This question is a bit vague, but I was wondering if someone might have an insightful answer.

Let $f_1$ and $f_2$ be irreducible polynomials in $\mathbb{Q}[x]$. Is there an easy criterion for knowing when the splitting fields of $f_1$ and $f_2$ yield the same field extensions of $\mathbb{Q}$?

Here is a related question. Let $L/\mathbb{Q}$ be a finite field extension. Assume both $f_1$ and $f_2$ remain irreducible in $L$. Given such an $L$, is there a way to determine when the splitting fields of $f_1$ and $f_2$ over $L$ are the same? (It is possible that the splitting fields of $f_1$ and $f_2$ over $\mathbb{Q}$ are different, but their splitting fields over $L$ are the same.)

  • $\begingroup$ I like the question, but for the time being the only trivial observation I have is that if someone tells you the field extension(s) they generate is abelian (over $\mathbb{Q}$) there might be a way. Supoose so and both are monic with integer coefficients (too easy?) First, test if they have real or complex roots (by using calculus to determine whteher they have one, and hence all, real root). Now you know ramification at infinity. Secondly, look at the discriminants: they are the discriminants of the orders $\mathbb{Z}(\alpha_i), i=1,2$ and let $D$ be the l.c.m. Now you need to look at... $\endgroup$ – Filippo Alberto Edoardo Apr 3 '13 at 23:46
  • $\begingroup$ ...the real/complex extensions of degree $N=\mathrm{deg}(f_i)$ inside $\mathbb{Q}(\zeta_D)$. If $D$ has many factors, I have no idea on how to proceed, but if it is prime there is a unique one. I this case you still need to test if $\mathhb{Z}(\alpha_i)$ are the maximal orders (in which case, if $D$ is a proper divisor of the discriminants, the extensions are different) or not...Very vague, and a bad treatment of a very special case, I am sorry ;-( EDIT: In the previous comment, I should have $D=$g.c.d and not l.c.m. $\endgroup$ – Filippo Alberto Edoardo Apr 3 '13 at 23:51

Without loss of generality we may assume that $f_1$ and $f_2$ are monic with integral coefficients.

Let $P_i$ be the set of primes $p$ such that the image of $f_i$ in $\mathbb F_p[x]$ factors into linear factors.

Then, as a consequence of Chebotarev's density theorem (actually the weaker Frobenius density theorem is good enough here) the splitting fields of $f_1$ and $f_2$ are the same if and only if the sets $P_1$ and $P_2$ differ only by finitely many elements.

Using effective versions of Chebotarev, one can make a finite (yet impracticable) criterion from that.

  • $\begingroup$ I'm looking for something with more insight, I'm afraid... I suppose if there is none then there is none. But I'm still holding out hope that someone will tell me that there is some paper that I should read about it, or that it somehow has to do with cohomology, or whatever insight might come this way... (BTW, the third question was a statement, not a question: "It is possible" rather than "Is it possible".) $\endgroup$ – James D. Taylor Apr 3 '13 at 19:30
  • $\begingroup$ @James D. Taylor: I changed my answer, so your comment probably does not apply anymore. $\endgroup$ – Peter Mueller Apr 4 '13 at 12:52

Krasner's Lemma could help. http://planetmath.org/krasnerslemma

  • $\begingroup$ How could it help here? The base fields are ^number fields^. Even over $p$-adic fields, the Krasner lemma only gives a sufficient, but not a necessary criterion for equality of fields. $\endgroup$ – Peter Mueller Apr 3 '13 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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