- Thread starter
- #1

- Mar 10, 2012

- 834

ATTEMPT:

If $p(x)$ is not irreducible in $F[x]$ then the splitting field of $p(x)$ over $F$ can have degree $2!=2$ over $F$.

Thus $p(x)$ should have no rational roots.

Now.

Claim: $p(x)$ should not have any complex root (in the field of complex numbers).

Proof: Suppose it did. Say $\alpha$ is a complex root of $p(x)$. Now since $p(x)$ is a polynomial of degree three it has at least one real root say $a$. Let $E$ be the splitting field of $p(x)$ over $F$. Then $[F(a):F]$ divides $[E:F]=3$. $[F(a):F] \neq 1$ since $p(x)$ is irreducible in $F[x]$. Thus $[F(a):F]=3$. Clearly $\alpha \not \in F(a)$. Thus $[F(a, \alpha):F] > [F(a):F]$. Since $[F(a, \alpha):F]$ divides $[E:F]$, it follows that $[E;F]> [F(a):F]=3$.

So all three roots of $p(x)$ in the field of complex numbers are real.

Now what do I do?