Let $L/\mathbb{Q}$ an Abelian Galois extension of degree $4\cdot7\cdot9\cdot13$. Prove that there exists no polynomial $f\in \mathbb{Q}[x]$ of degree $4\cdot7\cdot13$ such that $L$ is the splitting field of $f$.
I don't know how to translate the assumptions into something useful.
What can I conclude if $f$ must have a degree of $4\cdot7\cdot13$? How do I use the fact that the Galois group is abelian?
I'd appriciate your help on this. Thank you.
There is such a polynomial. As $L$ is Abelian, it is the compositum of extensions $L_4$, $L_9$, $L_7$ and $L_{13}$ of degrees $4$, $9$ etc., over $\Bbb Q$. There are polynomials $f_4$, $f_9$ etc., of degrees $4$, $9$ etc., over $\Bbb Q$ with splitting fields $L_4$ etc. Then $F=f_4f_9f_7f_{13}$ is a polynomial with splitting field $L$. Its degree is $33$ which is a lot less than $4\times7\times13$. You can bulk it up with linear factors to make its degree up to this number.
Oh, wait, did you want an irreducible polynomial whose splitting field is $L$? That's a completely different matter!