Past exam of the algebraic number theory course of the University of Oxford / 2023 / 1 / c

Put $\alpha =\sqrt{m_1}$, $\beta =\sqrt{m_2}$ and $F=\mathbb{Q}(\alpha )$. The hypotheses make $L/\mathbb{Q}$ a biquadratic extension. To check the degree of a field extension is four, if $\beta =x+y\alpha$ with rational $x,y$, squaring gives $xy=0$. The case $y=0$ would make $m_2$ a rational square; the case $x=0$ would make $m_2/m_1$ a rational square. Neither can occur for the specified distinct coprime square-free integers. Hence $[L:F]=2$.

The nonidentity field automorphism of $L/F$ fixes $\alpha$ and sends $\beta$ to $-\beta$. Thus, for $\gamma =(r+s\alpha )+(t+u\alpha )\beta$, the field norm and field trace areIf $\gamma$ is an algebraic integer, so is its conjugate, and their sum and product are algebraic integers. Conversely, put $T={\mathrm{Tr}}_{L/F}(\gamma )$ and $N=N_{L/F}(\gamma )$. The identityshows that $\gamma$ is integral over $\mathbb{Z}[T,N]$. If $T,N$ are algebraic integers, the ring generated by algebraic integers $\mathbb{Z}[T,N]$ is integral over $\mathbb{Z}$. Since integral dependence is transitive, $\gamma$ is an algebraic integer. This proves the trace and norm integrality criterion for a quadratic extension, including elements already in $F$.

Again put $\alpha =\sqrt{m_1}$ and $\beta =\sqrt{m_2}$. The ring of integers of each of the three quadratic subfields has integral basis $1,\alpha$, $1,\beta$, or $1,\alpha \beta$, respectively: their square-free radicands are respectively congruent to $3,2,2$ modulo four. The field traces of the algebraic integer $\gamma$ to those subfields areThey are algebraic integers, so comparison with each integral basis givesThis use of all three subfields is needed to control $d$ as well as the other coefficients.

The field norm to $\mathbb{Q}(\alpha )$ isIts two coefficients must be integers. The second condition says $ab$ is even, since $m_2$ is even. Reducing the numerator of the first coefficient modulo four givesReduction modulo two first gives $a\equiv b(\mathrm{mod}2)$. Since $ab$ is even, both $a$ and $b$ are even. The same congruence now gives $c^2+d^2\equiv 0(\mathrm{mod}2)$, or $c\equiv d(\mathrm{mod}2)$. Therefore every algebraic integer has the formwith four integer coefficients.

For the converse, $1,\alpha ,\beta$ are algebraic integers, andHere $4\mid 1+m_1$ and $2\mid m_2$, so both coefficients are integers. The monic polynomial $X^2-w^2$ and integral dependence is transitive show that $w$ is an algebraic integer. The four displayed generators are rationally linearly independent, since their coefficient matrix in $1,\alpha ,\beta ,\alpha \beta$ has determinant $1/2$. Hence the integral basis for coprime biquadratic radicands congruent to three and two isIn this formula $\sqrt{m_1{m}_2}$ denotes the product $\alpha \beta$ of the chosen square roots; this convention also handles negative radicands.

Put $\alpha =\sqrt{m_1}$, $\beta =\sqrt{m_2}$, andThe proposed integral basis for coprime biquadratic radicands congruent to two and one isEach entry is an algebraic integer, by the ring of integers of a quadratic field formula and closure under products. Their coefficient matrix in $1,\alpha ,\beta ,\alpha \beta$ has determinant $1/4$, so they are rationally linearly independent. It remains to show that they span every algebraic integer.

Write $\gamma =(a+b\alpha +c\beta +d\alpha \beta )/2$ initially with rational $a,b,c,d$. The field traces to $\mathbb{Q}(\alpha )$ and $\mathbb{Q}(\alpha \beta )$ are $a+b\alpha$ and $a+d\alpha \beta$. Since $m_1\equiv m_1{m}_2\equiv 2(\mathrm{mod}4)$, the corresponding integral basis in each subfield gives $a,b,d\in \mathbb{Z}$. The field trace to $\mathbb{Q}(\beta )$ is $a+c\beta$. Membership in $\mathbb{Z}[\omega ]$ says that $2a,2c$ are integers of equal parity. As $a$ is already an integer, this forces $c\in \mathbb{Z}$ too.

Now the field norm to $\mathbb{Q}(\beta )$ isBy the ring of integers of a quadratic field formula, $A/2$ is an integer and has the same parity as $ac-m_{1}bd$. Thus $A$ is even, giving $a\equiv c(\mathrm{mod}2)$, andUsing $m_1\equiv 2$, $m_2\equiv 1$ and $a\equiv c(\mathrm{mod}2)$, this becomesHence $b\equiv d(\mathrm{mod}2)$. Finally,an expression with integer coefficients. This proves the asserted integral basis without assuming a general theorem about products of elements of an integral basis.