# Thread: Splitting fields and irreducible polynomials

1. ## Splitting fields and irreducible polynomials

Let f(x) be an irreducible cubic polynomial in Q[X]. Show that the degree of its splitting field over Q is 6 if f ' (x) > 0 for all x.

This seems like quite an interesting question but I have no idea how we are meant to go about solving it - it seems quite strange to have a condition relating to the derivative in "splitting field" type questions.

2. Originally Posted by Amanda1990
Let f(x) be an irreducible cubic polynomial in Q[X]. Show that the degree of its splitting field over Q is 6 if f ' (x) > 0 for all x.

This seems like quite an interesting question but I have no idea how we are meant to go about solving it - it seems quite strange to have a condition relating to the derivative in "splitting field" type questions.
Think of the graph of $f(x)$. If $f'(x) > 0$ it means the polynomial is always increasing. Thus, it cross the x-axis precisely at one point. Let $\alpha$ be the real root of $f(x)$. Now if $\beta$ is the other root of $f(x)$ then it has to be complex*. Thus, the final root is $\bar \beta$. The splitting field is $K=\mathbb{Q}(\alpha, \beta, \bar \beta)$. Note that $\sigma : K\to K$ by complex conjugation is an automorphism. Now there are three possibilities to send $\alpha$ i.e. $\alpha \to \alpha, \alpha \to \beta, \alpha \to \bar \beta$ by isomorphism extension theorem. Call these maps $\tau_1,\tau_2,\tau_3$. Then we see that $S_3\subseteq \left< \sigma,\tau_1,\tau_2,\tau_3 \right>$.

*)You may want to say, but maybe $f$ has only real roots. But then it forces $f(x) = (x-a)^3$ for some $a\in \mathbb{R}$. However, $f'(a) = 0$ is not larger than $0$. So this is not possible.

3. OK thanks for the hints but I really can't see how this helps very much! I haven't come across many actual theorems relating to splitting fields - does your answer rely on some "standard" result?

4. Originally Posted by Amanda1990
OK thanks for the hints but I really can't see how this helps very much! I haven't come across many actual theorems relating to splitting fields - does your answer rely on some "standard" result?
Define, $\sigma : K \to K$ to be $\sigma (x) = \bar x$ i.e. complex conjugation.
Define, $\tau_1 : K \to K$ to be $\tau_1 (\alpha) = \alpha$ (i.e. the identity mapping).
Define, $\tau_2 : K \to K$ to be $\tau_2 (\alpha) = \beta$.
Define, $\tau_3 : K \to K$ to be $\tau_3 (\alpha) = \bar \beta$.
Notice that $\tau_2,\tau_3$ exist by the isomorphism extension theorem.
We know that $\tau_1,\tau_2,\tau_3, \sigma$ are all distinct automorphism of $K$.
But that means, $\sigma \tau_2, \sigma \tau_3$ are distinct automorphism too.
Thus, we have shown we have at least $6$ automorphism of $K$.
Howevever, $|\text{Gal}(K/\mathbb{Q})| = [ K : \mathbb{Q} ] \leq 3! = 6$.
Thus, we found all the automorphisms
This means $[K:\mathbb{Q}] = 6$.

Do you have any other questions?