Regulator of a number field

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a number field of signature $(r_1,r_2)$ (Number field, Archimedean embeddings and signature) and unit rank $r=r_1+r_2-1$, and let $({\epsilon }_1,\dots ,{\epsilon }_r)$ be a system of fundamental units of $K$ (System of fundamental units). Let $A$ be the $(r_1+r_2)\times r$ real matrix

$A=(\lambda ({\epsilon }_1)\cdots \lambda ({\epsilon }_r))$

whose columns are the logarithmic vectors $\lambda ({\epsilon }_i)$ in the doubled convention of the logarithmic embedding (Logarithmic embedding of a number field). For $k=1,\dots ,r_1+r_2$ let $A_k$ be the $r\times r$ matrix obtained from $A$ by deleting row $k$, and let $\det A_k$ be its determinant (For $n\ge 1$, the determinant over a commutative ring by the Leibniz formula, and $\mathrm{\mid }\det A\mathrm{\mid }$ for a real matrix). The regulator of $K$ is

$R_K:=\mathrm{\mid }\det A_k\mathrm{\mid }\mathrm{.}$

For $r=0$ the matrices $A$ and $A_k$ are empty, and the empty determinant is defined to be $1$; thus $R_{\mathbb{Q}}=1$ and $R_K=1$ for imaginary quadratic $K$.

Why a deleted row gives a well-defined number. The columns of $A$ are linearly independent over $\mathbb{R}$: the full-lattice theorem says that $\lambda ({\mathcal{O}}_K^{\times })$ is a full lattice in $H$ of rank $r=r_1+r_2-1={\dim }_{\mathbb{R}}H$. Thus a $\mathbb{Z}$-basis has $r$ vectors spanning $H$, hence is also an $\mathbb{R}$-basis of $H$. The vectors $\lambda ({\epsilon }_1),\dots ,\lambda ({\epsilon }_r)$ are another $\mathbb{Z}$-basis of the same group, and their change-of-basis matrix lies in ${\mathrm{GL}}_r(\mathbb{Z})$, hence is invertible over $\mathbb{R}$. Therefore these columns are $\mathbb{R}$-linearly independent and $A$ has rank $r$, so $r_1+r_2=r+1\ge 1$ (Row space, column space, nullspace, row rank, column rank and matrix rank, The logarithmic unit image is a full lattice). Each column lies in the hyperplane $H$, so its coordinates sum to zero (Unit logarithms lie in the trace-zero hyperplane). The deleted-row minors of a rank-$r$ real matrix with $r+1$ rows and zero column sums therefore satisfy $\det A_k=(-1{)}^{k-1}\det A_1$ and are all nonzero (Deleted-row minors of a zero-column-sum matrix agree up to sign); in particular $\mathrm{\mid }\det A_k\mathrm{\mid }$ does not depend on the deleted row $k$.

Independence of the fundamental system. The number $R_K$ above is defined from one chosen system of fundamental units; that the absolute deleted-row determinant does not depend on this auxiliary choice, so that $R_K$ depends on $K$ alone, is the content of The regulator is well defined ↗, which also shows $R_K>0$ in the rank-$r$ case. A change of fundamental system multiplies $A$ on the right by a matrix in ${\mathrm{GL}}_r(\mathbb{Z})$, which is why the absolute determinant, and not the signed one, is the invariant.

Normalization. The factor $2$ on the complex coordinates of $\lambda$ is part of the doubled convention fixed in the definition of the logarithmic embedding; with it, the regulator of a real quadratic field is $\log \epsilon$ for its fundamental unit $\epsilon >1$, and mixing conventions changes the value. Ordering the coordinates differently permutes rows of $A$, which leaves every $\mathrm{\mid }\det A_k\mathrm{\mid }$ unchanged, so the definition is insensitive to the ordering of the embeddings.