In a talk he gave at the MSRI in 2011, Manjul Bhargava sketches a proof of why the Hasse principle holds for all bivariate cubic forms. It goes as follows:
- Suppose the bivariate cubic forms $f$ has no root in $\mathbb{P}_{\mathbb{Q}}^1$.
- Then $f$ factors over a cubic field.
- By the Chebotarev density theorem, this implies that $f$ is irreducible mod $p$ for some prime $p$.
- Thus $f$ has no root in $\mathbb{P}_{\mathbb{Q}_p}^1$ for some prime $p$.
I regret to say that I am having trouble following this proof. My main difficulties lie in understanding how the Chebotarev density theorem comes into it, seeing as I thought that to be a statement about the density of primes in non-Abelian extensions.
Also, step (2): Why does the non-existence of a root in $\mathbb{P}_{\mathbb{Q}}^1$ imply that $f$ must factor over some cubic field extension?
If a textbook proof of the above statement exists, I would much appreciate it if someone could point it out.
Lastly: He also mentions that the reason the Hasse principle holds for bivariate cubic forms is "basically the same as for bivariate quadratic forms". But surely, the Hasse principle for bivariate quadratic forms simply follows from the fact that any element $a$ in an algebraic number field $K$ is a square in $K$ if and only if it is a square in every completion of $K$, which is a much simpler argument, is it not?
Many thanks.
Best Answer
Let $f(x,y) = ax^3 + bx^2y + cxy^2 + dy^3$ for $a,b,c,d \in \mathbf Q$. If $a$ or $d$ were $0$ then $f(1,0) = a$ would be $0$ or $f(0,1) = d$ would be $0$. Therefore the assumption that $f$ has no zeros in $\mathbf P^1(\mathbf Q)$ implies $a$ and $d$ are not $0$.
Now consider $$ F(x) = f(x,1) = ax^3 + bx^2 + cx + d, $$ which is cubic in $\mathbf Q[x]$ with no rational root: if $F(r) = 0$ for some $r \in \mathbf Q$ then $f(r,1) = 0$, but $f$ has no rational zero. A cubic in $\mathbf Q[x]$ with no root in $\mathbf Q$ is irreducible over $\mathbf Q$ (an accident of small-degree polynomials!), so $F(x)$ is irreducible over $\mathbf Q$. The following theorem explains why $F(x)$ has no root in some $\mathbf Q_p$.
Theorem. If $F(x) \in \mathbf Q[x]$ is irreducible over $\mathbf Q$ and has a root in $\mathbf Q_p$ for all $p$ then $F(x)$ is linear.
The same result and proof hold with $\mathbf Q$ replaced by an arbitrary number field (and the $\mathbf Q_p$'s replaced by all the nonarchimedean completions of that number field), but I stick to coefficients in $\mathbf Q$ to keep the notation simple.
Proof. Let $K = \mathbf Q(\alpha)$ where $F(\alpha) = 0$ and let $L$ be the Galois closure of $K$ over $\mathbf Q$.
The discriminant ${\rm disc}(F)$ is nonzero since $F(x)$ is irreducible over $\mathbf Q$. Let $S$ be the primes $p$ that appear in the factorization of ${\rm disc}(F)$ or that ramify in $L$ or that appear in a coefficient of $F(x)$. That is a finite set of primes. For $p \not\in S$, $F(x)$ is $p$-integral and $F(x) \bmod p$ is separable since ${\rm disc}(F) \not\equiv 0 \bmod p$.
For $p \not\in S$, $F(x)$ has a root $r$ in $\mathbf Q_p$ by hypothesis, and the root is in $\mathbf Z_p$ since $F(x)$ has $p$-integral coefficients and its leading coefficient (in fact each coefficient) is a $p$-adic unit. Therefore we can reduce the equation $F(r) = 0$ to $\overline{F}(\overline{r}) = 0$ in $\mathbf Z_p/(p) = \mathbf F_p$. We have $K \cong \mathbf Q(r)$, so some prime ideal factor of $p$ in $K$ has residue field degree $1$: the embedding of $K$ into $\mathbf Q_p$ by mapping $\alpha$ to $r$ puts a non-archimedean absolute value $|\cdot|_{\mathfrak p}$ on $K$ with residue field $\mathbf F_p$, so $f(\mathfrak p|p) = 1$.
