Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $K$ be a number field, with ring of integers $O_K$, and let $\alpha\in O_K$ be a primitive element for the extension $K/Q$, with minimal monic polynomial $f(x)\in Z[x]$. If $p$ is a prime number, then the factorisation of the principal ideal $(p)$ in the Dedekind ring $O_K$ can be described in terms of the reduction mod $p$ of $f(x)$ as long as $p$ does not divide the (finite) index $[O_K:Z[\alpha]]$. This is well known and rather elementary, as one can say.

On the other hand, I have never seen in the literature a more general statement relating the factorisation of $(p)$ with the reduction mod $p$ of $f(x)$ without the assumption that $p$ did not divide $[O_K:Z[\alpha]]$. In this more general setting, the sole mod $p$ reduction of $f(x)$ does not give enough information about the splitting of $(p)$, one does need some extra "characteristic zero" information. Would the $p$-adic valuation of the discriminant of $f(x)$ be enough? At least if one knows that $p$ is tamely ramified in $K$. Have you ever seen this discussed or have you ever thought about it? Many thanks.

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

The discriminant of the polynomial is equal to the discriminant of the field times the square of the index $[\mathcal{O}_K: \mathbf{Z}[\alpha]]$. Hence, if the $p$-adic valuation of the discriminant of $f(x)$ is $0$ or $1$, then one already has an integral basis $p$-adically. Yet already the examples $x^2 - p^2$ and $x^2 - a p^2$ where $a$ is a quadratic non-residue modulo $p$ (and $p$ is odd) show that the situation is hopeless in general, even if one knows that $p$ is unramified.

Since, I imagine, you are trying to determine the behavior of Hecke algebras from computing characteristic polynomials of small Hecke operators, you might be able to get mileage out of the following observation, which says that "small" discriminants must come from ramification rather than index. Suppose, for example, that $f(x)$ modulo $p$ is exactly divisible by $a(x)^2$ for some irreducible polynomial $a(x)$ over $\mathbf{F}_p$. By Hensel's lemma, $f(x)$ is divisible by a lift $A(x)$ of $a(x)^2$. There are the following possibilities:

  1. $A(x)$ is irreducible, and corresponds to an unramified prime of residue degree $2d$.
  2. $A(x)$ has two irreducible factors, corresponding to a pair of unramified primes of residue degree $d$.
  3. $A(x)$ is irreducible, and corresponds to a ramified prime of residue degree $d$ and ramification index $2$.

In the first two cases, the index of $\mathbf{Z}[\alpha]$ in $\mathcal{O}_K$ is divisible by $p^d$, and so the discriminant is divisible by $p^{2d}$. Hence, if you know that the valuation of the discriminant is less than $2d$, you can deduce that $A(x)$ corresponds to a ramified prime.

share|improve this answer
    
Thanks for your answer. I do not fully understand what you mean with your example in the third line. I guess I was looking at irreducible polynomials but probably that's not the point. Let me be slightly more optimist and let me point out that there is the following criterion (which probably I should have put in the question itself to begin with). Let f(x) be a monic, square free polynomial polynomial of degree n and with integer coefficients, and let p be a fixed prime. Let F be the number of ring homomorphisms from Z[x]/f(x) to $\bar F_p$, and let V be the p-adic valuation of the –  Tommaso Centeleghe Jan 15 '11 at 1:58
    
of the discriminant of f(x). THEN, if F = n - V we have that p does not divide the index of Z[x]/f(x) in its integral closure in Q[x]/f(x). Moreover in this case p is tamely ramified in each component of Q[x]/f(x). (A possible proof: arxiv.org/abs/1008.2059 Prop 5.6). I was then wondering if one can say something when the equality F = n - V does not hold (> always holds). And yes, of course I wanted this for computing congruences between modular forms!! ;-). –  Tommaso Centeleghe Jan 15 '11 at 2:04
    
I did not understand what you meant when you said in the third line: "Yet already the examples x2−p2 and x2−ap2 where a is a quadratic non-residue modulo p (and p is odd) show that the situation is hopeless in general". As to your second point, I am interested in getting as much information as possible on the p-adic properties of the number field K (coefficient field of a modular form of level one and weight k say) starting from the char poly of T_2 by looking at its reduction mod p and its discriminant. –  Tommaso Centeleghe Jan 15 '11 at 2:36
    
Also, a funny thing that happens is that for p<2600 and k<p+2, there is no congruence between two distinct classical mod forms of weight k and level one that can be justified by the fact that p divides the index of the Hecke algebra in its integral closure. In other words every congruence in the above range comes from ramification. (I posted another question on MO on this at some point but got no answer mathoverflow.net/questions/21323/…). May be there is a reason for this. –  Tommaso Centeleghe Jan 15 '11 at 2:39
    
Dear Esperantis, thanks for the clarification. I now understand what you meant with your example (sorry it's 6 am here..:-)), that clearly shows the impossibility of determining the spitting of p in terms of the reduction mod p of f(x) and of its discriminant (I guess technically I was only looking at irreducible polynomials, therefore we may consider replace $x^2-p^2$ by $x^2-bp^2$ where b is not a square but it is a quadratic residue mod p). This helps, I should have probably given another thought to my question before formulating it. –  Tommaso Centeleghe Jan 15 '11 at 4:44
show 6 more comments

Your Answer

 
discard

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.