Past exam of the algebra and analysis II for mathematics and philosophy course of the University of Oxford

Source paper: Mathematics Part A FOI release, member FOI-20210708-1 - Data/4308.pdf. The intact cover identifies Algebra and Analysis II for Mathematics and Philosophy, AC2(P), formal code 4308, Tuesday 18 June 2013, 9.30am to 11.30am. The five-page source has six questions, in Algebra and Analysis sections, with no printed part marks.

The minimal polynomial of a linear transformation $T$ is the unique monic polynomial of least degree among the nonzero polynomials $p\in F[x]$ satisfying $p(T)=0$:Such a polynomial exists because the endomorphism space is finite-dimensional, so sufficiently many powers $I,T,T^2,\dots$ are linearly dependent. Monic normalization is possible over the field $F$. Two monic polynomials annihilating $T$ of the same least degree have a difference of smaller degree annihilating $T$, so their difference is zero. For the zero vector space the convention is $m_T=1$.

Apply polynomial division in $F[x]$:Evaluation at $T$ gives $r(T)=f(T)-q(T)m_T(T)=0$. A nonzero remainder could be rescaled to a monic polynomial annihilating $T$ of smaller degree, contradicting the defining minimality. Hence $r=0$ andIn particular, the annihilating polynomials form the principal ideal $(m_T)$.

Since $p$ and $q$ are coprime, Bézout's identity for polynomials gives $a,b\in F[x]$ with $ap+bq=1$. For every $v\in V$,The first summand belongs to $U$: applying $p(T)$ gives $b(T)p(T)q(T)v=0$. The second belongs to $W$ by the same calculation. Here all polynomials in $T$ commute, and $p(T)q(T)=f(T)=0$. Thus $V=U+W$.

If $v\in U\cap W$, both terms on the right of the identity vanish, so $v=0$. ThereforeFinally, $p(T)Tv=Tp(T)v=0$ whenever $v\in U$; hence $T(U)\subseteq U$. Replacing $p$ by $q$ proves $T(W)\subseteq W$. These are invariant subspaces; this is the coprime-factor case of the primary decomposition theorem for linear maps.

Write $m_1,m_2$ for the minimal polynomials of the two restrictions. Since $p(T)$ vanishes on $U$ and $q(T)$ vanishes on $W$, part (b) givesHence $m_1,m_2$ are coprime. By the direct-sum decomposition, a polynomial annihilates $T$ exactly when it annihilates both restrictions, which happens exactly when it is divisible by both $m_1$ and $m_2$. ConsequentlyIf $m_1=p$ and $m_2=q$, this immediately gives $m_T=pq=f$. Conversely, if $m_T=pq$, the monic polynomials $p/m_1$ and $q/m_2$ have product one, so each is one. ThusThis also handles a zero summand, whose minimal polynomial is one; a nonconstant factor cannot be its minimal polynomial.

Part (b) requires $m_T$ to be a monic divisor of $x^2(x-1)$. The constant divisor one cannot annihilate a nonzero vector space. Every remaining divisor can occur in dimension three:Here are corresponding matrices, all satisfying $T^3-T^2=0$:The last matrix contains a size-two nilpotent Jordan block, requiring the factor $x^2$, and an eigenvector with eigenvalue one, requiring $x-1$. Thus its minimal polynomial is exactly $f$. The zero matrix has minimal polynomial $x\ne f$, providing the other requested example.

Let $p={\sum }_{j=1}^m⟨u,v_j⟩v_j$. Orthonormality gives $⟨u-p,v_k⟩=0$ for every $k$, so $u-p$ is orthogonal to $p$. The Pythagorean theorem for inner product spaces givesSince a squared norm is nonnegative,This is Bessel's inequality, with equality exactly when $u$ belongs to the span of the orthonormal set.

Use the Euclidean inner product on ${\mathbb{R}}^n$, with $a=(a_1,\dots ,a_n)$ and $b=(b_1,\dots ,b_n)$. If $b=0$, both sides of the desired inequality are zero. Otherwise take the single unit vector $v=b/\Vert b\Vert$ in part (a). ThensoThis is the real Cauchy-Schwarz inequality. Equality holds when the two vectors are linearly dependent.

In part (i), set every $b_i=1$. Its squared norm is $n$, givingThe equality condition is $a_1=\cdots =a_n$, because $a$ must be a scalar multiple of $(1,\dots ,1)$.

