How should I construct $\color{Green}{\text{explicitly}}$ the associated torus to the general quadratic field $ \mathbb{Q} \sqrt d $?

Question

In the introduction of the article An elliptic curve test for Mersenne primes by Benedict H. Gross, it is only mentioned that:

$T$ is the one-dimensional algebraic torus $T$ over $\mathbb{Q}$, associated to the real quadratic field $ k= \mathbb{Q} \sqrt3 $ .

1. First question: Could you explain $T$ for me in more details or could you construct it for me $\color{Red}{\text{explicitly}}$?

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If $l \geq 3 $ is a prime, and $p = 2^l − 1$ is the corresponding Mersenne number, then $$ p \overset {24} {\equiv} 7 .$$ $$ %% p {≡} 7 (mod 24).$$

Let $ A = \mathbb{Z} + \mathbb{Z} \sqrt3 $ be the ring of integers, of discriminant $12$, inside the real quadratic field $k = \mathbb{Q} (\sqrt3)$. Let $\sigma$ be the non-trivial automorphism of $k$, for which $\sigma(\sqrt3)=-\sqrt3$. The ring $A$ has class number $1$ and fundamental unit $$\varepsilon=2+\sqrt3 \ .$$ The unit $\varepsilon$ is totally positive and satisfies $\varepsilon . \varepsilon^{\sigma}=1$. It ($\varepsilon$) provides an integral point on the algebraic torus $T$ mentioned in the introduction.

Let $q$ be a prime number, and let $T(q)$ be the subgroup of $(A/q)^*$ consisting of elements of norm $1$ to $(Z/q)^*$. By reduction (mod $q$), we may consider $\varepsilon$ as an element of the finite group $T(q)$.

2. Second question: Why the norm map from $(A/q)^*$ to $(Z/q)^*$ is surjective?

3. Third question: What does it mean by: It ($\varepsilon$) provides an integral point on the algebraic torus $T$ mentioned in the introduction.

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4. Fourth question: This question is the natural generalization of the First question. How should I construct $\color{Blue}{\text{explicitly}}$ the associated torus to the general qudratic field $ \mathbb{Q} \sqrt d $?

Answer 1

Let $T$ be the algebraic curve defined by $x^2 - dy^2 = 1$ over ${\mathbb Q}$, where $d$ is a nonsquare. Over the quadratic extension $K = {\mathbb Q}(\sqrt{d})$, $T$ becomes isomorphic to $K^\times$, since $$ 1 = x^2 - dy^2 = (x - y\sqrt{d}) \cdot (x + y\sqrt{d}). $$ In fact, given any $t \in K^\times$ we have $\frac1t \in K^\times$ and $\frac1t \cdot t = 1$. Set $x = \frac12(t + \frac1t)$ and $y = \frac1{2\sqrt{d}}(t - \frac1t)$. Then the map $t \mapsto (x,y)$ is an isomorphism of abelian groups. This shows that, by definition, $T$ is a $1$-dimensional torus over the rationals since over $K$ we have $T \simeq K^\times$.

The point $(2,1)$ is an element of the torus for $d = 3$ since $2^2 - 3 \cdot 1^2 = 1$, and under the isomorphism above this element becomes $2 + \sqrt{3}$ when viewed over ${\mathbb Q}(\sqrt{3})$.

The very same argument holds over any field in which $d$ is not a square, in particular over finite fields with $p$ elements with $(\frac dp) = -1$. If you then set $q = p^2$, the norm map $N: {\mathbb F}_q^\times \longrightarrow {\mathbb F}_p^\times$ is surjective. In fact, the generating automorphism is the Frobenius maps, which raises everything to the $p$-th power, so the norm in your quadratic extension is $N(x) = x^{p+1}$. Its kernel, the elements of norm $1$, are the roots of the polynomial $x^{p+1} - 1$, and since we are working in a field, there are at most $p+1$ elements in the kernel. Thus the image of the norm has at least $p-1$ elements, and since ${\mathbb F}_p$ does have $p-1$ elements, we have equality.

If you replace the field with $p^2$ elements by a field with $p^3$ elements, then the elements of norm $1$ will give you a $2$-dimensional torus over ${\mathbb F}_p$.

Answer 2

In your paper p.1-6 formulates the Lucas-Lehmer test in the words of Franz's answer. The elliptic curve part starts p.7. In short :

• Look at the elliptic curve $E : y^2=x^3-12x, (x,y) \in \mathbb{Q}$. With its group law define $[n](x,y)$, and also $[i](x,y)= (-x,iy)$ so $E$ has complex multiplication by $\mathbb{Z}[i]$. Also (with the Schoof algorithm ?) one can show $E$ is generated by two points $P= (-2,4)$ of infinite order and $Q = (0,0)$ of order $2$.

• Let $p = 2^l-1$, assume it is prime, and set $E(p) : y^2 = x^3-12x, (x,y) \in \mathbb{F}_p$, still having CM by $\mathbb{Z}[i]$. The CM theory says $E(p) \simeq \mathbb{Z}[i]/(F-1)$ where $F = a+ib$ is equal to the Frobenius $(x,y) \mapsto (x^p,y^p)$, and $F^2 =-p$.

• Thus $-p = (F-1)^2-1+2(F-1)$ and $F-1$ is coprime with every prime $\ne 2$ and $E(p)$ is cyclic with $2^l$ elements with $P$ as generator.

• If $p$ is not prime and $ord(P) = 2^l$, then take $q$ the least prime factor of $2^l-1$, we still have $ord(P) = 2^l$ in $E(q)$. But (by the Hasse bound) this group has at most $q+1+2\sqrt{q} < 2^l$ points, a contradiction.

I'm not sure why you need to study it if you don't know the (CM) theory of elliptic curves, you should better start with Silverman's, the arithmetic of elliptic curves.