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Thread: Splitting fields and irreducible polynomials

  1. #1
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    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.
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    Quote Originally Posted by Amanda1990 View Post
    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 $\displaystyle f(x)$. If $\displaystyle f'(x) > 0$ it means the polynomial is always increasing. Thus, it cross the x-axis precisely at one point. Let $\displaystyle \alpha$ be the real root of $\displaystyle f(x)$. Now if $\displaystyle \beta$ is the other root of $\displaystyle f(x)$ then it has to be complex*. Thus, the final root is $\displaystyle \bar \beta$. The splitting field is $\displaystyle K=\mathbb{Q}(\alpha, \beta, \bar \beta)$. Note that $\displaystyle \sigma : K\to K$ by complex conjugation is an automorphism. Now there are three possibilities to send $\displaystyle \alpha$ i.e. $\displaystyle \alpha \to \alpha, \alpha \to \beta, \alpha \to \bar \beta$ by isomorphism extension theorem. Call these maps $\displaystyle \tau_1,\tau_2,\tau_3$. Then we see that $\displaystyle S_3\subseteq \left< \sigma,\tau_1,\tau_2,\tau_3 \right> $.


    *)You may want to say, but maybe $\displaystyle f$ has only real roots. But then it forces $\displaystyle f(x) = (x-a)^3$ for some $\displaystyle a\in \mathbb{R}$. However, $\displaystyle f'(a) = 0$ is not larger than $\displaystyle 0$. So this is not possible.
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    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?
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    Quote Originally Posted by Amanda1990 View Post
    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, $\displaystyle \sigma : K \to K$ to be $\displaystyle \sigma (x) = \bar x$ i.e. complex conjugation.
    Define, $\displaystyle \tau_1 : K \to K$ to be $\displaystyle \tau_1 (\alpha) = \alpha$ (i.e. the identity mapping).
    Define, $\displaystyle \tau_2 : K \to K$ to be $\displaystyle \tau_2 (\alpha) = \beta$.
    Define, $\displaystyle \tau_3 : K \to K$ to be $\displaystyle \tau_3 (\alpha) = \bar \beta$.
    Notice that $\displaystyle \tau_2,\tau_3$ exist by the isomorphism extension theorem.
    We know that $\displaystyle \tau_1,\tau_2,\tau_3, \sigma$ are all distinct automorphism of $\displaystyle K$.
    But that means, $\displaystyle \sigma \tau_2, \sigma \tau_3$ are distinct automorphism too.
    Thus, we have shown we have at least $\displaystyle 6$ automorphism of $\displaystyle K$.
    Howevever, $\displaystyle |\text{Gal}(K/\mathbb{Q})| = [ K : \mathbb{Q} ] \leq 3! = 6$.
    Thus, we found all the automorphisms
    This means $\displaystyle [K:\mathbb{Q}] = 6$.

    Do you have any other questions?
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