# Hasse’s principle

The most important and the oldest equations in number theory are perhaps the Diophatine equations. In elementary mathematics,  one of the techniques to solve the Diophatine equations is to module some prime number $p$, and to see if the reduced equation has a solution or not. As a strategy, this techique is used most often to show that the original Diophainte equation does not have any integer or rational solutions. In other words, the other way around does not work: even if the reduced equation module every prime number $p$ has some solutions, we can not deduce that the original Diophatine equation has a solution.

In some sense, Hasse’s principle says the same thing as above: suppose that $f\in \mathbb{Z}[X,Y,Z,...]$ is an irreducible polynomial with integer coefficients, it is clear that if $f=0$ has an integer solution, then so does $f=0(mod p)$. And since $f$ is irreducible, we can say, under some further mild conditions, that $f=0$ will also have some solutions in $\mathbb{Q}_v$ where $v$ runs over all the places of the number field $\mathbb{Q}$(here thus $v$ is just $2,3,5,7,...,\infty$)(this lifting solution property is also very important in number theory, it is called the Hensel’s property). Yet coversely, we can not garantie that it is true. If it is true, then we say that this polynomial satisfies the Hasse’s principle.

Using Hilbert’s symbol over the rationals, we can show that, all the homogenuous polynomials of degree $2$ of any number of variables satisfy Hasse’s principle.

Yet, things become complicated when the degree of the polynoial gets larger than $2$.

Another way of fomulating this principle is to use the adelic language. We write $\mathbb{A}_{\mathbb{Q}}=\mathbb{A}$ for the adelic ring associated to $\mathbb{Q}$, that is the restricted product $\mathbb{A}=\prod'_v\mathbb{Q}_v$ where $v$ runs over all the places of $\mathbb{Q}$. Then using the embedding of $\mathbb{Q}$ into $\mathbb{A}$, we can eextend $f$ to $\mathbb{A}$, and $f$ has some solutions in $\mathbb{A}$ if and only if $f$ has some solutions in $\mathbb{Q}_v$ for all $v$. So, in this situation, $f$ satisfies Hasse’s principle if and only if we have an implication, $f$ has solutions in $\mathbb{A}$ implies that it has solutions in $\mathbb{Q}$.

Yet even for polynomials of degree $3$, this principle can not hold. One classic example is due to Reichardt and Lind. The polynomial $2Y^2=X^4-17Z^4$ does not satisfy Hasse’s principle.

To show that this polynomial admits solutions module every prime number $p$(the case that $v=\infty$ is easily seen to be true), we can simply put $Z=1$, and try to find solutions to the polynomial $2Y^2=X^4-17$. One way to show that it indeed has some solutions is to use the character theory and the inequality $|J(\xi,\lambda)|\leq \sqrt{q}$. Using these methods, we can show the existence of solutions for $p>9$, including the case $p=17$. And using the Hensel’s property of the local fields $\mathbb{Q}_p$, we can show that the solutions in $\mathbb{F}_p^2$ can be lifted to $\mathbb{Q}_p$, since both $2Y^2,X^4-17$ have simple roots for all $p>2$, which is particularly true for the cases $p>9$. So it remains for the cases $p=2,3,5,7$. For the case $p=7$, we see that $(X,Y,Z)=(3,2,1)$ is a solution, and as above, can be lifted to $\mathbb{Q}_7$. For the case $p=5$, we see that $(X,Y,Z)=(0,2,1)$ is a solution, and can be lifted to $\mathbb{Q}_5$. For the case $p=3$, one solution module $p=3$ is $(X,Y,Z)=(1,1,1)$, and can also be lifted to $\mathbb{Q}_3$. For the case $p=2$, this is a little tricky, yet we can still show that the root module $8$,$(X,Y,Z)=(3,0,1)$ can be lifted to $\mathbb{Q}_2$. So, all in all, we have that the polynomial $2Y^2=X^4-17Z^4$ has solutions for each place.

