Take the 2-minute tour ×

When I am testing conjectures I have about number fields, I usually want to control the ramification, especially minimize to a single prime with tame ramification. Hence, I usually look for fields of prime discriminant (sometimes positive, sometimes negative).

I get the feeling that I cannot be the only one who does this...

And so, are there families of number fields of prime discriminant for each degree? Or at least degree 3 and 4? (They are the coolest. Except quadratics. Of course.) What about: given a prime - can I find a polynomial of degree d with the prime as its discriminant?

share|improve this question
hum...but it seems we only have good formulas of discriminant for good (simple) field extensions. If we want to good (simple) discriminants, then the field extension itself would be quite complicated...no? –  natura Jan 4 '10 at 21:59
I might be missing something, but doesn't Stickleberger's theorem imply that your final question is false for all primes of the form 4k+3? –  Ben Linowitz Jan 4 '10 at 22:10
@basic: No. For instance $\mathbb{Q}(\zeta_p)/\mathbb{Q}$ (here $\zeta_p$ is a primitive $p$th root of unity) is a very simple extension for which a very precise formula for the discriminant is available. In particular, it is only ramified at $p$ and is tamely ramified there. It doesn't quite answer Dror's question, because there is at most one such extension of any given degree. –  Pete L. Clark Jan 4 '10 at 22:11
@Ben Linowitz: sure, as written. Maybe he means $\pm$... –  Pete L. Clark Jan 4 '10 at 22:11
@Pete, Yes that's what I mean. Good simple extensions like cyclotomic extensions have good formula of Discriminants, but if we want to have a discriminant as simple as a prime number, then the extension itself probably won't be very simple. –  natura Jan 4 '10 at 22:22

5 Answers 5

Klueners Malle online might be just the thing you're looking for. Make your own lists! And here's some they made themselves, if you run out of ideas.

share|improve this answer
so e.g. reh.math.uni-duesseldorf.de/cgi-klueners/… gives you a list of number fields of degree 5 with prime power discriminant, and many of the later entries have prime discriminant. –  Kevin Buzzard Jan 4 '10 at 22:26

A recent paper of Bhargava and Ghate discusses the enumeration of quartic fields of prime discriminant (see section 7).

share|improve this answer
I was not familiar with Bhargava or his thesis work, but I must say, as I am reading his series of papers on composition laws, that it is awesome! So glad I posted this question... –  Dror Speiser Jan 5 '10 at 20:04

As for your second question: if $f$ is an irreducible polynomial woth degree $n$ and prime discriminant (actually, squarefree discriminant is sufficient), then the roots of $f$ generate an extension with Galois group $S_n$ (thus these examples are not necessarily the best candidates for testing conjectures since you miss out on all the more interesting Galois groups). Since $S_n$ is not solvable for $n \ge 5$, class field theory is your friend only for $n = 2, 3, 4$.

  • For $n = 2$, the situation is trivial.
  • For $n = 3$, there is a number field with prime discriminant $p$ if and only if the
    quadratic field with discriminant $p$ has class number divisible by $3$.
  • For $n = 4$, there is an $S_4$-extension with prime discriminant if and only if there is a quadratic number field with discriminant $p$ and class number divisible by $3$ such that one of its unramified cubic extensions has class number divisible by $2$ (and thus necessarily by $4$).

There's a very nice article by Shanks (A survey of quadratic, cubic and quartic algebraic number fields (from a computational point of view), Proc. 7th southeast. Conf. Comb., Graph Theory, Comput.; Baton Rouge 1976, 15-40 (1976)) where you will find more.

Scholz, during the 1930s, showed how to construct (using class field theory, which means you will not get generators, just the existence) of Galois groups with small solvable groups; in his construction, the number of ramified primes can be controlled.

share|improve this answer
Then, who is my friend at $n>=5$ and $G=S_n$? Also, for 3 or 4, is there a fast computational method to find a class of order 3 in the quadratic field? Would be interesting only if it didn't involve computing the whole class group. –  Dror Speiser Feb 16 '10 at 20:49
"if $f$ is an irreducible polynomial woth degree $n$ and prime discriminant (actually, squarefree discriminant is sufficient), then the roots of $f$ generate an extension with Galois group $S_n$" How does one prove this? –  Timo Keller Apr 14 '10 at 18:37
This goes back to Arnold Scholz (1934); a recent reference is Nakagawa, "On the Galois group of a number field with square free discriminant", Comm. Math. Univ. St. Pauli 37 (1988), 95-99 –  Franz Lemmermeyer Apr 26 '10 at 12:58
@Franz: Just reading the review on Nakagawa's paper on Mathscinet, I see he did the case when the discriminant of the field is a prime. However it is not clear to me that when the polynomial has prime discriminant, its splitting field would have prime discriminant. (while I am waiting for my interlibrary requested article) Would you be a little more precise as did Nakagawa/Scholz do that? –  Ying Zhang May 2 '10 at 18:01

John Jones' Tables http://hobbes.la.asu.edu/NFDB/ are my favorite. I think his data tables are the most complete online. Check for example in Klueners-Male tables for cubic fields of prime discriminant -3299, and you'll see that there are no results shown. However Jones' tables contains the 4 cubic fields with such discriminant.