In ${\mathbb{R}}^{2n}$, take the two unit vectorsTheir supports are disjoint, so they are orthonormal. Apply part (a) to $a=(a_1,\dots ,a_{2n})$:Multiplying by $n$ gives

Equality in part (a) requires $a$ to lie in the span of the two chosen unit vectors. Equivalently, there are real constants $r,s$ such thatOne can also see the necessity directly by writing the nonnegative deficit aswhere ${\bar{a}}_e$ and ${\bar{a}}_o$ are the means of the even and odd entries. The deficit vanishes exactly when each group is constant; the two group constants need not be equal.

The first isomorphism theorem for rings states that the mapis a ring isomorphism. It is well defined because two representatives differ by an element of the kernel exactly when their images agree. It preserves addition and multiplication, is onto its stated codomain, and has trivial kernel. In particular, if $f$ is surjective then $R/\ker f\cong S$.

Define $\Phi :R\to R/I\times R/J$ by $\Phi (r)=(r+I,r+J)$. It is a ring homomorphism withSince $I+J=R$, choose $i\in I,j\in J$ such that $i+j=1$. For arbitrary $a,b\in R$, the element $r=aj+bi$ satisfies $r-a=i(b-a)\in I$ and $r-b=j(a-b)\in J$. Hence $\Phi (r)=(a+I,b+J)$, proving surjectivity. The first isomorphism theorem now givesThis is the two-ideal Chinese remainder theorem for ideals. The hypothesis is comaximality of the two ideals; no assertion that every pair of ideals has this property is needed.

Coprimality gives polynomials $a,b\in \mathbb{Q}[x]$ with $ag+bh=1$, so the ideals $(g)$ and $(h)$ are comaximal. AlsoIndeed, if $u=gr=hs$, multiplying $ag+bh=1$ by $u$ writes $u=aghs+bhgr$, a multiple of $gh$; the reverse inclusion is immediate. Part (b), with $f=gh$, givesThe explicit isomorphism takes the class of a polynomial $r$ to its classes modulo $g$ and $h$.

Because $g,h$ are nonconstant, their classes in $\mathbb{Q}[x]/(gh)$ are nonzero: each has smaller degree than $gh$, so neither is a multiple of $gh$. Their product is zero. Thus the quotient has zero divisors.

If both factors have degree one, each quotient by one factor is isomorphic to $\mathbb{Q}$, by evaluation at its rational root. Coprimality ensures the two roots are distinct. Part (c) therefore identifies the ring with $\mathbb{Q}\times \mathbb{Q}$. If $(r,s{)}^2=(0,0)$, then $r^2=s^2=0$ in a field, so $r=s=0$.A ring may therefore have zero divisors without having nonzero nilpotent elements.

Factorization over $\mathbb{Q}$ givesFor $R_1$, the factors $x-1$ and $x+1$ are coprime. The nonzero elements $(1,0)$ and $(0,1)$ multiply to zero, so the ring is neither an integral domain nor a field. Squaring coordinatewise shows it has no nonzero square-zero element.

For $R_2$, the polynomial $x^2+1$ has no rational root and is irreducible of degree two over $\mathbb{Q}$. The quotient is consequently a field, hence an integral domain, and has no nonzero square-zero element.

For $R_3$, the class $\epsilon$ is nonzero by its degree, but ${\epsilon }^2=0$. Thus this ring is neither an integral domain nor a field, and does have a nonzero square-zero element.No two are isomorphic. Being a field separates $R_2$ from the other two; existence of a nonzero square-zero element separates $R_1$ from $R_3$. Both properties are preserved by ring isomorphisms.

A function is holomorphic on an open set when it is complex differentiable at every point of that set. Explicitly, for every $z\in U$, the limitexists as a finite complex number. The limit is independent of the direction in which the complex increment approaches zero; existence of derivatives only along the real axis would not suffice.

Cauchy's integral theorem says that if $f$ is holomorphic on an open set $U$ and a closed, piecewise smooth contour $\gamma$ is null-homotopic in $U$, thenIn particular, this holds for every closed contour when $U$ is simply connected, and for a simple closed contour whose boundary and enclosed region lie in $U$. For the later arguments it suffices to use the triangular version: the integral around a triangle is zero if the triangle and its interior lie in $U$ and $f$ is holomorphic there.

Both the holomorphicity condition and a suitable condition excluding holes inside the contour matter. Continuity alone, which is the starting hypothesis for this question, does not imply the theorem's conclusion; nor does holomorphicity on an arbitrary punctured open set imply zero integral around every loop.

Morera's theorem states that a continuous function on an open subset of $\mathbb{C}$ is holomorphic if its integral around every triangular contour, with triangle and interior contained in that open set, is zero.

