MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Nancy Childress - "Class Field Theory", is a good book for class field theory if you are interested (but possibly you have already gone through this book or something equivalent I am guessing), specifically Chapter $6$ deals with the explicit statements of the main theorems of Class Field Theory. – Vinoth Dec 31 '09 at 9:29
This is a great question. Can someone please post a non-elementary example of a Shimura variety of some CM field and a related F that splits at primes with certain modular congruences? I have never seen an example that isn't a quadratic field of class number 1, or a cyclotomic field. – Dror Speiser Dec 31 '09 at 19:24

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.

share|cite|improve this answer
One should exclude the prime 7 ( = F(2) ), as Shimura notes. – Emerton Dec 31 '09 at 3:53
Please correct it to Q[x]/F (with square brackets, not Q(x)/F, with round brackets). – Chandan Singh Dalawat Dec 31 '09 at 4:29
Thanks for the correction regarding the brackets. – David Speyer Dec 31 '09 at 13:27
I didn't add anything about 7, because I thought it would take a little too long to say. The prime 7 ramifies completely in Q(alpha): it is P^3 for a prime ideal P. The polynomial F is (x-2)^3 mod 7. The latter fact means that 7 does divide F(k) for some k, although one doesn't use the term "split" to refer to a totally ramified prime. – David Speyer Dec 31 '09 at 13:33
For the curious, the issue which sometimes occurs for small primes, but doesn't here, is the following: Consider the polynomial x^2+9. This is sometimes divisible by 3, even though the prime 3 does not split or ramify in Q(\sqrt{-9}). (And, modulo 3, this polynomial factors as x^2.) This is because the powers of \sqrt{-9} are not an integer basis for the ring of integers in Q(\sqrt{-9}). That is the behavior that I don't think happens in Shimura's example; the factorization of F mod p always reflects the factorization of p in Q(alpha) . – David Speyer Dec 31 '09 at 13:36

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 ${\mathbf Q}(\zeta_7)$.

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.

share|cite|improve this answer
is it like to find the polynomial equation of the j-invariants? – natura Jan 4 '10 at 3:59
Yes. If $K$ is a quadratic imaginary extension of $\mathbf Q$, then the polynomial over $K$ whose roots are the $j$-invariants of those elliptic curves with CM by the ring of integers of $K$ has splitting field equal to the Hilbert class field (i.e. maximal everywhere unramified abelian extension) of $K$, and so gives one example of such an $F$. The theory of Shimura and Taniyama extends this kind of result to the case of general CM abelian varieties, rather than just CM elliptic curves. – Emerton Jan 4 '10 at 4:06

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 $Q$.

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

share|cite|improve this answer
"I finally found an explicit example of cyclic Galois extension of ...(which I had been wondering for a while...)" For any prime $p$, $\mathbb{Q}(e^{\frac{2 \pi i}{p}})/\mathbb{Q}$ is a cyclic extension of order $p-1$. – Pete L. Clark Jan 4 '10 at 4:18
ah, my typo... it should be cyclic of order 3...:) thank you. – natura Jan 4 '10 at 5:05

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.

share|cite|improve this answer

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.

share|cite|improve this answer
One approach is via class field theory plus Chebotarev. If you ask this as a question, I can give a more detailed answer. (Comments don't really allow enough space for a proper answer.) – Emerton Jan 13 '10 at 19:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.