Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 166 / 2

Let $K$ be a number field. Its places consist of its real embeddings, conjugate pairs of complex embeddings, and the finite places associated with nonzero prime ideals $\mathfrak{p}$ of its ring of integers of a number field. At a real or complex place use the usual absolute value. If $\mathfrak{p}$ lies over the prime number $p$ with ramification index $e_{\mathfrak{p}}$, normalize its absolute value byThese normalizations extend the standard absolute values on $\mathbb{Q}$.

Write $d_v=[K_v:{\mathbb{Q}}_v]$ for the local degree of a place. Thus $d_v$ is $1$ at a real place, $2$ at a complex place, and $e_v{f}_v$ at a finite place. The Absolute multiplicative Weil height isThe product formula shows that this is unchanged when $K$ is replaced by a larger number field containing $\alpha$.

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The product formula says that every $x\in K^{\times }$ satisfiesFirst suppose that $x$ is an algebraic integer. Its principal ideal has the prime ideal factorizationTaking the ideal norm givesOn the other hand, the field norm is the product over embeddings, soEquating these expressions proves the formula for algebraic integers. Every nonzero element of $K$ is a quotient of two algebraic integers, and multiplicativity completes the proof.

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For every $k\in \mathbb{Z}$,For $k\ge 0$, this follows directly fromat each place of a number field. The product formula gives $H({\alpha }^{-1})=H(\alpha )$, becauseand this handles negative $k$ as well.

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If $\alpha \ne 0$ has degree $d$, then at every Archimedean embeddingIndeed, the factor of the defining product belonging to that Archimedean place shows $\max (1,|\alpha |{)}^{1/d}\le H(\alpha )$; using the exact local degree only improves this estimate. This proves the upper bound. Apply it to ${\alpha }^{-1}$ and use $H({\alpha }^{-1})=H(\alpha )$ to obtain the lower bound. This is the Liouville height inequality.

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Write $\mathcal{L}(R)$ for the polynomial length. The height bound for a polynomial evaluation isprovided the denominator is nonzero. At non-Archimedean places the integral coefficients and ultrametric inequality give the local estimate without an extra constant; at Archimedean places the triangle inequality gives the polynomial length. Multiplication over every place of a number field and the product formula produce the displayed bound.

Taking $P(X,Y)=X+Y$ and $Q=1$ givesTaking $P(X)=X+1$ gives

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For coprime integers $p,q$ with $q>0$,This follows either directly from the real and p-adic absolute values, or from the height-Mahler measure formula applied to the primitive minimal polynomial $qX-p$.

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Choose distinct prime numbers $p,q$ so large that $pq>C(p+q)$, and setThe fraction $(p+q)/(pq)$ is reduced, because neither $p$ nor $q$ divides $p+q$. The height of a rational number therefore givesThe choice of $p,q$ proves the required strict inequality.

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For an integer $m>1$, let ${\alpha }_m$ be a root ofIts roots areThe polynomial is irreducible over $\mathbb{Q}$: its discriminant is $4m(m-1)$, and the product of the coprime consecutive integers $m$ and $m-1$ cannot be a square unless both are squares, which is impossible for consecutive positive squares beyond $0,1$.

The height-Mahler measure formula givesBoth algebraic conjugates of ${\alpha }_m+1$ exceed $1$, soConsequentlyFix, for example, $\delta =1/10$. For all sufficiently large $m$, the ratio is larger than $1+\delta$ by a fixed margin, while $H({\alpha }_m)\to \infty$. Hence, for every $C>0$, some sufficiently large $m$ satisfies

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