Integral basis and discriminant of a coprime-discriminant compositum

By [F1] the given bases are $\mathbb{Q}$-bases of $K$ and $L$, and [F2] gives $[KL:L]=m$ and $[KL:K]=n$.

Every embedding $\sigma :K\to \mathbb{C}$ extends to an embedding $\tilde{\sigma }:KL\to \mathbb{C}$ with $\tilde{\sigma }{\mathrm{\mid }}_L={\mathrm{i}\mathrm{d}}_L$. Indeed, by [F9] write $K=\mathbb{Q}(\alpha )$ with minimal polynomial $f$ of degree $m$; then $KL=L(\alpha )$, and $[L(\alpha ):L]=m=\mathrm{deg}f$ shows that $f$ has no factor of intermediate degree over $L$, so $f$ is irreducible over $L$ and $KL\cong L[t]\mathrm{/}(f)$ as $L$-algebras. The evaluation homomorphism $L[t]\to \mathbb{C}$, $t\mapsto \sigma (\alpha )$, kills $f$ because $\sigma$ fixes $\mathbb{Q}$ and $f(\alpha )=0$; it therefore induces the required $\tilde{\sigma }$ under the isomorphism $L[t]\mathrm{/}(f)\cong KL$.

Put $V:={\sum }_{i=1}^{m}L{\alpha }_i\subseteq KL$. Since ${\alpha }_1,\dots ,{\alpha }_m$ are a $\mathbb{Q}$-basis of $K$, each product ${\alpha }_i{\alpha }_j$ is a $\mathbb{Q}$-linear combination of them; the same structure constants lie in $L$, so $V$ is closed under multiplication. Also $1\in V$, and $V$ contains both $L$ and $K$. It is finite-dimensional over $L$. For each nonzero $x\in V$, multiplication by $x$ is an injective $L$-linear map $V\to V$, because $V$ lies in the field $KL$; finite dimensionality makes the map surjective, so some $y\in V$ satisfies $xy=1$. Thus $V$ is a field containing $K$ and $L$, hence $V=KL$. By [F2] it has $L$-dimension $m$. Its $m$ elements ${\alpha }_i$ span $V$; if they were dependent, a nonzero coefficient in a dependence relation could be inverted in $L$ to express one ${\alpha }_i$ in terms of the others, so the remaining $m-1$ vectors would still span $V$. But an $m$-element basis of $V$ is an independent subset of a space with that $(m-1)$-element spanning set, contradicting [F7]. Thus the ${\alpha }_i$ are independent and spanning, hence an $L$-basis by [F7]. The symmetric argument with $W:={\sum }_{j=1}^{n}K{\beta }_j$ shows that ${\beta }_1,\dots ,{\beta }_n$ form a $K$-basis of $KL$ by the same finite argument.

Applying [F6] to the tower $\mathbb{Q}\subseteq L\subseteq KL$ with the $\mathbb{Q}$-basis ${\beta }_1,\dots ,{\beta }_n$ of $L$ and the $L$-basis ${\alpha }_1,\dots ,{\alpha }_m$ of $KL$, the products ${\alpha }_i{\beta }_j$ form a $\mathbb{Q}$-basis of $KL$; in particular they are $\mathbb{Q}$-linearly independent and there are $mn$ of them.

Let $\gamma \in {\mathcal{O}}_{KL}$. By step 2.1 there are unique $x_1,\dots ,x_m\in L$ with $\gamma ={\sum }_i{x}_i{\alpha }_i$. Each $x_i$ lies in $L=\mathrm{Frac}({\mathcal{O}}_L)$, so there are integers $a_{ij}$ and a positive integer $r$ with $x_i={\sum }_j(a_{ij}\mathrm{/}r){\beta }_j$; dividing out common factors, we may assume that no prime divides $r$ and all $a_{ij}$ simultaneously.

Let ${\sigma }_1,\dots ,{\sigma }_m$ be the distinct embeddings of $K$ into $\mathbb{C}$ and let ${\tilde{\sigma }}_k$ extend ${\sigma }_k$ as in step 1.2. Applying ${\tilde{\sigma }}_k$ to $\gamma ={\sum }_i{x}_i{\alpha }_i$ and using ${\tilde{\sigma }}_k{\mathrm{\mid }}_L=\mathrm{i}\mathrm{d}$ together with $x_i\in L$ gives the linear system ${\sum }_i{\sigma }_k({\alpha }_i)x_i={\tilde{\sigma }}_k(\gamma )$, $k=1,\dots ,m$. Its coefficient matrix $M:=({\sigma }_k({\alpha }_i))$ has determinant $D$ with $D^2=d_K\mathrm{\ne }0$ by [F5]; the entries ${\sigma }_k({\alpha }_i)$ of $M$ are algebraic integers, and so are the entries ${\tilde{\sigma }}_k(\gamma )$, by [F3] applied to the integral elements ${\alpha }_i\in {\mathcal{O}}_K$ and $\gamma \in {\mathcal{O}}_{KL}$.

Cramer's rule applied to the system of step 4.1 gives $D{x}_i=D_i$, where $D_i$ is the determinant of a matrix with algebraic-integer entries; hence $D$ and every $D_i$ are algebraic integers by [F3], and $\mathrm{\Delta }{x}_i=D\cdot D_i$ is an algebraic integer, where $\mathrm{\Delta }:=D^2=d_K$ is a nonzero integer.

