Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1

For an odd prime $p$, the Legendre symbol $(a/p)$ is $0$ if $p\mid a$, $1$ if $a$ is a nonzero square modulo $p$, and $-1$ otherwise. Euler's criterion statesIf $p\mid a$ both sides vanish. Otherwise write $a\equiv g^j$ for a primitive root $g$. The squares are precisely the even powers of $g$, while $g^{(p-1)/2}$ has order two and therefore equals $-1$. Hence $a^{(p-1)/2}=(-1{)}^j$, proving the criterion and the useful special case $(-1/p)=(-1{)}^{(p-1)/2}$.

For odd $n$, $n^2+4\equiv 5(\mathrm{mod}8)$. Every prime divisor $p$ of this number is odd, and $p\nmid n$, so $(n/2{)}^2\equiv -1(\mathrm{mod}p)$. Euler's criterion forces $p\equiv 1(\mathrm{mod}4)$, hence $p\equiv 1$ or $5(\mathrm{mod}8)$. A product of such primes can be $5(\mathrm{mod}8)$ only if a prime $5(\mathrm{mod}8)$ occurs to an odd power.

Suppose the primes $5(\mathrm{mod}8)$ were $p_1,\dots ,p_r$. Choose the odd integer $n=p_1\cdots p_r$ (or $n=1$ for an empty list). None of these primes divides $n^2+4$, since the latter is $4$ modulo each $p_i$. Yet the preceding argument gives a prime divisor $5(\mathrm{mod}8)$, outside the list. This contradiction proves infinitely many primes congruent to $5$ modulo $8$.

The closure is closed and bounded in ${\mathbb{R}}^2$, hence compact by the Heine-Borel theorem. It is nonempty. The extreme value theorem therefore gives a point of the closure at which the continuous function attains its maximum. The closure, rather than the open domain, is essential.

At an interior local maximum, the second derivatives in both coordinate directions are nonpositive: restrict the function to each coordinate line and apply the one-variable second-derivative test. Thus ${\nabla }^2\varphi \le 0$ there, contradicting ${\nabla }^2\varphi \ge ϵ>0$. Since a maximum exists on the closure, a maximum lies on the boundary.

For $\eta >0$, use the weak maximum principle for elliptic operators perturbation ${\varphi }_{\eta }(x)=\varphi (x)+\eta |x{|}^2$. Its Laplacian is at least $4\eta$, so the preceding argument puts its maximum on the boundary. If $R^2={\max }_{\bar{\Omega }}|x{|}^2$, then, for any $x\in \bar{\Omega }$,Letting $\eta \downarrow 0$ gives $\varphi (x)\le {\max }_{\partial \Omega }\varphi$. The boundary is compact, so this maximum is attained. This proves the weak weak maximum principle for elliptic operators even when the Laplacian may vanish.

Apply the weak weak maximum principle for elliptic operators to the harmonic function $\varphi$ and to $-\varphi$. Their boundary maxima are both zero, giving $\varphi \le 0$ and $\varphi \ge 0$ on the closure. Consequently $\varphi \equiv 0$.

For a block code, the minimum distance of a code is the minimum Hamming distance between distinct codewords, where Hamming distance counts positions in which two equal-length binary strings differ. We use equal-length codewords, so the fact that the codomain allows arbitrary finite strings creates no ambiguity. Minimum distance $\delta$ means every two distinct codewords differ in at least $\delta$ positions and some pair differs in exactly $\delta$.

Changing at most $\delta -1$ bits cannot turn a codeword into another codeword. Testing membership therefore detects every such nonzero error. For $t=\lfloor (\delta -1)/2\rfloor$, suppose a received word lies within distance $t$ of two codewords. The triangle inequality would put their mutual distance at most $2t<\delta$, a contradiction. Hence the radius-$t$ Hamming balls are disjoint and nearest-codeword decoding corrects all errors of weight at most $t$.

