- #1

- 4

- 0

Problem: Let f be monic irreducible polynomial over a field F, k be monic irreducible polynomial over a field K, deg f = deg k. Let u be common root of f and k. Prove (or disprove by counterexample), that f=k over field (F intersection K), i.e. polynomials f and k are identical.

Proof would be easy, if polynomial f would be irreducible over some field which includes fields F and K (with using "Abel's lemma", i.e. for any element u which is algebraic over some field, there is exactly one monic irreducible polynomial f over this field, f(u)=0).

Unfortunately, polynomial f could be reducible over smallest field which includes F and K.

(x^2-6 is irreducible over Q(sqrt(2)) and Q(sqrt(3)) but reducible over Q(sqrt(2), sqrt(3)).

On the other side, if polynomial f-k would be nontrivial, deg (f-k) < deg f=deg k, f-k has u as root. But, we cannot use Abels lemma again, because of polynomial f-k is over larger field than F and K a so can be reducible over this field. It doesn't follow f=k.

How make the right proof (or find counterexample - but I think, that claim is true, i.e. f=k)?

Proof would be easy, if polynomial f would be irreducible over some field which includes fields F and K (with using "Abel's lemma", i.e. for any element u which is algebraic over some field, there is exactly one monic irreducible polynomial f over this field, f(u)=0).

Unfortunately, polynomial f could be reducible over smallest field which includes F and K.

(x^2-6 is irreducible over Q(sqrt(2)) and Q(sqrt(3)) but reducible over Q(sqrt(2), sqrt(3)).

On the other side, if polynomial f-k would be nontrivial, deg (f-k) < deg f=deg k, f-k has u as root. But, we cannot use Abels lemma again, because of polynomial f-k is over larger field than F and K a so can be reducible over this field. It doesn't follow f=k.

How make the right proof (or find counterexample - but I think, that claim is true, i.e. f=k)?

Last edited: