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Let $f$ and $g$ be irreducible polynomials over a field $K$ with $\deg f=\deg g =3$ and let the discriminant of $f$ be positive and the discriminant of $g$ negative.

Does it follow that the splitting fields of $f$ and $g$ are linearly disjoint? If yes, why?

(Def.: Two intermediate fields $M_1, M_2$ of an algebraic field extension $L|K$ are called linearly disjoint, if every set of elements of $M_1$, that is linearly independent over $K$, is also linearly independent over $M_2$. Here, this is equivalent to $M_1 \cap M_2 = K.$)

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    Sorry, I forgot to mention, that the splitting fields should be real. $K$ should be real, so the case $i \in K$ can not occur.2012-05-22
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    IIRC the discriminant of a cubic has a square root in the splitting field. So both splitting fields can't be real? Please edit the question so that all the assumptions are included.2012-05-22
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    Ok, then we omit the asspumtion that all splitting fields are real. But $K \subseteq \mathbb R$ should be still valid. Is it now true, that the splitting fields of $f$ and $g$ are linearly disjoint? Or have to be made more assumptions?2012-05-22
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    What makes you think that linear disjointness of $M_1$ and $M_2$ is equivalent to $M_1\cap M_2= K$ ?2012-05-23
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    This is why the splitting fields $F_f$ of $f$ and $F_g$ of $g$ over $K$ are finite galois extensions by assumption. And therefore linear disjointness is equivalent to $F_f \cap F_g = K$.2012-05-23

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