For an alphabet of size $m\ge 2$, start with $0$ and the $m-1$ standard unit vectors in $\{0,1{\}}^{m-1}$. Their minimum distance of a code is one. Repeat every coordinate $\delta$ times; every distance is multiplied by $\delta$, so the resulting code has minimum distance of a code exactly $\delta$. A one-letter alphabet has no pair of distinct codewords and therefore no finite minimum distance of a code under this definition; the construction concerns the usual nontrivial coding situation.

The standard normal distribution function $\Phi$ givesThe observations are independent Bernoulli random variables, because the original errors are independent. Thus the probit regression has intercept $\alpha =(\mu -\log c)/\sigma$ and coefficient vector $\gamma =\beta /\sigma$. With the specified known $\sigma$ and $c$, recovery isThe scale would not be recoverable from threshold data alone if $\sigma$ were unknown.

Substitution of the separated exponential profile into the age-structured population equation gives $r^{\prime }=-(\gamma +\mu (a))r$. ThereforeFor $C\ne 0$, the birth boundary condition cancels $C$ and becomes the Euler-Lotka equationThe survival probability to age $a$ is $e^{-\mu a}$ for constant death rate. Hence the expected lifetime offspring count, or net reproduction rate, isThe Gamma integral evaluates the Euler-Lotka equation as $Bp!/(\lambda +\mu +\gamma {)}^{p+1}=1$, requiring $\lambda +\mu +\gamma >0$. Its real solution and profile areThus $B^{\ast }=(\lambda +\mu {)}^{p+1}/p!$, and $B>B^{\ast }$ gives positive growth. The profile is a nonzero formal separated solution for all positive $B$; a finite total population additionally requires $\gamma +\mu >0$, equivalently $(Bp!{)}^{1/(p+1)}>\lambda$. A decaying-in-time mode need not have an integrable age profile.

Introduce the Mellin transform integral $J(s)={\int }_0^{\infty }{x}^{s-1}/(1+x^2)dx$, valid for $0<\mathrm{Re}s<2$. A keyhole contour for $z^{s-1}/(1+z^2)$ with $0<\arg z<2\pi$ has residues $i^{s-1}/(2i)$ and $(-i{)}^{s-1}/(-2i)$, with the latter argument $3\pi /2$. The two sides of the positive-axis cut give $(1-e^{2\pi i(s-1)})J(s)$. The small and large arcs vanish in this strip; simplifying the residue theorem identity yieldsDifferentiation under the integral is valid on compact substrips because the extra $\log x$ is dominated at both endpoints. ThereforeAt $s=3/2$, the requested values are

The principle of stationary action says that the physical path makes the action stationary under smooth variations $q+\eta h$ with fixed endpoints $h(t_1)=h(t_2)=0$. It need not minimize the action. Differentiating and integrating by parts givesThe fundamental lemma of the calculus of variations then gives the Euler-Lagrange equation

Use $r\in (0,l)$, the bead's distance from the pivot, and $\theta$, the rod angle from downward vertical. The kinetic energy is $m({\dot{r}}^2+r^2{\dot{\theta }}^2)/2$. If the identical springs have common natural length $r_{\ast }$, their potential is $k[(r-r_{\ast }{)}^2+(l-r-r_{\ast }{)}^2]/2$. Its $r$-dependent part is $k(r-l/2{)}^2$; the unspecified common natural length changes only an additive constant. Gravity contributes $-mgr\cos \theta$. Consequently the Lagrangian may be taken asThis uses the illustrated symmetric two-spring arrangement without assuming an unstated natural length.

Apply the Euler-Lagrange equation separately to $r$ and $\theta$:The first equation balances radial inertia, centrifugal motion, gravity and the net spring force. The second is the angular-momentum equation $d(m{r}^2\dot{\theta })/dt=-mgr\sin \theta$. For $r>0$ it is equivalently $\ddot{\theta }+2\dot{r}\dot{\theta }/r+g\sin \theta /r=0$.

