A problem of Shimura and its relation to class field theory

Question

In Chapter II.10 of The Map of My Life, Goro Shimura mentions a certain problem:

The second topic concerns a polynomial $F(x)$ with integer coefficients. Take $$ F(x) = x^3 + x^2 - 2x - 1, $$ for example. For an integer $n$, we consider the decomposition of $F(n)$ into the product of prime numbers. We can allow $n$ to be negative, but let us assume $n$ to be positive here. Thus $$ F(1) = -1, F(2) = 7, F(3) = 29, F(4) = 71, F(5) = 139, $$ $$ F(6) = 239, F(7) = 13 \cdot 29, F(8) = 13 \cdot 43, F(9) = 7 \cdot 113, \ldots $$ The prime numbers appearing as factors of $F(n)$ form a sequence $$ 7, 13, 29, 43, 71, 113, 139, 239, \ldots $$ Now the question is: What are these prime numbers? In fact, we can prove that every such prime number $p$, excluding 7, has the property that $p+1$ or $p-1$ is divisible by 7. Conversely, every such prime number appears as a factor $F(n)$ for some positive integer $n$.

While learning class field theory on my own, I realized that the main theorem in easier cases can be formulated in terms of prime factors of $F(n)$ as above, and at that moment I was very happy. The polynomial $F$ cannot be taken arbitrarily. Actually, the equation $F(x) = 0$ has $2 \cos (2\pi/7)$ as a root, and that fact is essential. If $F(x) = x^2 - a$ with an integer $a$, the problem can be solved by the quadratic reciprocity law. In fact, my later work on the so-called complex multiplication is closely connected with this question of finding $F$ for which the sequence corresponding to [the sequence of integers above] can be determined.

My question concerns the arguments in the final excerpted paragraph. Namely, how exactly are easier cases of the Main Theorem of Class Field Theory related to this problem? Additionally, how does complex multiplication help in finding a polynomial $F$ given a sequence of primes as above?

Answer 1

Set $\alpha = 2 \cos (2 \pi/7))$. As Shimura says, $\mathbb{Q}[x]/F(x) \cong \mathbb{Q}(\alpha)$.

To say that $p$ divides $F(k)$ for some $k$ is to say that $F$ has a root in $\mathbb{F}_p$. This is basically the same as saying that $p$ splits in $\mathbb{Q}(\alpha)$. (There could be some issues regarding small primes, although I think there are not in this case.)

A special case of the main results of Class Field Theory is that $p$ factors (and, in fact splits) in $\mathbb{Q}(\alpha)$ if and only if $p \equiv \pm 1 \mod 7$. In general, Class Field Theory tells us that the factorization of $p$ in a number field $K$ is determined by congruence conditions whenever $\mathrm{Gal}(K/\mathbb{Q})$ is abelian. In this case, $\mathrm{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q}) \cong \mathbb{Z}/3$.

This is not to say that you need Class Field Theory to establish this result. Because $\mathbb{Q}(\alpha)$ is a subfield of a cyclotomic field, you can also establish all of these statements in a more elementary way using computations with roots of unity.

Answer 2

Let $K$ be a number field and $F$ a polynomial with coefficients in $K$. Class field theory shows that the following properties of $F$ are equivalent:

(1) There is subgroup $H$ of a generalized ideal class group $C_{\mathfrak m}$ (of some appropriately chosen conductor $\mathfrak m$) and a finite set of prime ideals $S$ (including all primes involved in the denominators of $F$) such that $F$ splits completely modulo $\wp$ (for $\wp \not\in S$) if and only if the class of $\wp$ in $C_{\mathfrak m}$ lies in $H$.

(2) The splitting field of $F$ over $K$ is an abelian extension.

In Shimura's example, the field $K$ is $\mathbf Q$, the conductor is $7$, the set $S$ is $\{7\}$, and the splitting field is the degree 3 extension of $\mathbf Q$ contained in ${\mathbf Q}(\zeta_7)$.

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So in general, to construct $F$ of the type considered by Shimura, one must construct abelian extensions of number fields $K$. The theory of complex multiplication is one tool that allows one to do this.

Answer 3

I just saw this and would like to add that Shimura's example is the smallest case ($a=-1$) of Shank's simplest cubic $$F(x)=x^3-ax^2-(a+3)x-1$$ with discriminat $P^2$, $P=a^2+3a+9$. When $P$ is prime say eg. $P=13$ ($a=1$) also noted above, $19$, $37$, $79$, $97$, $139$, $163$,...,then the primes dividing $F(m)$ for some $m$ are again $P$ and primes $q$ which are cubic residues mod $P$. The Galois group is again $\mathbf Z/3$ so the proof above still works -- An explicit way to see $\mathbf Z/3$ is to verify the relation $F(x)=-(1+x)^3 F(g(x))$ where $g(x)=-1/(x+1)$ is of order $3$ in $\mathrm{PSL2}(\mathbf Z)$, so the roots are $\alpha$, $g(\alpha)$, $g^2(\alpha)$ and hence must be all real.

Answer 4

I would like to write some kind of summary of the above two answers. There is nothing new here.

Consider $\mathbb{Q} \subset \mathbb{Q}(\zeta_7)$,(and we fix $\zeta_7=e^{\frac{2\pi i}{7}}$) this is an abelian Galois extension, and the Galois group is $(Z/7Z)^{\times}$. Frobenius element over $p$ for $p \neq 7$ (7 is the ramification) acts on $\zeta_7$ by sending it to its $p$-th power.

Now consider $\alpha=\zeta_7+\zeta_7^{-1}$, which is in $\mathbb{Q}(\zeta_7)$, consider all its Galois conjugates, which are precisely $\alpha_2 = \zeta^2 +\zeta^{-2}$, $\alpha_3=\zeta^3+\zeta^{-3}$. So we have $\mathbb{Q} \subset \mathbb{Q}(\alpha)$ is Galois.

Now $Frob_p$ maps $\alpha$ to $\zeta^p + \zeta^{-p}$, which is equal to $\alpha$ if and only if $p \equiv 1, -1 (mod 7)$. And they are precisely those primes that are totally split in $\mathbb{Q}(\alpha)$. For those primes, $Z/p = \mathbb{O}_{\mathbb{Q}(\alpha)}/p$, thus we can always find some $n$, s.t., $\alpha_i \equiv n(mod p)$, which is equivalent to say that $F(x)$ totally split over $Z/p =F_p$.

I am glad to know this problem and answer since I finally found an explicit example of cyclic Galois extension of $\mathbb{Q}$...(which I had been wondering for a while...)

We can also do the similar things for p=13. just take $\beta=\theta +\theta^5+\theta^8+\theta^{12}$,where $\theta$ is the 13-th root of unity, then $Q(\beta)$ is again a cyclic Galois extension of $Q$.

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when I said cyclic above, I meant for cyclic of order 3. Thank Peter for pointing out.

Answer 5

I want to add that one gets such results only for abelian extensions ("congruence conditions do not suffice"). How can one prove this? I'd start with the norm limitation theorem.