Squares with prescribed trace in p-adic field extension

Question

Let K be a finite extension of some p-adic field $\mathbb{Q}_p$, and $L|K$ the unramified cubic extension. I have a statement about a certain $K$-algebra being split by this extension which I can prove by general formalism of Brauer groups. Checking the down-to-earth equations I noticed that it would also follow from the following:

For every $y \in L$, there is a square $x^2\in L$ such that $Tr(x^2y) = 0$,

where Tr is the trace of $L|K$. So I wondered if this is true. In other words, does the (K-)two-dimensional kernel of $Tr( \cdot y)$ contain squares?

Note if $y$ is a square itself, we are just asking for a square in $L$ with trace 0. However plausible this seems to me, I can only see one if the extension is a Kummer extension (i.e. $K$ contains third unit roots), in which case there is a generator $x$ of $L|K$ with minimal polynomial $X^3-a$, $a\in K$, and $x^2$ will do as it has minimal polynomial $X^3-a^2$. Since my extension is unramified, some expression of unit roots might do in general, but I don't see it.

Answer 1

When $p \neq 2$, we can prove every $K$-linear map $L \to K$ has a nonzero square in its kernel. We can, in fact, prove the stronger fact that every one-dimensional $K$-subspace of $L$ contains a square, by using

Theorem: If $p \neq 2$ and $\alpha \in L^\times$, then there exists $\beta \in K^\times$ such that $\alpha \beta$ is square.

The abelian group $L^\times$ can be factored into three components

$$ L^\times \cong \mathbb{Z} \times \mu^{(p)}(L) \times U_L^{(1)} $$

where $\mathbb{Z}$ records the valuation, $\mu^{(p)}(L)$ is the group of roots of unity of order relatively prime to $p$, and $U_L^{(1)}$ is the multiplicative group of elements of $L$ that reduce to $1$ in the residue field. We can construct $\beta$ by adjusting each of the three components in turn:

• Since $L$ is unramified, any uniformizer of $K$ has odd valuation. (actually, since $[L:K]$ is odd, I think that would remain true even if $L$ is ramified) Thus, we can make $\alpha \beta$ have even valuation.

• Let $q$ be the order of the residue field of $K$. Since the index $[\mu^{(p)}(L) : \mu^{(p)}(K)] = \frac{q^3-1}{q-1} = q^2 + q + 1$ is odd, all nonsquares in $\mu(K)$ are also nonsquare in $\mu(L)$. Since the product of two nonsquares is square and units have zero valuation, we can additionally ensure the component of $\alpha \beta$ in $\mu^{(p)}(L)$ is square.

• When $p \neq 2$, then every element of $U^{(1)}$ is already a square. We can see this, for example, by Hensel's lemma. Thus, we need do nothing further.

Answer 2

Disclaimer: I don't actually understand, much less have read, most of the contents of the papers I've mentioned below, just enough to understand they contain answers to the questions posed above.

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The questions posed above are interesting. I think I know the answer but I don't think I can give an elementary proof.

The short answer is as follows. Counterexamples exist when $p = 2$, and you should be able to write one down explicitly using the explicitly determination of the Hilbert symbol when $K = \mathbb{Q}_2$.

The long answer is as follows. Here is a more sophisticated way to think about your question. For now, let $K$ be any field of characteristic $0$ and let $f(X) = X^3 + a X + b$ be an irreducible polynomial over $K$. Let $L = K[X] / (f(X))$ denote the cubic extension described by $f(X)$.

If $z$ is any element in $L$, I can consider the quadratic form on $L$ given by $x \mapsto \text{Tr}_{L/K} ( x^2 z / f'(X))$. This is the form you wrote down if you make the substitution $y = z / f'(X)$.

This form plays a role in the arithmetic of the elliptic curve given by the equation $E : Y^2 = f(X)$. More precisely, there is an injective map$$E(K) / 2 E(K) \to \left( L^\times / (L^\times)^2 \right)_{\ker N},$$where $\ker N$ denotes the set of elements in $L^\times$ which have norm in $K^\times$ equal to a square.

You can define an interesting subset of $( L^\times / (L^\times)^2 )_{\ker N}$ as the set of elements $z$ in $L$ for which the quadratic form $\text{Tr}_{L/K} ( x^2 z / f'(X))$ on $L$ represents $0$; i.e. for which there exists $x$ in $L^\times$ with $\text{Tr}_{L/K} ( x^2 z / f'(X)) = 0$. This is exactly your question, but we are now restricting to those $y$ of the form $z / f'(X)$, where $z$ lies in $\ker N$. Call this subset $S$.

When you are studying the arithmetic of the elliptic curve $E$ this way, one of the first things you prove is that $E(K) / 2 E(K)$ lies in the subset $S$. You are asking whether the set $S$ equals the whole of $( L^\times / (L^\times)^2 )_{\ker N}$. (In fact you are asking something stronger than this, because you consider the whole of the group $L^\times / (L^\times)^2$.)

So we want to know if the set $S$ equals the whole of $( L^\times / (L^\times)^2 )_{\ker N}$, or not. The answer, which involves some Galois cohomology (so at the level of Serre's book on local fields, say), is described in section 6.2 of the paper available here:

http://www.math.harvard.edu/~gross/preprints/stable23.pdf.

In fact, this paper studies hyperelliptic curves and not just elliptic curves; this means one considers extensions $L / K$ of any odd degree (not just cubic extensions).

The upshot is that the set $S$ equals $( L^\times / (L^\times)^2 )_{\ker N}$ if we are working in odd residue characteristic (and then both sets have size $1$). In even residue characteristic, and if $K / \mathbb{Q}_2$ is unramified, the set $( L^\times / (L^\times)^2 )_{\ker N}$ has size $4$ and the set $S$ has size $1$. So the answer to your question is no in this case.

It isn't necessary to introduce elliptic curves to carry out this argument, but one does need Galois cohomology, Tate duality, etc., i.e. basically all the hard parts of local class field theory. The key point that makes the whole thing work is the sentence in the paper linked above:

One can show that the resulting map$$η : \text{H}^1(k, D) \to \text{H}^1(k, \text{SO}(W)) \cong \mathbb{Z}/2\mathbb{Z}$$is an even quadratic form whose associated bilinear form is the cup product on $\text{H}^1(k, J[2])$ induced from the Weil pairing.

Another reference is the following paper of Serre, which studies related (although different) questions:

Serre, Jean-Pierre, L'invariant de Witt de la forme $\text{Tr}(x^2)$, Comment. Math. Helv. 59 (1984), no. 4, 651–676.