0
$\begingroup$

Let $f,g\in \mathbb{Z}[x]$, $(f,g)=h\in \mathbb{Z}[x]$ (if this is not true, can we construct $h$ in another way?). Let $K_1,K_2$ be the splitting fields of $f,g$ over $\mathbb{Q}$ respectively. Is there any relation between $K_1\cap K_2$ and the splitting field over $\mathbb{Q}$ of $h$?

Example: $f(x)=x^7-12$, $g(x)=x^7-11$. Then $(f,g)=1$, $K_1\cap K_2=\mathbb{Q}[\zeta]$, where $\zeta$ is a nontrivial root of $x^7=1$. Is there any theorem in general?

Thanks.

$\endgroup$
1
$\begingroup$

I don't think one can say much in general. Let $f$ be any irreducible polynomial, let $g(x)=f(x+1)$, then you'll have $h=1$ with splitting field the rationals, but $K_1=K_2=K_1\cap K_2$ will just be the splitting field of $f$. At the other extreme, it's easy to find irreducible $f$ and $g$ such that $K_1\cap K_2$ is just the rationals.

Maybe if you make your question a bit more precise?

$\endgroup$
1
$\begingroup$

As Gerry has pointed out, the two don't need to be equal in general. But there is still a theorem: the splitting field of $(f,g)$ is always contained in the intersection of the splitting fields of $f$ and $g$.

It should be fairly clear why: any root of $(f,g)$ must be a root of $f$ and also a root of $g$.

(Note: I'm assuming that we're working within a specified algebraic closure, say $\overline{\mathbb{Q}}\subset\mathbb{C}$. Otherwise, I would need to say that there is a "natural injection", since there is no a priori notion of containment.)

$\endgroup$

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.