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Let $f(x)\in F[x]$ be a polynomial of degree $n$. Let $K$ be a splitting field of $f(x)$ over $F$. Then [K:F] must divides $n!$.

I had this mis-concept, since $f(x)$ can have at most n roots. then its splitting field degree can at most be $n$

to make this more intuitive, take cubic polynomial for example, can someone give me an example of cubic polynomial which has splitting filed degree equal to 6. does this mean this polynomial now have 6 roots, as splitting filed in my understanding is $Q(\alpha_1, \alpha_2,\alpha_3)$

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The splitting degree can at most be $n!$ and not $n.$ –  Reader Nov 1 '12 at 16:43

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up vote 2 down vote accepted

Just take the case of $f(x)=x^3-2$. The field $K=\mathbb Q[\sqrt[3]{2}]$ is degree $3$, but it doesn't split $f(x)$ - it does not contain the complex roots.

The polynomial doesn't have six roots, it just fails to split. In $K$, $f(x)$ factors as $(x-\sqrt[3]2)(x^2+\sqrt[3]2x + \sqrt[3]2^2)$, and the quadratic component does not factor in this field.

But $[K:\mathbb Q]=3$, and the splitting field is "bigger" than $K$.

If $f(x)$ is a prime polynomial in the ring $F[x]$, then $K=F[x]/\left<f(x)\right>$ is a field extension of $F$ with $[K:F]=\deg f$, but $K$ only provides one root, $x$, in general, so $f$ does not "split" in $K$ except in very specific cases.

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thanks, seems I didn't get clear the definition of field extension degree. –  zinking Nov 2 '12 at 6:16

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