Relative positions and velocities measured from galaxy $A$ are $r_B-r_A$ and $v_B-v_A$. Spatial homogeneity says that the same position-to-relative-velocity rule applies from every galaxy. Thus $v(r_B-r_A)=v(r_B-r_O)-v(r_A-r_O)$. Taking $r_O=0$ and putting $r_B=x+y$, $r_A=y$ gives additivity $v(x+y)=v(x)+v(y)$. For a continuous physical velocity field, additivity implies rational homogeneity and, by continuity, real homogeneity. Thereforefor a position-independent matrix $H$. Regularity is necessary to exclude pathological discontinuous additive functions.

Decompose the given matrix as $H=(5D/t)I+(D/t)W$, whereIts symmetric part is isotropic: no cosmological shear, expansion scalar $\nabla \cdot v=15D/t$, and linear scale factor proportional to $t^{5D}$ for $t>0$. Its antisymmetric part is rigid rotation with angular velocityThe vorticity is $2\mathbf{\omega }$. Comoving separations obeyThus all distances grow by the same power, volumes grow as $t^{15D}$, and the pattern rotates about the fixed axis $(1,-2,1)$ through angle $\sqrt{6}D\log (t/t_0)$. Homogeneity alone permits this rotation; full cosmological isotropy would not.

For $r\ge 2$, let $H$ be the $r\times (2^r-1)$ binary matrix whose columns are all distinct nonzero vectors of ${\mathbb{F}}_2^r$. The binary Hamming code is $C=\ker H$. Its parity-check matrix has rank $r$ since its columns include the standard basis. Hence $C$ is linear of length $2^r-1$ and dimension $2^r-1-r$.

No word of weight one or two lies in $C$, since columns are nonzero and distinct; three columns $u,v,u+v$ sum to zero, so the minimum distance of a code is exactly three. The syndrome of a single-bit error is its column of $H$, uniquely identifying the erroneous position. The $2^r$ possible syndromes correspond to no error or exactly one of $2^r-1$ single-bit errors. Alternatively, every radius-one Hamming ball has $2^r$ words andDisjoint balls therefore cover the whole word space. This proves linearity, one-error correction and perfection. For $r=3$, the parameters are $[7,4,3]$, giving the 16 messages needed below.

Independence of individual bit errors gives $P=(1-p{)}^{4N}$. Independent retransmissions have a geometric distribution with success probability $P$, so the expected total number of transmissions, including the successful one, is $1/P$. At the supplied values, $P\approx 0.018279$ and $1/P\approx 54.71$.

A Hamming code decoder succeeds on a seven-bit block exactly when there are zero or one errors. Its radius-one decoding regions are disjoint, so an error of weight at least two cannot decode back to the original codeword. ThusNumerically $Q\approx 0.97929$, so only about $1.0212$ transmissions are expected. Even accounting for seven rather than four bits per letter, expected bits sent are $7N/Q\approx 7148$, compared with $4N/P\approx 218830$. The Hamming method is substantially better under the stated error detection and independent retransmission assumptions.

The Poisson regression uses independent counts with mean $m_i$ and logarithmic link function. Writing $c$ for concentration, the fitted predictors for strains $0,1,2$ are respectivelyThe interaction formula includes concentration, strain indicators and their products. The intercept is the baseline log mean at zero toxin; exponentiating it gives about $87.5$ fleas. Strain coefficients compare log means at zero concentration. Interaction coefficients compare concentration slopes with the common strain. A unit concentration increase multiplies the three means by about $0.751$, $1.057$ and $0.769$ respectively.

For design matrix $X$, the fitted expected Fisher information is $X^{T}WX$, $W=\mathrm{diag}({\hat{m}}_i)$, since a Poisson count has variance equal to its mean. Standard errors are square roots of diagonal entries of $(X^{T}WX{)}^{-1}$. Each reported Wald test divides its coefficient estimate by its standard error and compares with an approximate standard normal distribution under a zero-coefficient null, conditional on the other terms. Concentration has strong evidence of a negative baseline effect. At zero concentration strain 1 is not significant at 5% ($p=0.087$), while strain 2 is ($p=0.030$). The strain-1 slope differs strongly from the baseline ($p=0.000193$); the strain-2 slope has no detected difference ($p=0.808$). A nonsignificant difference is not proof of equality. The very significant intercept is a test of mean one at zero toxin, with little scientific relevance.

