Let $K=\mathbb{Q}[\alpha]$ where $\alpha$ is integral over $\mathbb{Z}$ such that the Galois hull of $K$ can be embedded in $\mathbb{R}$. Let $S=\mathbb{Z}[\alpha]$. Let $x_1, \ldots , x_n$ be a $\mathbb{Q}$-basis of $K$ and let $f: K \to \mathbb{Q}$ be a $\mathbb{Q}$-linear map such that $f(x_i \cdot x_j)=\delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta.

Is it true that the $\mathbb{Z}$-module generated by $x_1, \ldots, x_n$ is closed under multiplication by elements of $S$?

  • $\begingroup$ Why do you make the assumption that the Galois closure (what I suppose you meant by Galois hull) of $K$ over $\mathbf Q$ is in the real numbers? $\endgroup$
    – KConrad
    Apr 15, 2014 at 16:50
  • $\begingroup$ You are specifying $f$ at all $x_i\cdot x_j$ which is more than $n$ in number, exceeding the degree of $K$; how can you guarantee such a linear map exists? $\endgroup$ Apr 17, 2014 at 10:37
  • $\begingroup$ @KConrad: I just need to consider this case, so I thought this assumption might help. @P Vanchinathan: I can't. The existence is meant as an assumption. $\endgroup$
    – Hans
    Apr 24, 2014 at 19:34
  • $\begingroup$ @P Vanchinathan: The conditions imposed via $f$ mostly constrain the choice of generators $x_i$, and these can be changed by an integer unimodular transformation without changing the lattice (or the field). Triangular matrices alone already give us $n(n-1)/2$ degrees of freedom. The remaining conditions then mostly say that the lattice is "nice". Working through examples, for fixed $n$, I always ended up with only one condition on (a defining polynomial for) $K$. $\endgroup$
    – GNiklasch
    Apr 25, 2014 at 7:32

1 Answer 1


When $\alpha$ is fixed at the beginning, this is not always true, and it doesn't matter whether $K$ is or isn't totally real.

Note first that any full-rank lattice in any number field admits some order of the field as endomorphisms. (Proof: The action of any algebraic integer $y$ generating the field on the lattice $\sum x_i \mathbb{Z}$ by multiplication takes each $x_j$ to a rational linear combination of the $x_i$. Let $N$ be the common denominator of the $n^2$ rational coefficients. Then $Ny$ takes each $x_j$ to an integer linear combination of the $x_i$, and thus all of $\mathbb{Z}[Ny]$ takes the lattice into itself. Of course, a larger order might still work, too.)

So if you want to conclude that a particular given $\mathbb{Z}[\alpha]$ works, you'll need to impose a condition on the lattice which explicitly depends on $\alpha$, not merely on the multiplication in $K$.

The condition $f(x_i \cdot x_j) = \delta_{ij}$ does admit nontrivial examples. I found one by looking at quadratic fields and trying $x_1=1$ and algebraic integer $x_2$; then $f(1\cdot x_2)=0$ and $f(1^2)=f(x_2^2)=1$ force $x_2^2-ax_2-1=0$ for some $a$, and $a=8$ produces (the unit ideal of) the non-maximal order $\mathbb{Z}[\sqrt{17}]$. This is not a fractional ideal of the maximal order: Multiplication by the algebraic integer $\alpha=(1+\sqrt{17})/2$ does not preserve the lattice. Had we started from $\alpha=\sqrt{17}$ instead, the same lattice and the same $f$ would have worked.

In higher degrees, a power base $(x_1,x_2,x_3,\ldots) = (1,x,x^2,\ldots)$ for the lattice never admits such an $f$ when $n\ge3$, because we'd get $0 = f(1\cdot x_3) = f(1\cdot x^2) = f(x_2\cdot x_2) = 1$, but this obstacle can be avoided by passing to another base of the same lattice. When I start from $x_1=1$ and algebraic integer $x_2$, I always seem to end up with one relation among the coefficients of the defining polynomial of $x_2$. This puts a mild constraint on $K$, but not much of an additional constraint on the lattice. And it implies nothing about the relationship between the lattice and a given $\alpha$.


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