Past exam of the options II course of the University of Oxford / 2009 (XMAT 4304, AO2)

Options II, Honour School of Mathematics Part A, Trinity Term; Thursday 25 June 2009, 9.30am–12.30pm. The private editable transcription preserves the intact cover and all nineteen questions across seventeen pages. The source was supplied through the past-papers disclosure, in original Part A bundle pages 141–157. No exact worked-answer advertisement was found in the bounded check. The Classical Mechanics revision advertisement names other years, so it does not establish availability for this year's question.

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

If the positive characteristic were composite, say $r=uv$ with $1<u,v<r$, then $(u{1}_K)(v{1}_K)=0$ while neither factor is zero, by minimality of $r$. This contradicts a field having no zero divisors. Thus the characteristic is a prime $p$.

Regard $K$ as a vector space over its prime subfield ${\mathbb{F}}_p$. Since $K$ is finite, a maximal linearly independent list terminates and is a finite basis, of some dimension $n\ge 1$. Each field element has a unique coordinate vector in ${\mathbb{F}}_p^n$, so

For a simple field extension, the powers $1,\lambda ,\dots ,{\lambda }^{d-1}$ form a basis when the minimal polynomial has degree $d$: reduction modulo that polynomial spans the extension, and a relation of smaller degree would contradict minimality. Hence $[F(\lambda ):F]=d$. The field is $K$, whose dimension is $n$, giving $d=n$.

Define $\vartheta (x)=x^p$ for every $x\in K$. In characteristic $p$, all intermediate binomial coefficients vanish, so $\vartheta (x+y)=\vartheta (x)+\vartheta (y)$; multiplicativity and preservation of one are immediate. If $x^p=y^p$, then $(x-y{)}^p=0$ and the field property gives $x=y$. Thus this field homomorphism is injective, and finiteness makes it surjective. It is therefore a field automorphism, sending $\lambda$ to ${\lambda }^p$ as required.

Here the iteration index is a positive integer: $1\le m<n$. If ${\vartheta }^m(\lambda )=\lambda$, then ${\vartheta }^m$ fixes $F$ and the generator, so it fixes all of $K$. Every element of $K$ would then solve $T^{p^m}-T=0$. A nonzero polynomial of degree $p^m$ over a field has at most $p^m$ roots, contradicting $|K|=p^n>p^m$. Thus ${\vartheta }^m(\lambda )\ne \lambda$. The printed inequality alone omits positivity of $m$; at $m=0$ the claimed inequality would be false.

The multiplicative group $K^{\times }$ has order $p^n-1$, so Lagrange's theorem for finite groups gives $x^{p^n-1}=1$ for $x\ne 0$. Including zero, $x^{p^n}=x$ for every $x\in K$. Therefore ${\vartheta }^n$ is the identity. No positive smaller power is the identity by part (iii), soFor $n=1$ this says that Frobenius itself is the identity, as it should be.

A group action assigns to each $g\in G$ and $x\in \Omega$ a point $g\cdot x$ so that $1\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. The orbit of $x$ is $Gx=\{g\cdot x:g\in G\}$. The action is transitive if there is exactly one orbit, equivalently any point can be carried to any other by some group element.

Let $G$ act by left multiplication on its five left cosets of $H$. This gives a group homomorphism $\phi :G\to S_5$. Its kernel is the core of a subgroup, ${\bigcap }_{g\in G}gH{g}^{-1}$: an element fixes every coset precisely when all its conjugates lie in $H$. It is a normal subgroup contained in $H$, so the hypothesis forces the kernel to be trivial. Thus $G$ embeds in $S_5$.

The image has order $60$ and index two, hence is the kernel of a surjection $S_5\to C_2$. All transpositions are conjugate and generate $S_5$, so under any such surjection they must all have the same nonidentity image. This determines the map uniquely as the sign homomorphism. Its kernel is $A_5$, proving $G\cong A_5$.

To prove simplicity directly, the nonidentity conjugacy classes in $A_5$ have sizes $15,20,12,12$. A double transposition has centralizer of order four in $A_5$; a three-cycle has centralizer of order three; a five-cycle has centralizer of order five, yielding the two classes of size twelve. The class sizes together with the identity account for all sixty elements. A normal subgroup must be a union of these classes containing the identity, and its order must divide sixty. None of the proper nontrivial sums of these sizes with one is a divisor of sixty. Therefore $A_5$, and hence $G$, is simple.