Now about your question on the primes, as Ben mentioned p must be 1 mod 4 so the question has some hope. Even for p that are "allow" to be discriminants there might not exist a field of fix degree d of discriminant p. For example, class field theory tells you that there is a cubic field of discriminant p if and only if the 3-Sylow part of $Cl(\mathbb{Q}(\sqrt{p}))$ is non-trivial. In fact you can even tell how many of them there are! As an example of this we can conclude that there are no cubic fields of discriminant p=-3, even though p is a fundamental discriminant. A similar analysis can be done for quartic fields, and I think for them the behavior is related to the 2-Sylow part of $Cl(\mathbb{Q}(\sqrt{p}))$.

Also you might want to look at this paper of Jone's which I think is very close to your question http://hobbes.la.asu.edu/papers/OnePrimeJR.pdf

share|improve this answer
very interesting links! do you mind explaining this statement? thank you! "For example, class field theory tells you that there is a cubic field of discriminant p if and only if the 3-Sylow part of Cl(Q(\sqrt{p})) is non-trivial." –  natura Jan 5 '10 at 4:25
@basic: I have written a paper that uses this, so instead of copy and paste, I better give you the link to it. math.wisc.edu/~mantilla/PagProb/pruebas.pdf What you are wondering can be found at the beginning of section 4. –  Guillermo Mantilla Jan 5 '10 at 4:59
this sounds real cool, I didn't know of the relation to the 3 part... Btw, I mention in my question that positive and negative is fine, so negative 3 mod 4 is also fine. –  Dror Speiser Jan 5 '10 at 5:45
yes -3 mod 4 is the same as 1 mod 4. Now, the impossibility I was referring to is p=2,3 mod 4. The point that Ben was trying to make in his comment is that the discriminant of a number field is always congruent to 0 or 1 mod 4(this is known as Stickleberger's theorem). In particular if a prime p is a discriminant, so the question has some hope to be answered, then p has to be 1 mod 4 (regardless if p is positive or negative). Hence, p = 2,3 mod 4 is never a possibility. –  Guillermo Mantilla Jan 5 '10 at 17:19
Usually p=2,3 mod 4 refers to a positive prime, and it really does seem that that is what Ben meant, in which case, negative of a prime 3 mod 4 is fine. Hence my comment, as well as Pete's. –  Dror Speiser Jan 5 '10 at 19:11

I am going to annoy all the people who answered above, but I am pretty sure the answer to Dror's question is basically no. In particular, is it known that there are infinitely many cubic fields of prime discriminant? I have not heard of such a result -- if one is out there then I would be extremely grateful if someone would share the appropriate references with me.

As is pointed out above, there is a classical correspondence between such fields and subgroups of $Cl(\mathbb{Q}(\sqrt{p}))$ of index 3. However, I'm not aware that this makes the question easier to answer.

There is also the work of Bhargava and Ghate, Delone-Faddeev, Davenport-Heilbronn, etc. which says that cubic (and quartic and quintic) fields are parameterized by integral orbits on nice prehomogeneous vector spaces which meet certain local conditions. For example, in the cubic case, cubic rings are parameterized by integral binary cubic forms up to $GL_2(\mathbb{Z})$ equivalence, and maximal cubic orders are those cubic rings which meet a certain local condition at each prime.

This allows you to prove formulas for the number of cubic fields with $Disc(K) < X$ with good error terms, and this works if you ask for the condition $d | Disc(K)$. This allows you to run a sieve.

However sieves are notoriously bad at finding primes! The information above is also essentially available in the twin prime problem, but all we can prove there is that there are infinitely many primes $p$ so that $p + 2$ has at most two prime factors.

You can use this argument to find cubic fields with three (I think) prime factors -- there is a paper of Belabas and Fouvry that does this. Maybe you could push their arguments a little bit better. But one cannot hope to find primes this way.

Of course there are excellent computational results, and I don't want to take anything away from these. But I feel like the question is asking if there are infinite families, and I'm pretty sure this is widely expected but not at all known.

share|improve this answer
You can find infinitely many cubic fields of prime discriminant if you assume a conjecture of Hardy and Littlewood on primes in quadratic progressions -- just take the "simplest cubic fields" where the quadratic function giving the discriminant represents a prime value. –  Cam McLeman Nov 17 '10 at 2:07
Simplest cubic fields have a square discriminant. –  Dror Speiser Nov 17 '10 at 7:00
Oops. . –  Cam McLeman Nov 17 '10 at 12:43
@Frank: I wonder how far Heath-Brown's results on primes represented by binary cubic forms can be extended to proving infinitely many primes represented by the discriminant form (of a general cubic). Sure - the degree goes up by one, but the number of variables goes up by two! –  Dror Speiser Nov 17 '10 at 20:02
Dror - interesting suggestion! I'm not familiar enough with Heath-Brown's results to hazard a guess. –  Frank Thorne Nov 18 '10 at 1:28

Your Answer