By step 3.2, $\mathrm{\Delta }{x}_i={\sum }_j(\mathrm{\Delta }{a}_{ij}\mathrm{/}r){\beta }_j$ lies in $L$, so $\mathrm{\Delta }{x}_i\in L\cap \bar{\mathbb{Z}}={\mathcal{O}}_L$ by step 5.1 and [F4]. As ${\beta }_1,\dots ,{\beta }_n$ is an integral basis, the rational numbers $\mathrm{\Delta }{a}_{ij}\mathrm{/}r$ are integers; equivalently $r\mid \mathrm{\Delta }{a}_{ij}=d_K{a}_{ij}$ for all $i,j$.

The denominator was reduced in step 3.2, so no prime divides $r$ and every $a_{ij}$; therefore their common gcd is $1$. By [F8], choose integers $c,b_{ij}$ with $cr+{\sum }_{i,j}{b}_{ij}{a}_{ij}=1$. Multiplying by $d_K$ and using $r\mid d_K{a}_{ij}$ from step 6.1 shows that every term on the left is divisible by $r$, hence $r\mid d_K$.

Regroup the same expansion from step 3.2 as $\gamma ={\sum }_j{y}_j{\beta }_j$, where $y_j={\sum }_i(a_{ij}\mathrm{/}r){\alpha }_i\in K$. The symmetric trace-matrix argument, now using the $K$-basis ${\alpha }_i$ and the $L$-basis ${\beta }_j$ from step 2.1, gives $r\mid d_L{a}_{ij}$ for every same coefficient $a_{ij}$. The same reduced-denominator identity of [F8] therefore gives $r\mid d_L$. Together with step 7.1 and $\gcd (d_K,d_L)=1$, this implies $r=1$. Thus $\gamma ={\sum }_{i,j}{a}_{ij}{\alpha }_i{\beta }_j\in {\mathcal{O}}_K{\mathcal{O}}_L$ for every $\gamma \in {\mathcal{O}}_{KL}$, proving ${\mathcal{O}}_{KL}\subseteq {\mathcal{O}}_K{\mathcal{O}}_L$.

Conversely ${\mathcal{O}}_K{\mathcal{O}}_L\subseteq {\mathcal{O}}_{KL}$: products and sums of integral elements are integral by [F3], and they lie in $KL$. With step 8.1 this gives ${\mathcal{O}}_{KL}={\mathcal{O}}_K{\mathcal{O}}_L$, and since the products ${\alpha }_i{\beta }_j$ are $\mathbb{Q}$-linearly independent by step 3.1 and span ${\mathcal{O}}_K{\mathcal{O}}_L$ over $\mathbb{Z}$ by construction, they form an integral basis of ${\mathcal{O}}_{KL}$.

The restrictions of the $mn$ distinct embeddings of $KL$ into $\mathbb{C}$ to $K$ and to $L$ define a map $\mathrm{Emb}(KL)\to \mathrm{Emb}(K)\times \mathrm{Emb}(L)$; it is injective, because an embedding of $KL$ is determined by its restrictions to $K$ and to $L$, which together generate $KL$ over $\mathbb{Q}$. Both sides have $mn$ elements by [F5], so the map is a bijection. Order rows by pairs $(k,l)$ and columns by pairs $(i,j)$, with the first index slow in each order. The embedding matrix of the integral basis ${\alpha }_i{\beta }_j$ of step 9.1 then has entries ${\sigma }_k({\alpha }_i){\tau }_l({\beta }_j)$.

The embedding matrix of step 10.1 is $K_1{K}_2$: summing over the intermediate pair $(a,b)$ gives ${\sum }_{a,b}{\sigma }_k({\alpha }_a){\delta }_{lb}{\delta }_{ai}{\tau }_b({\beta }_j)={\sigma }_k({\alpha }_i){\tau }_l({\beta }_j)$. Here $(K_1{)}_{(k,l),(i,j)}={\sigma }_k({\alpha }_i){\delta }_{lj}$ and $(K_2{)}_{(k,l),(i,j)}={\delta }_{ki}{\tau }_l({\beta }_j)$. By [F10], $\det (K_1{K}_2)=\det (K_1)\det (K_2)$. In the first-index-slow ordering of step 10.1, $K_2$ is block diagonal with $m$ blocks $N=({\tau }_l({\beta }_j))$, so $\det (K_2)=\det (N{)}^m$. For $K_1$, reorder both its rows and its columns by the same perfect shuffle from $(k,l)$ to $(l,k)$ and from $(i,j)$ to $(j,i)$. These are the same permutation of $mn$ positions, so their determinant signs multiply to $1$; in the resulting ordering $K_1$ is block diagonal with $n$ blocks $M=({\sigma }_k({\alpha }_i))$. Hence $\det (K_1)=\det (M{)}^n$, using [F10] and the Leibniz formula for block diagonal matrices.

By [F5], $d_{KL}=\det ({\sigma }_k({\alpha }_i){\tau }_l({\beta }_j){)}^2=(\det M{)}^{2n}(\det N{)}^{2m}=d_K^n{d}_L^m=d_K^{[L:\mathbb{Q}]}{d}_L^{[K:\mathbb{Q}]}$, using $\det (M{)}^2=d_K$ and $\det (N{)}^2=d_L$ from [F5] and $m=[K:\mathbb{Q}]$, $n=[L:\mathbb{Q}]$. This completes the proof.