Choose an inscribed regular tetrahedron from one of the regular compounds of five tetrahedra in the dodecahedron, and let $H$ be its rotational stabilizer. Its twelve tetrahedral rotations preserve the dodecahedron. Concretely, with $\varphi =(1+\sqrt{5})/2$, use dodecahedron vertices $(\pm 1,\pm 1,\pm 1)$ together with cyclic coordinate placements of $(0,\pm {\varphi }^{-1},\pm \varphi )$. The tetrahedron has the four cube vertices whose coordinate product is $+1$. Cyclic coordinate permutations and even numbers of coordinate sign changes preserve both vertex sets, have determinant $+1$, and give twelve distinct rotations. No other rotation can preserve this tetrahedron, because a regular tetrahedron has only twelve orientation-preserving symmetries.

Their action on its four vertices is faithful and consists exactly of the even permutations: the eight rotations through $\pm 120^{\circ }$ induce three-cycles, the three half-turns induce double transpositions, and there is the identity. Thus $H\cong A_4$, independently of part (ii).

Finally a nonidentity element of $H$ has order two or three. In the dodecahedron's full rotation group, all fifteen half-turns about opposite-edge axes form one conjugacy class, and all twenty third-turns about opposite-vertex axes form one conjugacy class. These geometric conjugacies follow by carrying the corresponding oriented axes to one another by rotations. A normal subgroup containing such an element must contain its whole class, which already has more than twelve elements. Hence no nonidentity normal subgroup of $G$ is contained in $H$, verifying the required extra property.

For an odd prime number $p$, the Legendre symbol $(a/p)$ is zero if $p\mid a$, is $+1$ if $a$ is a nonzero square modulo $p$, and is $-1$ otherwise.

Euler criterion states thatWhen $p\nmid a$, the power is congruent to $+1$ or $-1$, according as $a$ is or is not a quadratic residue.

Gauss lemma for quadratic residues states: for $p\nmid a$, take the least positive residues of $a,2a,\dots ,((p-1)/2)a$ modulo $p$, and let $s$ be the number greater than $p/2$. Then

For $a=2$, the least positive residues in Gauss lemma for quadratic residues are simply $2,4,\dots ,p-1$. The number exceeding $p/2$ isFor $p\equiv 1,3,5,7(\mathrm{mod}8)$, respectively, this count is even, odd, odd, even. Substituting these parities into the lemma gives

By multiplicativity of the Legendre symbol and Euler criterion,Combining this with part (iv) yields

Suppose all primes congruent to $3$ modulo eight are $p_1,\dots ,p_r$, and put $N=p_1\cdots p_r$, an odd integer. The integer $N^2+2$ is odd and congruent to three modulo eight. If a prime $q$ divides it, then $N^2\equiv -2(\mathrm{mod}q)$ and $q\nmid N$, since a common divisor of $N$ and $N^2+2$ would divide two. Thus $(-2/q)=1$, so part (v) forces $q\equiv 1$ or $3(\mathrm{mod}8)$.

If all prime factors congruent to three had even multiplicity, their product, including the factors congruent to one, would be one modulo eight. Since $N^2+2$ is three modulo eight, at least one prime factor is congruent to three and occurs to odd multiplicity. But it is none of the $p_i$, since those divide $N$. This contradiction proves infinitely many primes are congruent to three modulo eight.

The function is nonnegative and measurable, so Lebesgue integrability is equivalent to finiteness of its integral. For $\alpha \ne -1$,Letting $\epsilon \downarrow 0$ gives a finite limit exactly when $\alpha >-1$; at $\alpha =-1$ the integral is $-\log \epsilon$ and diverges. By the monotone convergence theorem for truncations,

For $\alpha \ne -1$,The limit as $R\to \infty$ is finite exactly when $\alpha <-1$; the case $\alpha =-1$ diverges logarithmically. Positivity and the monotone convergence theorem identify these limits with the Lebesgue integral. Hence

The function is measurable and negative on this interval. Its absolute value obeysThe final integral follows by integration by parts and $x^2\log x\to 0$ at zero. The comparison test for integrals proves integrability.