Now we bring in Frobenius elements. Set $G = {\rm Gal}(L/\mathbf Q)$ and $H = {\rm Gal}(L/K)$. Since $p$ is unramified in $L$ (by $p \not\in S$), we can talk about a Frobenius element at a prime $\mathfrak P$ over $p$ in $L$. Since $F(x)$ is $p$-integral and $F(x) \bmod p$ is separable, the permutation action of ${\rm Frob}(\mathfrak P|p)$ on the roots of $F(x)$ has a cycle structure that matches the degrees of the irreducible factors of $F(x) \bmod p$. (The actual choice of $\mathfrak P$ over $p$ doesn't matter, since all Frobenius elements at primes over $p$ are conjugate to each other in $G$ and thus permute the roots of $F(x)$ with the same cycle structure.)
We know $F(x) \bmod p$ has a root in $\mathbf F_p$, so it has a linear irreducible factor mod $p$. Thus ${\rm Frob}(\mathfrak P|p)$ has a fixed point in its permutation action on the roots of $F(x)$: this Frobenius element fixes a root of $F(x)$ in $L$, so this Frobenius element fixes one of the subfields of $L$ conjugate to $K = \mathbf Q(\alpha)$. The subgroup of $G$ corresponding to that subfield is conjugate to $H$. Hence ${\rm Frob}(\mathfrak P|p) \in \bigcup_{\sigma \in G} \sigma H\sigma^{-1}$.
Now we bring in the Chebotarev density theorem: every element of $G$ is a Frobenius element at infinitely many primes, in fact at a positive density of primes. So each $g \in G$ is a Frobenius element at a prime ideal lying over a prime outside of $S$. The above argument proves $g \in \bigcup_{\sigma \in G} \sigma H\sigma^{-1}$. Thus $$ G = \bigcup_{\sigma \in G} \sigma H\sigma^{-1}. $$
A finite group is never the union of conjugates of a proper subgroup, so $H = G$ and thus $K = \mathbf Q$. Therefore $\mathbf Q(\alpha) = \mathbf Q$, so $F(x)$ has a root in $\mathbf Q$. Since $F(x)$ is irreducible over $\mathbf Q$ with a rational root, $F(x)$ is linear. QED
In this proof, we did not need $F(x)$ to have a root in $\mathbf Q_p$ at every prime $p$. It is okay to say $F(x)$ has a root in $\mathbf Q_p$ outside a set $S$ of primes of density $0$, since Chebotarev implies that each $g \in G$ is a Frobenius at a prime outside $S$. For example, if $F(x)$ has a root in $\mathbf Q_p$ for all but perhaps finitely many $p$ then we can still deduce that $F(x)$ has a root in $\mathbf Q$.
Therefore an irreducible $F(x) \in \mathbf Q[x]$ of degree greater than $1$ has no root in $\mathbf Q_p$ for infinitely many $p$. When $F(x)$ is cubic, not having a root in a field is equivalent to being irreducible over that field. Therefore a cubic irreducible in $\mathbf Q[x]$ has no root in $\mathbf Q_p$ for infinitely many $p$. Returning to the binary cubic form, $f(x,1)$ has no root in $\mathbf Q_p$ of infinitely many $p$, so for infinitely many $p$ we can't solve $f(x,y) = 0$ with $[x,y] \in \mathbf P^1(\mathbf Q_p)$ with $y \not= 0$. Of course we can't solve it with $y = 0$ either, since that would be the point $[x,0] = [1,0]$ and $f(1,0) = a \not= 0$.