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Let $k$ be an arbitrary field, $d\in k$, and $d$ is not a square in $k$. Consider the binary quadratic form $f(x,y)=x^2-d y^2$ (it is the norm from $k(\sqrt{d})$ to $k$). I am looking for a reference to a proof of the following fact:

There exist a field extension $K/k$ and a non-zero element $a\in K$ such that $f$ does not represent $a$ over $K$ (that is, there is no $x,y\in K$ such that $a=x^2-d y^2$).

Edited question (which I really meant): Let $l/k$ be a separable quadratic field extension (of fields of any characteristic). Prove that there exists a field extension $K/k$ such that the norm map $N\colon (l\otimes_k K)^\times\to K^\times$ is not surjective.

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Shouldn't be $k$ a number field? –  Wadim Zudilin Jun 6 '10 at 13:25
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I assume the characteristic of $k$ is not $2$. Take $K = k((t))$ and apply Proposition V.2.3 of Serre's Local Fields to the unramified quadratic extension $K(\sqrt{d})/K$. In particular, the element $t$ is not represented. –  Pete L. Clark Jun 6 '10 at 14:07
    
@Pete: Many thanks! I edited my question. Will the same proof be valid for the edited question also in char. 2? If you post your comment as an answer, I will be happy to accept it! –  Mikhail Borovoi Jun 6 '10 at 15:14
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up vote 3 down vote accepted

Take $K = k((t))$ (formal Laurent series field) and apply Proposition V.2.3 of Serre's Local Fields to the unramified quadratic extension $Kl/K$. (Note that separability of $l/k$ is needed so that $Kl/K$ is unramified.) Then the image of the norm map consists precisely of elements of even $t$-adic valuation: in particular $t$ is not in the image. Having done the local case, it becomes clear that $K = k(t)$ would also work.

[Note that I am leaving the answer as community wiki. This is because I am supposed to be "on vacation" from MO: i.e., concentrating on my own work! I made an exception here because this is something that I knew off the top of my head and because Mikhail is such an eminent mathematician that it is my privilege to help him out, in however small a way. But I better not get back into the reputation game...]

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Many thanks indeed! –  Mikhail Borovoi Jun 6 '10 at 18:36
    
Uh oh. The vacation police are gunna get you. –  Steven Gubkin Jun 9 '10 at 18:26
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