Relabelling factor levels merges strains 0 and 2 into one group and leaves strain 1 separate. Model $M_2$ retains three intercepts but allows only two slopes: strain 1 versus the common slope of strains 0 and 2. Thus it has five parameters, compared with six in $M_1$. Model $M_3$ merges both intercepts and slopes for strains 0 and 2, retaining only four parameters. The natural nesting is $M_3\subset M_2\subset M_1$.

For these nested generalized linear models, the difference in Poisson deviance is an asymptotic likelihood-ratio test statistic with degrees of freedom equal to the parameter difference. Comparing $M_2$ with $M_1$ gives $56.93-56.87=0.06$ on one degree of freedom, far below $3.84$: the additional strain-2 slope is unnecessary. Comparing $M_3$ with $M_2$ gives $76.98-56.93=20.05$ on one degree of freedom, far above $3.84$: merging the strain-0 and strain-2 intercepts is rejected. Comparing $M_3$ with $M_1$ uses $20.11$ on two degrees of freedom and requires a two-degree-of-freedom threshold, about $5.99$, rather than the supplied $3.84$.

$M_2$ is the most appropriate of the three models: retain different strain baselines and the special strain-1 slope, but pool the unsupported strain-2 slope difference. Residual degrees of freedom are $74,75,76$; deviances of this magnitude give no obvious evidence of overdispersion from these summaries alone.

Write $y={\int }_C{e}^{xt}f(t)dt$ and move the factor $x$ onto $e^{xt}$ by integration by parts. The differential-equation residual isThus the Laplace integral solution is obtained fromup to a nonzero constant and branch choices. The endpoint factor is proportional to $t^b(1-t{)}^{a-b}$, so the given positive real parts ensure its vanishing at $0$ and $1$.

For $x>0$, choose the directed intervals $C_1=[0,1]$ and $C_2=(-\infty ,0]$; for $x<0$, choose $C_1=[0,1]$ and $C_2=[1,\infty )$. On each interval choose a continuous branch separately. Equivalently their integrands, after an irrelevant constant phase is removed, use $t^{b-1}(1-t{)}^{a-b-1}$ on $(0,1)$, $(-t{)}^{b-1}(1-t{)}^{a-b-1}$ on $t<0$, and $t^{b-1}(t-1{)}^{a-b-1}$ on $t>1$. Exponential decay at the infinite endpoint removes the remaining boundary term.

For large positive $x$, $C_1$ has a nonzero leading term proportional to $e^x{x}^{-(a-b)}$, while $C_2$ is proportional to $x^{-b}$. For large negative $x$, $C_1$ is proportional to $(-x{)}^{-b}$ and $C_2$ to $e^x(-x{)}^{-(a-b)}$. These distinct asymptotics prove independence in each case; the gamma factors are nonzero under the assumptions.

For $a=3,b=1$, integrate $e^{xt}(1-t)$ on $[0,1]$ to get $(e^x-1-x)/x^2$. A convenient independent pair on either side of zero isDirect substitution verifies both, and their ratio is not constant. Their singularities cancel in $y_2-y_1$. Since $e^x-1-x=x^2/2+O(x^3)$, the solution regular at zero with the required normalization isIts power series supplies the removable value and verifies the equation at the singular point.

Set $y=a^2$. Multiplication of the Friedmann equation by $4a^4$ givesFor the expanding branch emerging from $y(0)=0$, $\dot{y}(0)=2\sqrt{\Gamma }$. Differentiating where $\dot{y}\ne 0$ gives $\ddot{y}=(4\Lambda /3)y-2$, and continuity extends it through a regular turning point. Put $\kappa =\sqrt{4\Lambda /3}$. Solving this linear equation with both initial data givesInitially $a(t)\sim (2\sqrt{\Gamma }t{)}^{1/2}$. If $\lambda >1$, $y$ stays positive and strictly increasing; at late times $a\sim [3(\lambda -1)/(4\Lambda ){]}^{1/2}{e}^{\kappa t/2}$. This universe expands forever.