To prove this, work in any open disk $D$ whose closure lies in $U$, and fix $z_0\in D$. Definewhere the segment lies in $D$ by convexity. The zero integral on the triangle with vertices $z_0,z,z+h$ givesContinuity at $z$ implies that the last integral tends to $f(z)$ as $h\to 0$: its difference from $f(z)$ is bounded by ${\sup }_{0\le t\le 1}|f(z+th)-f(z)|$. Therefore $F^{\prime }=f$ on $D$. The function $F$ is holomorphic, and the allowed theorem that a holomorphic derivative is holomorphic implies that $f=F^{\prime }$ is holomorphic on $D$.

Every point of $U$ lies in such a disk, proving the result throughout $U$:The proof is local and does not require $U$ to be simply connected.

For two moving vertices $p(s),q(s)$, the integral along their joining segment isThe integral factor is continuous in $s$ by the allowed fact, and the other factor is continuous because the vertices are continuous. Thus $J_{p,q}$ is continuous, including when its two endpoints coincide.

The integral around the triangle is exactlyEach term is continuous, so their sum is continuous at zero andNo holomorphicity is used here. Compactness of the fixed containing triangle also gives a direct proof of the allowed fact through uniform continuity of $f$.

First take $z\in U_{-}$ and put $w=\bar{z}\in U_{+}$. Reflection symmetry of the function givesHence $f$ is holomorphic on $U_{-}$, with $f^{\prime }(z)=\overline{f^{\prime }(\bar{z})}$ there.

It remains to cross the real axis. Fix a real point $p\in U$ and a small disk about $p$ with closure in $U$. Consider a triangle in a still smaller disk. Intersect its filled interior with the closed upper and lower half-planes and triangulate the resulting convex polygons. This gives finitely many triangles lying on one side of the real axis, possibly with edges on that axis. Degenerate pieces contribute zero.

For an upper piece, translate its three vertices by $i\epsilon$ with $\epsilon >0$. For sufficiently small $\epsilon$ the translated triangle lies entirely in $U_{+}$ and remains in $U$, so its boundary integral is zero by Cauchy's theorem. Part (d), or its segment-integral proof, shows that the integral converges to that around the untranslated piece as $\epsilon \downarrow 0$. Thus the original piece has zero integral. Translate lower pieces by $-i\epsilon$ to obtain the same conclusion in $U_{-}$.

When all the oriented triangle integrals are added, their internal edges cancel, including the segments on the real axis. The original triangle therefore has zero boundary integral. Continuity of $f$ on $U$ now allows Morera's theorem in the disk, proving holomorphicity near $p$. Together with the two half-plane arguments this givesThis is a Morera gluing lemma across a straight line; the original continuity hypothesis is essential to the limiting step.

For an isolated singularity at $a$, write the Laurent expansion in a punctured disk as $f(z)={\sum }_{k=-\infty }^{\infty }{c}_k(z-a{)}^k$. Its residue isThe sine in the denominator has simple zeros precisely at the integers, and the numerator is nonzero there. Consequently $g(z)=\pi \cot (\pi z)$ has simple poles exactly at $n\in \mathbb{Z}$, withThere are no other finite singularities.

The residue theorem states that if a function is holomorphic on and inside a positively oriented simple closed contour except at finitely many isolated singularities $a_1,\dots ,a_N$ in its interior, thenNo singularity may lie on the contour. More generally, a closed contour has a winding-number-weighted sum of residues, provided it and its enclosed winding regions lie in the domain on which the isolated singularities are specified.

A subset $B\subseteq \mathbb{C}$, with its subspace topology, is compact when every open cover of $B$ has a finite subcover. The Heine-Borel theorem in the complex plane, identified with ${\mathbb{R}}^2$, saysClosedness is in the ambient complex plane. For a continuous real-valued function on a nonempty compact set, the supremum and infimum are finite and attained.

Periodicity reduces all the indicated vertical segments to the compact segment $1/2+i[-1,1]$, which misses every pole. Continuity therefore gives the first uniform bound. In fact an explicit identity gives a stronger estimate for every real $y$:Thus we can take $D_1=\pi$.

