11
$\begingroup$

Say that $L/K$ is a quadratic extension of number fields with $K$ totally real and $L$ totally imaginary.

Then the Hasse norm theorem says that an element of $K$ that is everywhere a local norm is the global norm of something in $L$. (In fact, this is more generally true, whenever $L/K$ is cyclic.)

Is it also the case that an element of $\mathcal{O}_K^*$ that is everywhere a local norm is the global norm of something in $\mathcal{O}_L^*$?

$\endgroup$
1
  • $\begingroup$ In the question, do you mean a local norm of a unit, or just a local norm ? $\endgroup$ May 31, 2011 at 12:29

1 Answer 1

17
$\begingroup$

No: there are units that are norms of elements but not norms of units. The simplest example are real quadratic number fields $Q(\sqrt{m}\,)$ with $m$ a sum of two squares such that the fundamental unit has positive norm, for example $m = 34$.

For finding other examples, one may look at the ambiguous class number formulas for cyclic extensions $L/K$ of prime degree $p$: $$ Am(L/K) = h_K \frac{\prod e(P)}{p(E_K : E_K \cap NL^\times)}, \qquad Am_s(L/K) = h_K \frac{\prod e(P)}{p(E_K : N E_L)}. $$ Here $Am$ denotes the subgroup of ideal classes fixed by the Galois group $G$, $Am_s$ the subgroup of classes generated by ideals fixed by $G$, and $e(P)$ is the ramification index of the prime $P$. The index $$ (Am:Am_s) = (E_K \cap NL^\times : NE_L) $$ is the obstruction to the local-global principle for units.

Edit. Let me, however, point out that D. Folk (When are global units norms of units?, Acta Arith. 76 (1996), 145-147) has proved the following result: if $L/K$ is normal and if $H$ denotes the Hilbert class field of $L$, then a unit from $K$ that is a local norm in all completions of $H/K$ is the norm of a unit from $L$. This suggests the following question: given a cyclic extension $L/K$, is there an unramified abelian extension $E/L$ with the property that a unit from $K$ is the norm of a unit from $L$ if and only if it is a local norm in all completions of $E/K$?

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.