If $0<\lambda <1$, expansion stops when $\tanh \kappa t_{\max }=\lambda$, givingThe physical solution expands from zero and then symmetrically recollapses to zero; negative $a^2$ beyond this interval is not a continuation of a real scale factor.

The completeness theorem for propositional logic states that $S\models \varphi$ if and only if $S⊢\varphi$. Soundness follows by checking that axioms are tautologies and inference rules preserve truth. For completeness it suffices to prove that every syntactically consistent set has a model.

The deduction theorem for propositional logic used here is: $S\cup \{\psi \}⊢\chi$ if and only if $S⊢\psi \Rightarrow \chi$, in classical propositional logic. A proof is finite, so the union of a chain of consistent sets remains consistent: a contradiction would already occur in one member. By Zorn's lemma, extend a consistent $S$ to a maximal consistent set $T$. If $\psi \notin T$, adding $\psi$ yields a contradiction; the deduction theorem gives $T⊢\neg \psi$, and maximal consistency puts $\neg \psi$ in $T$. Conversely $\psi$ and $\neg \psi$ cannot both belong to $T$. Maximal consistency also makes $T$ deductively closed.

Assign each primitive proposition the truth value true precisely when it belongs to $T$. The maximal consistent set truth lemma follows by induction on formulas. For negation use the preceding dichotomy. For implication, if $\neg \psi \in T$ or $\chi \in T$, the classical implication axioms put $\psi \Rightarrow \chi$ in $T$; if $\psi \in T$ and $\neg \chi \in T$, modus ponens prevents that implication from belonging to $T$. This is exactly its truth table. Other connectives are defined from these or handled by the same induction. The resulting valuation satisfies all of $T$, hence all of $S$.

If $S⊬\varphi$, the deduction theorem and classical contradiction rules show $S\cup \{\neg \varphi \}$ is consistent. Its model falsifies $\varphi$, so $S/\models \varphi$. This proves completeness.

The propositional compactness theorem says $S$ is satisfiable if every finite subset is satisfiable. Otherwise completeness would make $S$ inconsistent, and a finite proof of contradiction would involve only finitely many assumptions, contradicting their satisfiability. Equivalently, every semantic consequence of $S$ is a consequence of a finite subset. The decidability theorem for propositional logic says there is an algorithm deciding whether a formula is provable with no assumptions: enumerate all truth assignments to its finitely many primitive propositions, check whether it is a tautology, and use soundness and completeness to identify tautologies with theorems. This does not claim an algorithm for arbitrary infinitely presented theories.

Finally, the valuation assigning false to every $p_n$ satisfies all the implications in $S$ and falsifies $p_1$. By soundness, $S⊬p_1$.

A strongly regular graph with parameters $(k,a,b)$ is a simple graph in which each vertex has degree $k$, adjacent vertices have exactly $a$ common neighbours, and distinct nonadjacent vertices have exactly $b$ common neighbours. Its order $n$ is often included separately as a fourth parameter.

Let $A$ be the adjacency matrix of a graph, $J$ the all-ones matrix and $1$ the all-ones vector. Counting two-step walks givesBecause $b\ge 1$, any two nonadjacent vertices have a common neighbour, so the graph is connected. The eigenvalue $k$ is simple: for a real eigenvector at $k$, a maximal entry and the averaging equation force its neighbours, and then every vertex, to have the same value. Since $A$ is real symmetric, its other eigenvectors lie in $1^{\perp }$, where $J$ vanishes. Their eigenvalues are thereforeHere $d>0$ for the present incomplete connected graph. Indeed $k\ge b$; if $d=0$, then $k=b=a$, which conflicts with $a\le k-1$ for an edge. If $r,s$ have multiplicities $m_r,m_s$, dimension and zero trace give $m_r+m_s=n-1$ and $k+m_{r}r+m_{s}s=0$. ThusThese are dimensions of eigenspaces, hence integers. A multiplicity may be zero; that does not affect the conclusion.

