Quintic polynomials generating cyclic extensions We know that a cubic equation generates a cubic cyclic extension iff it has a perfect square discriminant. Now I am wondering if there is a similar condition for quintic polynomials. So I am trying to find "are there any conditions on the coefficients of a quintic polynomials under which it generates a cyclic quintic extension?". Or anything that will help in this direction.
 A: Robin French gave a talk at the American Math Society meeting last week, "A new algorithm for Galois groups of quintic polynomials." If I recall correctly, he constructs a resolvent of degree 24, and the degrees of the irreducible factors of that resolvent are all you need to know to get the Galois group of the original irreducible quintic. 
EDIT: This degree 24 resolvent is also discussed in Awtrey and Shill, "Absolute resolvents and masses of irreducible quintic polynomials," pages 31-42 of the book, Collaborative Mathematics and Statistics Research, Topics from the 9th Annual UNCG Regional Mathematics and Statistics Conference. Bits of the book are accessible on Google Books. 
A: An answer can be found in the beautiful book by Cox Galois Theory, Theorem 13.2.6 page 372.
The result is as follows.

Theorem. Assume that $f \in F[x]$ is monic, separable and irreducible of degree $5$ and that $\textrm{char}(F) \neq 2$. Let $L$ be the splitting field of $f$ over $F$, and set $\textrm{Gal}(L/F)=G \subset S_5$. Then the following holds.
(a) $G \subset A_5$ if and only if the discriminant $\Delta(f)$ is a square in $F;$
(b) $G$ is conjugate to a subgroup of $\textrm{AGL}(1, \, \mathbb{F}_5)$ if and only if the sextic resolvent $\theta_f(x)$ of $f$ has a root in $F;$
(c) $G$ is conjugate to $\langle (12345) \rangle$ (hence cyclic) if and only if $F$ splits completely over $F(\alpha)$, where $\alpha$ is a root of $f.$

In particular, the fact that the discriminat of $f$ is a perfect square is a necessary, but not sufficient condition in order to have a cyclic quintic extension. 
Moreover, if $\textrm{char}(F)=0$ then $f$ is solvable by radicals over $F$ if and only if $G$ is isomorphic to a subgroup of $\textrm{AGL}(1, \, \mathbb{F}_5)$, i.e. if and only if its sextic resolvent $\theta_f(x)$ has a root in $F$.
Over the rationals, a more precise criterion useful to compute the Galois group of $f$ (and in particular to check if it is cyclic) can be found in the paper 
D. S. Dummit, Solving solvable quintics, Math. Comp. 57 (1991), 387-401, 
see in particular Theorem $2$. The conditions given 
are the following: $G$ is the cyclic group of order $5$ if and only if $\Delta(f)$ is a perfect square, the sextic resolvent $\theta_f(x)$ has a rational root and two quadratic polynomials (constructed in a somewhat complicate way from the coefficients of $f$) are reducible over $\mathbb{Q}$.   
