness of these quotients follows from an integral basis; a nonzero ideal contains a nonzero principal ideal, whose index is the absolute field norm of its generator.

For the computation put $s=\sqrt{-6}$. The Kummer-Dedekind theorem says that when $p$ does not divide the power-basis index, a factorization $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ gives $(p)={\prod }_i(p,g_i(s){)}^{e_i}$, with each factor having norm $p^{\mathrm{deg}{g}_i}$. The given full power basis has index one. Since $X^2+6\equiv X^2(\mathrm{mod}2)$ and $X^2+6\equiv (X-1)(X+1)(\mathrm{mod}7)$,Here $N(P)=2$ and $N(Q)=N(Q^{\prime })=7$. A prime ideal dividing an ideal of norm $98=2\cdot 7^2$ lies above $2$ or $7$, and these are all the prime ideals above them. Unique factorization and norm multiplicativity therefore leave exactlyThey are distinct by their prime-ideal valuations, and each has norm $98$.

Equality $(\alpha )=(\beta )$ gives $\alpha =u\beta$ and $\beta =v\alpha$ for some $u,v\in {\mathcal{O}}_K$. Substitution and cancellation of the nonzero $\alpha$ yield $uv=1$. Thus $u$ is a unit and $\alpha =u\beta$.

For the ideal-power assertion, suppose first that all ideals are nonzero. A prime ideal divides at most one $Q_i$ because the $Q_i$ are pairwise coprime. Taking its valuation in the given equality givesSince $\gcd (n_i,k)=1$, every exponent $v_P(Q_i)$ is divisible by $k$. Dividing all its prime-factorization exponents by $k$ constructs an integral ideal $I_i$ with $Q_i=I_i^k$. If a $Q_i$ is zero, pairwise coprimality forces all other $Q_j$ to be the unit ideal; taking $I_i=0$ and the other $I_j={\mathcal{O}}_K$ covers that degenerate case too.

For the first Diophantine equation, $x$ cannot be odd: then $x^2+17\equiv 2(\mathrm{mod}8)$, while a cube modulo eight is $0$ or odd. Hence $x$ is even and $y$ is odd. Also $17\nmid x$, since otherwise $v_{17}(x^2+17)=1$, impossible for a cube.

In $\mathcal{O}=\mathbb{Z}[s]$, $s=\sqrt{-17}$, factorThe two principal ideals are coprime. A common prime ideal would divide their sum $2x$ and difference $2s$, and hence lie above $2$ or $17$; odd norm excludes $2$, while $17\nmid x$ excludes $17$. Their coprime product is the cube $(y{)}^3$, so the ideal-power assertion makes $(x+s)=I^3$. The class group has order $4$, while $[I{]}^3=1$, so $[I]=1$. Write $I=(a+bs)$ for integers $a,b$.

The only units are $\pm 1$, since their norm equation is $a^2+17b^2=1$. Equality of principal ideals therefore gives $x+s=\pm (a+bs{)}^3$. Absorb the sign into $a+bs$. Comparing the coefficient of $s$ yieldsThus $b=1$ or $-1$. The first gives $a^2=6$ and the second gives $3a^2=16$, both impossible for an integer $a$. Consequently the first equation has no integral solutions.

For the product equation, odd $x$ gives $(x^2+17)(9x^2+17)\equiv 4(\mathrm{mod}8)$, again not a cube. If $17\mid x$, each factor has $17$-adic valuation one, so the product has valuation two, also impossible. Thus any hypothetical solution has $x$ even and $17\nmid x$. The two positive factors are odd, and their greatest common divisor dividesIt must therefore be one. A product of coprime positive integers is a cube only if each factor is a cube. This would make $x^2+17$ a cube, contradicting the result already proved. Hence the product equation also has no integral solutions.