Count edges from the $k$ neighbours of a fixed vertex to its $n-k-1$ nonneighbours. Each neighbour contributes $k-a-1$ such edges, and each nonneighbour receives $b$. With $a=0,b=3$,The graph is incomplete, so there are nonadjacent vertices and $k\ge 3$. Substituting into the eigenvalue multiplicity formula gives a multiplicity difference of $k^2/\sqrt{4k-3}$. This must be an integer. As $k>0$, $d=\sqrt{4k-3}$ is rational and hence an integer; it is positive and odd. Moreover $d\mid k^2$ and $4k=d^2+3$, so reducing $16k^2$ modulo $d$ gives $d\mid 9$. Thus $d\in \{1,3,9\}$. The value $d=1$ gives $k=1$, impossible here; the other two give $(k,n)=(3,6)$ and $(21,162)$. Consequently $|G|\in \{6,162\}$. This is a necessity argument, not a claim that integrality alone constructs either graph.

If the two-dimensional group representation were reducible, Maschke's theorem would split it into a direct sum of two one-dimensional group representations. In a basis respecting that decomposition, every representing matrix would be diagonal. All would therefore commute, contradicting the specified pair. Hence the group representation is irreducible. Complete reducibility over $\mathbb{C}$ is essential to this argument.

Call the two proposed matrices $A$ and $B$. Direct multiplication gives $A^{2n}=I$, $B^2={ϵ}^{n}I=A^n$, and $B^{-1}AB=A^{-1}$. Both are invertible since $ϵ\ne 0$. They therefore satisfy every defining relation, and the universal property of a group presentation gives the desired group representation.

Put $\zeta =e^{\pi i/n}$. The group representations from the preceding construction with $ϵ={\zeta }^j$, $1\le j\le n-1$, will be denoted ${\rho }_j$. Their diagonal matrices for $a$ have distinct eigenvalues. The only possible invariant lines would be these two eigenspaces, and the matrix for $b$ exchanges them. Thus each ${\rho }_j$ is irreducible. Equivalently their $a,b$ matrices do not commute, so part (a) applies. The choices $j$ and $2n-j$ give equivalent group representations by exchanging basis vectors. The displayed range contains exactly one of each pair, and distinct members have different traces at $a$ because $\cos (\pi j/n)$ is strictly decreasing there.

For a one-dimensional character of a representation, write $a\mapsto \alpha$, $b\mapsto \beta$. The conjugation relation forces ${\alpha }^2=1$, and ${\beta }^2={\alpha }^n$. Hence there are exactly four choices:For even $n$, $\beta =\pm 1$ for either $\alpha$; for odd $n$, $\alpha =1$ has $\beta =\pm 1$ and $\alpha =-1$ has $\beta =\pm i$. Finally,The sum of squares of irreducible representation dimensions proves that these four one-dimensional and $n-1$ two-dimensional group representations exhaust the irreducibles, including $n=1$ where the latter list is empty.

The conjugacy classes are $\{1\}$, $\{a^n\}$, $\{a^r,a^{-r}\}$ for $1\le r\le n-1$, and two classes containing respectively $b{a}^{2s}$ and $b{a}^{2s+1}$, each of size $n$. Conjugation by $b$ inverts powers of $a$, while conjugation by $a$ changes the exponent of $b{a}^r$ by two; these facts prove the list and sizes. The sizes total $4n$.

The complete character table is specified by the following array, with one column for every $r$ and one row for every $j$ in the indicated ranges:Here $1\le r,j\le n-1$; the four pairs $(\alpha ,\beta )$ are those in part (ii). In particular, this formula displays both parity cases without concealing the different values $\pm i$ when $n$ is odd. The zero entries follow because the two-dimensional matrices representing $b{a}^r$ are off-diagonal.