Yet, it can not have rational solutions. Si not, we can assume that une solution $(x,y,z)$ are three integers and $gcd(x,z)=1$. Then, if a prime $p\neq2$ divides $y$, then $17$ will be a square in $\mathbb{F}_p$, so according to the quadratic recirpocity’s law, $p$ will also be a sqaure module $17$. Moreover, $2$ is also a square module $17$($6^2=2(mod 17)$), so all the possible prime factors of $y$ are squares module $17$, this shows that $y$ is itself also a square module $17$. Suppose that $y=y'^2(mod 17)$, then from $2y^2=x^4-17z^4$, we deduce that module $17$, there is $2y'^4=x^4(mod 17)$. This means that $2$ is a fourth power in $\mathbb{F}_{17}$. Yet direct verification shows that this is not true, thus a contradiction.

Remark: in some litteratures, this polynomial is said directly to be an elliptic curve module $p\neq2,17$. This can be shown as follows: first we write $X^4-17=(X-a)(X-b)(X-c)(X-d)$ in $\bar{\mathbb{F}_p}$(note that for $p\neq2,17$, these four roots are distincts), and then use the transformation $X'=X/(X-a)$, we get that $2Y^2(X'-1)^4=d'(X'-a')(X'-b')(X'-c')$ where $a',b',c',d'$ are constants in function of $a,b,c,d$, so another transformation, $Y'=Y(X'-1)^2$ gives $2Y'^2=d'(X'-a')(X'-b')(X'-c')$. So, in this way, we get a polynomial of degree $3$, which determines an elliptic curve if it is irreducible(this can be easily verified). It is in this sense that we say that the polynomial $2Y^2=X^4-17Z^4$ is an elliptic curve for most of the $p$.

Note that, in the above arguement, one point is essential: we use the law of quadratic reciprocity. The elementary technique of module some prime number mentioned at the beginning of the post uses essentially the Chinese remainder lemma(中国剩余定理). This reflects one of the essential differences between the Chinese remainder lemma and the law of quadratic reciprocity: the Chinese remainder lemma is a local result. This means that, a system of equations, $x=a_i(mod p_i)(i=1,2,...,n)$ is always solvable for whichever $a_i,p_i$. We can change any parameter in this system of equations, and the resulting system of equations is still solvable. This shows that the Chinese remainder lemma is local. Yet for the law of quadratic reciprocity, this is not the case. For a system of equations, $x^2=a_i(mod p_i)(i=1,...,n)$, this is not always solvable. A mere change in power makes this system rather difficult to resolve. Why it is hard to resolve? There is a problem of compatibility or coherence in the system of equations. Consider the simplest case: $x^2=p(mod q),x^2=q(mod p)$. This system admits some solution if and only if $(\frac{p}{q})=1=(\frac{q}{p})$. Yet, here the law of quadratic reciprocity says that, this is not always the case, especially when $(p-1)/2\times (q-1)/2$ is an odd number. So, in this new case, we can not change arbitrarily the parameters in the system of equatios.

We have mentioned in some previous post that the Hilbert’s product formula is equivalent to the law of quadratic reciprocity. Thus, according to the above argument, Hilbert’s product formula is a global result.

There are of course other examples that do not satisfy Hasse’s principle, for example, the polynomial due to Selmer(the one in the name ‘Selmer’s group’), $3X^3+4Y^3+5Z^3=0$.

There are various results concerning for which polynomials Hasse’s principle can hold. For example, the Manin obstruction is something that measures the failure to satisfying Hasse’s principle. It says that if the Manin obstruction of a polynomial is trivial, then this polynomial satisfies Hasse’s principle(still, we can not say the converse). In the case where the variety determined by the polynomial is an abelian variety, the Manin obstruction is the same as Tate-Shafarevich group. In this case, we can say that, the Tate-Shafarevich group of a polynomial is trivial ifand only if the polynomial satisfies Hasse’s principle. This will be another subjet.

Advertisements