1. ## Algebra

Let
$f(x) = x^4+ax^3+bx^2+cx+d$
be a polynomial in $\mathbb Q[x]$ which has no rational root.

Show that $f(x)$ is irreducible in $\mathbb Q [x]$

if the cubic resolvent
$g(y)=y^3-by^2+(ac-4d)y-a^2d+4bd-c^2$ has no rational root.

2. Originally Posted by GaloisGroup
Let
$f(x) = x^4+ax^3+bx^2+cx+d$
be a polynomial in $\mathbb Q[x]$ which has no rational root.

Show that $f(x)$ is irreducible in $\mathbb Q [x]$

if the cubic resolvent
$g(y)=y^3-by^2+(ac-4d)y-a^2d+4bd-c^2$ has no rational root.
If the cubic resolvent has no rational roots then it is irreducible. Therefore, the Galois group of $f(x)$ must contain $A_4$ and so it splitting field is at least a $12$-th degree extension. Now if the original polynomial was reducible then its will factor as a product of two quadradics which would mean its Galois group would be at most a 4 degree extension. Which is a contradiction.

3. Originally Posted by ThePerfectHacker
Therefore, the Galois group of $f(x)$ must contain $A_4$
How can you conclude that the galois group contain $A_4?$ what is the reason for it?

4. Originally Posted by GaloisGroup
How can you conclude that the galois group contain $A_4?$ what is the reason for it?
I think that is a theorem about resolvents. If you never seen it then I do not know of anything else to say.

5. do you remember where you have seen this theorem in which book.