Let $c_1,\dots ,c_d$ be the coefficients of $f$. Each is an algebraic integer. The ring $R=\mathbb{Z}[c_1,\dots ,c_d]$ is a finitely generated $\mathbb{Z}$-module: reduce powers of each generator by its monic integral equation. As $f$ is monic, $R[\alpha ]$ is generated over $R$ by $1,\alpha ,\dots ,{\alpha }^{d-1}$, so is a finite $\mathbb{Z}$-module. It is torsion-free as a subgroup of a characteristic-zero field, hence free of finite rank. Multiplication by $\alpha$ on it is represented by an integer matrix. Its monic characteristic polynomial, by Cayley-Hamilton theorem, annihilates $\alpha$. Thus $\alpha$ is an algebraic integer. This proves the required transitivity of integrality in this setting.

Every root of the monic $h$ is also a root of the monic polynomial $h^n\in {\mathcal{O}}_K[x]$, and is therefore an algebraic integer. The coefficients of $h$ are elementary symmetric polynomials in these roots, hence algebraic integers because algebraic integers form a ring. They also belong to $K$, so belong to ${\mathcal{O}}_K$. Consequently $h\in {\mathcal{O}}_K[x]$.

The cubic has no rational integer root among the divisors of four, so is irreducible. Its polynomial discriminant is $4-27\cdot 16=-428=-4\cdot 107$. Thus the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$ is one or two, since its square divides the discriminant.

Set $\beta =({\alpha }^2+\alpha )/2$. Using ${\alpha }^3=\alpha +4$ givesand hence ${\beta }^3-{\beta }^2-3\beta -2=0$. Therefore $\beta$ is integral. The lattice spanned by $1,\alpha ,\beta$ contains $\mathbb{Z}[\alpha ]$ with index two. Changing basis divides the discriminant by four, giving $-107$, which is squarefree. An integral lattice of full rank with squarefree discriminant cannot have any further index in the full ring of integers. Henceis the required integral basis. The relations also give $\alpha ={\beta }^2-\beta -2$, so this ring can equally be generated by $\beta$.

For $\alpha \in L$ with $f(\alpha )=0$, evaluation $K[t]\to L$, $g\mapsto g(\alpha )$, has $(f)$ in its kernel. It therefore induces a $K$-algebra homomorphism ${\sigma }_{\alpha }:K[t]/(f)\to L$. Conversely, such a homomorphism is determined by $\alpha =\sigma (\bar{t})$, since every class is a polynomial in $\bar{t}$. Applying it to $f(\bar{t})=0$ shows $f(\alpha )=0$. These constructions are inverse, giving the root-embedding correspondenceIrreducibility ensures the quotient is a field, so each unital homomorphism is injective.

A splitting field of $f$ over $K$ is an extension generated over $K$ by all its roots in which $f$ splits into linear factors. A nonconstant irreducible polynomial is separable polynomial when its roots in a splitting field are distinct, equivalently when it is coprime to its formal derivative. For a general polynomial, the convention compatible with the printed assertion is that all its irreducible factors are separable; repeated occurrences of the same factor are allowed under this convention.

For a finite field of characteristic $p$, the finite-field Frobenius automorphism $x\mapsto x^p$ is injective and hence surjective. If an irreducible polynomial $g$ had zero derivative, every nonzero exponent would be divisible by $p$. Taking $p$th roots of its coefficients would give $g=h^p$, contradicting irreducibility. Thus $g^{\prime }\ne 0$, and irreducibility gives $\gcd (g,g^{\prime })=1$. Every irreducible factor is therefore separable: finite fields are perfect.

If instead one defines a separable polynomial to mean that the entire polynomial has no repeated roots, the printed claim for every $f$ is false: $(t-1{)}^2$ is a counterexample over any finite field. The proof above establishes the precise intended perfect-field assertion; a polynomial has entirely distinct roots if and only if it is also squarefree.