Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 166 / 3

For a linear form $L=a_1{X}_1+\cdots +a_N{X}_N$ over a number field $K$, define its height to be the projective height of its coefficient vector:The product formula makes this independent of multiplying $L$ by a nonzero scalar.

The Siegel lemma says the following. Let $D=[K:\mathbb{Q}]$, let $L_1,\dots ,L_M$ be linear forms in $N$ variables with $DM<N$, and suppose $H_p(L_i)\le \mathcal{H}$ for every $i$, where $\mathcal{H}\ge 1$. Then there is a nonzero $\mathbf{x}\in {\mathbb{Z}}^N$ annihilated by all the forms and satisfying

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Writeso there are $N=2(n+1)$ unknown integer coefficients. Let $t$ be the number of integers $r$ withFor each such $r$, impose the linear equation over $K=\mathbb{Q}(\alpha )$where $D_r$ is the normalized derivative of a polynomial. The coefficient of $p_j$ is $\left(\genfrac{}{}{0ex}{}{j}{r}\right){\alpha }^{j-r}$ and that of $q_j$ is $\left(\genfrac{}{}{0ex}{}{j}{r}\right){\alpha }^{j-r+1}$. The local definition of projective height, together with $\left(\genfrac{}{}{0ex}{}{j}{r}\right)\le 2^n$, givesfor a constant $C_0$ depending only on $\alpha$.

There are $M=t$ forms over the degree-$d$ field $K$, andMoreover $N-dM\ge \kappa n+2$. Applying Siegel lemma gives a nonzero integral coefficient vector withbecause the exponent is bounded in terms of $\kappa$ and every fixed power of $n$ is at most exponential in $n$. These $P,Q$ have all the required vanishing normalized derivatives.

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Suppose $P$ and $Q$ were linearly dependent over $\mathbb{Q}$. Then $P=cQ$ for some $c\in \mathbb{Q}$, with the zero-polynomial cases included. Since $d\ge 3$, $c+\alpha \ne 0$, and thereforeIf this has multiplicity $t$ at $\alpha$, then the minimal polynomial of $\alpha$ to the power $t$ divides $Q$ in $\mathbb{Q}[X]$. Hence $dt\le \mathrm{deg}Q\le n$.

The construction in part (b) givesWhen $\kappa <1$, this exceeds $n/d$ for all sufficiently large $n$ depending only on $\kappa$ and $d$, contradicting $dt\le n$. Thus $P,Q$ are linearly independent.

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The determinant in the hint is independent of $Y$ and equals the WronskianIt is nonzero because $P,Q$ are linearly independent. Also $\mathrm{deg}W<2n$, and coefficient convolution givesfor a constant $C_0$ depending only on $C$.

Fix $y$, and suppose $F(X,y)$ has multiplicity $r$ at $a=p/q$. Inthe two terms vanish to orders at least $r$ and $r-1$, so $W$ has multiplicity at least $r-1$ at $a$. The primitive polynomial $(qX-p{)}^{r-1}$ therefore divides $W$ in $\mathbb{Z}[X]$ by Gauss lemma for polynomials. Comparing leading coefficients givesIf $r>\delta n+1$, this implies $q^{\delta n}<C_0^n$ and hence $q<C_0^{1/\delta }$. Choosing $q$ larger than this bound proves that $r\le \delta n+1$, uniformly in $y$.

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It is enough to prove the result for $0<\kappa <1$, since a construction for a smaller positive value gives the weaker vanishing requirement for any larger one. Apply part (b) with $\kappa /2$. For large $n$, part (c) gives linearly independent polynomials $P_0,Q_0$, andsatisfies $H(G)\le C_1^n$ and has order at leastat $X=\alpha$ after setting $Y=\alpha$.

Set $\delta =\kappa /(4d)$. By the rational multiplicity bound for a linear auxiliary polynomial, once $q_1$ is sufficiently large, the one-variable polynomial $G(X,y)$ has multiplicity at most $\delta n+1$ at $p_1/q_1$. Consequently there is some integer $j\le \delta n+1$ such thatPut $F=D_j^{X}G$. The normalized derivative of a polynomial preserves integral coefficients and multiplies height by at most $2^n$, so $H(F)\le C_2^n$. Differentiation lowers the vanishing order at $(\alpha ,\alpha )$ by at most $j$; for sufficiently large $n$,Finally write $F=P-YQ$ by replacing the coefficient of $Y$ by its negative. Then $F(p_1/q_1,y)\ne 0$ and all the claimed bounds hold.

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If $\alpha$ is nonreal, its positive distance from $\mathbb{R}$ makes the assertion immediate, so assume $\alpha \in \mathbb{R}$. Fix $\epsilon >0$ and putChoose $\kappa >0$ so small thatSuppose for a contradiction that infinitely many reduced fractions satisfy $|\alpha -p/q|<q^{-\mu }$. Their denominators are unbounded.

Choose one such $p_1/q_1$ with $q_1$ arbitrarily large, and then a later one $p_2/q_2$ with $q_2$ arbitrarily large. Choose the integer $n$ so thatThen $n$ can be made sufficiently large for part (e). Apply it with $y=p_2/q_2$. Since $F$ has integral coefficients, degree at most $n$ in $X$, and degree at most one in $Y$, its nonzero rational value satisfies the denominator boundThe supplied upper estimate and the two approximation inequalities give, with $a=(2-\kappa )n/d$,Using the upper bound $q_2<q_1^{a+(2-\kappa )/d}$ in the denominator estimate and comparing yieldsAfter taking $n$th roots and letting the choice of $q_2$ make $n$ large, this bounds $q_1^{\eta }$ by a constant depending only on $\alpha ,\kappa$. That contradicts the ability to choose $q_1$ arbitrarily large. Hence only finitely many such rational approximations exist. This is the Thue-Siegel rational approximation bound.

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