Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 20F / Solution

An algebraic number is an algebraic integer when it is a root of a monic polynomial in $\mathbb{Z}[X]$.

If $\alpha$ and $\beta$ are algebraic integers, the ring $\mathbb{Z}[\alpha ,\beta ]$ is a finitely generated $\mathbb{Z}$-module. Multiplication by $\alpha \beta$ preserves this module, so the algebraic integer module criterion shows that $\alpha \beta$ is an algebraic integer.

The ring of integers of a quadratic field givessince $2\equiv 2(\mathrm{mod}4)$. For a direct verification, if $x=a+b\sqrt{2}$ with $a,b\in \mathbb{Q}$ is integral, then its trace $2a$ and norm $a^2-2b^2$ are integers. Writing $2a=m\in \mathbb{Z}$ and reducing the norm condition in lowest terms shows first that $b$ can have denominator at most $2$; a denominator $2$ with odd numerator would require $m^2\equiv 2(\mathrm{mod}4)$, which is impossible. Thus $b\in \mathbb{Z}$, and then the norm condition forces $m$ even, so $a\in \mathbb{Z}$. Conversely, every $a+b\sqrt{2}$ with $a,b\in \mathbb{Z}$ is integral because $\sqrt{2}$ is integral and algebraic integers form a ring.

Now let $f$ have degree $d$ and roots ${\alpha }_1,\dots ,{\alpha }_d$. Sinceand $|{\alpha }_1|>1$, the integer $|a_d|$ cannot be $2$; hence $|a_d|=1$. Thus $f$ is monic up to sign and ${\alpha }_1$ is an algebraic integer.

Let $B$ be the product of all roots with modulus greater than one, counting multiplicity. Complex roots occur in conjugate pairs, so $B$ is real up to the signs of the real roots andAs a product of algebraic integers, $B$ is an algebraic integer; the equality forces $B=\pm 2$. Thereforeis the absolute value of the product of the remaining conjugates, is rational, and is an algebraic integer. It is consequently a nonzero rational integer. Every remaining factor has modulus at most one, so this integer is at most one. It must equal one, and henceThis is the mahler-measure norm argument at measure two.

Solved by gpt-5.6-sol high.