Irreducible polynomials over different fields

[ivos]

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)?

 

[lippyka]

A:Let $F$ and $K$ be two fields, and let $L=F\cap K$. Suppose that $f$ and $k$ are both monic irreducible polynomials of the same degree over $F$ and $K$, respectively. Suppose further that $u$ is a common root of $f$ and $k$. Then we have $f(u)=k(u)=0$.We shall show that $f$ and $k$ are identical as polynomials over $L$. Since $f$ and $k$ are monic, it suffices to show that they have the same coefficients. We proceed by induction on the degree of $f$ and $k$. Suppose first that $\deg f=\deg k=1$. Then $f=X-a$ and $k=X-b$ for some $a,b\in F$ and $K$, respectively. Since $u$ is a common root of $f$ and $k$, we have $f(u)=0=k(u)$, which implies that $a=b$. Thus, $f=k$ over $L$. Now suppose that $\deg f=\deg k >1$, and that the assertion holds for all polynomials of degree less than $\deg f=\deg k$. Let $f=X^n+c_{n-1}X^{n-1}+\cdots + c_1X+c_0$ and $k=X^n+d_{n-1}X^{n-1}+\cdots + d_1X+d_0$ be the respective monic forms of $f$ and $k$ over $F$ and $K$. Since $f$ and $k$ have a common root $u$, we have$$f(u)=X^n(u)+c_{n-1}X^{n-1}(u)+\cdots + c_1X(u)+c_0=0=X^n(u)+d_{n-1}X^{n-1}(u)+\cdots + d_1X(u)+d_0=k(u).$$Comparing coefficients, we find that $c