For $x\ge 1$, $(1+x{)}^2\le 4x^2$, soThe integral of the lower bound from one to $R$ is $(\log R{)}^2/8$, which diverges. Thus the nonnegative function has infinite integral and is not integrable.

The function is continuous and nonnegative, bounded above by $x$ on $(0,1)$ and by $x{e}^{-x}$ on $(1,\infty )$, so it is integrable. For $x>0$ the geometric series givesThe series is absolutely summable in integral becauseEach integral follows by substituting $t=nx$ and integrating $t{e}^{-t}$ by parts. Tonelli's theorem applied to the absolute terms, followed by Fubini's theorem, therefore justifies termwise integration and gives

On a product of sigma-finite measure spaces, Tonelli's theorem says that a nonnegative measurable function has equal product and iterated integrals, allowing $+\infty$.

Fubini's theorem says that if a measurable function is absolutely integrable on the product, almost every section is integrable, the integrals of sections are integrable, and both iterated integrals equal the product integral. In particular, to establish absolute integrability one can first apply Tonelli's theorem to the absolute value and show its iterated integral is finite.

Both terms of each one-dimensional section are integrable by a positive scaling substitution, since $f\in L^1(0,\infty )$. For fixed $y>0$,For fixed $x>0$,The strict inequalities hold because $a>1$ and the measurable integrand is strictly positive on each nonempty interval. If $g$ were integrable on the stated product domain, Fubini's theorem would make both iterated integrals finite and equal. But the first order of integration would give a strictly negative value, and the other a strictly positive value. This contradiction proves $g$ is not integrable on that product.

The function is nonnegative and continuous on the positive quadrant. In polar coordinates its product integral is, by the nonnegative change of variables formula and Tonelli's theorem,The angular factor is at most one, while the supplied radial integral is $1/2$, so the product integral is at most $\pi /4$ and is finite. This establishes integrability before any signed interchange is used.

Let $A$ denote the angular integral. The polar evaluation is $A/2$. In Cartesian coordinates Fubini's theorem instead factors the integral asusing $u=x^2$, $v=y^2$. All factors converge: their powers exceed $-1$ at zero and the exponential controls infinity. Equating the evaluations gives

A subset $U$ of a metric space is open if every $x\in U$ has a radius $r>0$ with the open ball $B_d(x,r)\subseteq U$.

Continuity at a point means: for every $\epsilon >0$ there exists $\delta >0$ such that $d_X(x,x^{\prime })<\delta$ implies $d_Y(f(x),f(x^{\prime }))<\epsilon$.

The function is continuous if it is continuous at every point of $X$. Equivalently, the preimage of each open set in $Y$ is open in $X$.

Nonnegativity and symmetry follow from $d$, and $d_a(x,y)=0$ with $x\ne y$ would force both $x=a$ and $y=a$, a contradiction. For the triangle inequality, when the intermediate point $z$ is different from both endpoints,If $z$ equals an endpoint the inequality is immediate, and if $x=y$ its left target is zero. These cover every case, so $d_a$ is a metric.

This is the star metric centered at a point.

If $x\ne b$, then $d_a(b,x)=d(b,a)+d(a,x)\ge d(b,a)$, so no such $x$ lies in the strict open ball. Its centre always does. HenceThus every point other than $a$ is isolated.

The singleton calculation shows that every map out of $(X,d_a)$ is continuous at each point other than $a$: use the domain ball consisting only of that point. Thus global continuity in this metric is equivalent to continuity at $a$, proving $[\alpha ]⟺[\beta ]$.

For every $x$, including $x=a$, one has $d_a(a,x)=d(a,x)$. Therefore the open balls centred at $a$ are identical for $d_a$ and $d$. The epsilon-delta definitions of continuity at $a$ are identical, giving $[\beta ]⟺[\gamma ]$. Hence all three conditions are equivalent.

Every point of $A$ differs from $a$, so its singleton is an open ball by part (ii). The union of these singleton open sets is $A$. Thus $A$ is open whenever $a\notin A$.

If $A$ is open and contains $a$, it contains some $d_a$-ball about $a$, which is exactly the corresponding $d$-ball. Conversely, if $B_d(a,\epsilon )\subseteq A$, this is a $d_a$-neighbourhood of $a$, while every other point of $A$ has its open singleton inside $A$. Hence $A$ is open. Thus the stated ball-containment condition is necessary and sufficient.

Every point other than $a$ is isolated, so no such point can enter the closure unless it was already in $B$. The only possible extra point is $a$, and it lies in the closure precisely when every $d$-ball about $a$ meets $B$. Therefore the three answers, in the printed order, areIn the middle case, a ball of radius below $d(a,B)$ misses $B$; in the last case the infimum zero makes every positive-radius ball meet $B$. If $B$ is empty its closure is empty as usual, with the convention $d(a,\varnothing )=+\infty$.

A subset is connected when its subspace topology admits no separation into two disjoint nonempty open sets. Equivalently, its only subsets that are both open and closed in that subspace are the empty set and the entire subset.

A separation of $f(X)$ would pull back under the continuous map to two disjoint nonempty open sets covering $X$: nonemptiness follows because their points belong to the image. This contradicts connectedness of $X$. Hence a continuous image of a connected space is connected.

The underlying set of the quotient space is the set of equivalence classes $X/R$. With quotient map $q(x)=[x]$, its quotient topology is defined by declaring $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. This makes $q$ a continuous surjection and is the finest topology with that property.

Let $C$ be any equivalence class. Since $A$ is both open and closed, $C\cap A$ is both open and closed in the subspace topology on $C$. Its connectedness makes $C\cap A$ either empty or all of $C$. Thus an equivalence class meeting $A$ lies wholly in $A$. Hence $A$ is saturated, a union of whole equivalence classes.

If $X$ is connected, its continuous quotient image is connected by part (a). Conversely, suppose $X$ is disconnected. It has a nonempty proper subset $A$ both open and closed. Part (i) makes $A$ and its complement saturated, so $q^{-1}(q(A))=A$ and $q^{-1}(q(X\setminus A))=X\setminus A$. The quotient topology therefore makes these disjoint nonempty images open, and they cover $X/R$. Thus the quotient is disconnected. This proves the two spaces are connected simultaneously.

The difference map $(z,w)\mapsto z-w$ is continuous, and $X$ is its preimage of the open set $\mathbb{C}\setminus \{0\}$. Thus $X$ is open in the product.

The class of $(z,w)$ is $C_z=\{z\}\times (\mathbb{C}\setminus \{z\})$. A punctured plane is path-connected: two points can be joined by a polygonal path avoiding the deleted point. Consequently every class is connected.

The projection $\pi$ is onto, since $(z,z+1)\in X$ for every $z$. It is continuous as a restricted coordinate projection. If $U$ is open in $X$, it is open in the product because $X$ itself is open. For a point $(z,w)\in U$, the product-topology basis gives open sets $V,W$ with $(z,w)\in V\times W\subseteq U$. Then $V\subseteq \pi (U)$. Hence $\pi (U)$ is open.

The map is constant exactly on the equivalence classes, so it induces a bijection $\overline{\pi }:X/R\to \mathbb{C}$. It is continuous by the defining property of the quotient topology. It is also open: for quotient-open $V$, $\overline{\pi }(V)=\pi (q^{-1}(V))$ is open. ThusThe plane is connected, and the classes are connected, so part (b) now yields $X$ is connected.

The $i$th partial derivative is ${\partial }_{i}f(x_0)={\lim }_{t\to 0}[f(x_0+t{e}_i)-f(x_0)]/t$. The directional derivative in direction $v$ is the same limit with $v$ in place of $e_i$, using the specified vector rather than normalizing it. Differentiability means that there is a linear map $L:{\mathbb{R}}^n\to \mathbb{R}$ withFor $v=(v_1,v_2)$ with $v_1\ne 0$, the difference quotient at the origin is $t{v}_2^3/v_1$, tending to zero. If $v_1=0$, the function is identically zero on the directional line, again giving derivative zero. Thus every directional derivative exists and is zero.

Along the curved path $(x,y)=(t^3,t)$, however, $f(t^3,t)=1$ for $t\ne 0$, while $f(0,0)=0$. Hence the function is not continuous at the origin. Differentiability implies continuity, so it is not differentiable there either. Direction-by-direction limits alone do not control general approaches to a point.