Write $L=m+1/2$. On the vertical sides of the square the same bound holds, since both real coordinates are half-integers. On either horizontal side $|\mathrm{Im}z|=L\ge 1/2$. For $\mathrm{Im}z=y>0$, put $w=e^{2\pi iz}$, so $|w|=e^{-2\pi y}$. Usinggives $|g(z)|\le \pi (1+e^{-2\pi y})/(1-e^{-2\pi y})\le \pi \coth (\pi /2)$. Conjugation gives the same estimate below the real axis. Hence an explicit bound valid on every square isFor $L>|b|$, the contour has length $8L$ and every point on it has $|z|\ge L$. The reverse triangle inequality yields $|z^2-b^2|\ge L^2-b^2$, so the ML inequality givesOrient the square positively. Its enclosed poles at the integers $-m,\dots ,m$ have residues $1/(n^2-b^2)$. Since $b$ is nonintegral, the additional poles $b$ and $-b$ are distinct from those integers and from one another. Their residues are $g(b)/(2b)$ and $g(-b)/(-2b)$, both equal to $g(b)/(2b)$ because $g$ is odd. Their combined contribution is therefore $\pi \cot (\pi b)/b$.

The residue theorem givesThe series is absolutely convergent, since its terms are $O(n^{-2})$. Taking the limit, then including the $n=0$ term, yieldsThis is cotangent contour summation. Keeping the $n=0$ term separate until the last step determines the sign of the $1/(2b^2)$ contribution.

A conformal map preserves oriented angles between regular smooth curves through each point. For a holomorphic map with nonzero derivative, the derivative also describes the local scale and rotation.

Indeed, if two curves ${\gamma }_1,{\gamma }_2$ meet at $z$ and have nonzero tangent vectors $v_1,v_2$, the real chain rule and complex differentiability giveBecause $f^{\prime }(z)\ne 0$, the image tangents are nonzero andTaking arguments proves preservation of oriented angles. Multiplication by $f^{\prime }(z)$ rotates every tangent by $\arg f^{\prime }(z)$ and scales every length by $|f^{\prime }(z)|$. Thus the hypothesis gives conformality at every point of $U$; global injectivity is not needed for this local conclusion.

A Möbius transformation maps generalized circles to generalized circles on the Riemann sphere. A generalized circle is an ordinary circle or a straight line together with the point at infinity. Its image is a straight line precisely when it contains infinity, equivalently when the preimage of infinity lies on the extended real axis.

If $c=0$, the pole is infinity itself and the transformation is the affine map $(a/d)z+b/d$, taking the extended real axis to a line. If $c\ne 0$, the pole is $-d/c$, which is real exactly when $d/c$ is real. Thus the full condition isEquivalently, it is $\mathrm{Im}(d\bar{c})=0$, a condition also valid when $c=0$. If it fails, the image misses infinity and is therefore an ordinary circle. The nonzero determinant ensures that none of these generalized circles collapses to a point.

Liouville's theorem states that every bounded entire function is constant:For example, the Cauchy derivative estimate on a circle of radius $R$ gives $|f^{\prime }(z)|\le M/R$. Letting $R\to \infty$ gives $f^{\prime }=0$ everywhere, proving the conclusion.

It suffices to construct an injective holomorphic map ${\Phi }_j:U_j\to \mathbb{D}$ for each domain. Then ${\Phi }_j\circ f$ is a bounded entire function and is constant by Liouville's theorem; injectivity of ${\Phi }_j$ forces $f$ to be constant.

For the upper half-plane, use the Cayley transform between the half-plane and diskIts denominator is nonzero there, and $|z+i{|}^2-|z-i{|}^2=4\mathrm{Im}z>0$, so its image lies in the unit disk. The determinant is nonzero, hence it is injective and conformal.

On the plane slit along the nonnegative real axis, choose the branchThis is the holomorphic square root obtained from the corresponding logarithm branch. Its argument lies in $(0,\pi )$, so it maps $U_2$ bijectively onto the upper half-plane, with inverse $w\mapsto w^2$. Its derivative is $1/(2S(z))\ne 0$. Thereforeis an injective conformal map of $U_2$ onto the unit disk.

For the complement of the closed interval, first useIts only finite pole is the excluded point one, and it is injective with derivative $1/(1-z{)}^2\ne 0$. If $M(z)=t\ge 0$, solving gives $z=t/(1+t)\in [0,1)$, which is excluded. Hence $M(U_3)\subset U_2$, andis the required injective conformal map into the disk. More precisely, $M(U_3)=U_2\setminus \{-1\}$ and ${\Phi }_3(U_3)=\mathbb{D}\setminus \{0\}$, since $S(-1)=i$. No claim that the interval complement is conformally equivalent to the whole disk is required.

Applying Liouville to the three bounded compositions givesThis illustrates Liouville rigidity from a bounded holomorphic embedding: an explicit injective bounded coordinate is enough.