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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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Let me do the case where $K={\bf Q}$. Let $L_f,L_g$ be the splitting fields. Then $L_f\cap L_g$ must be a subfield of both $L_f$ and $L_g$. So, what are the subfields of those splitting fields?

Well, $L_f$ has three real cubic subfields and one real quadratic; $L_g$ has one real cubic subfield, two imaginary cubic subfields, and an imaginary quadratic subfield. So if the intersection isn't $\bf Q$, it must be the real cubic subfield. But then we have a single cubic subfield that has three real embeddings and also two complex embeddings, and that's absurd. QED.

I don't think I made much use of the hypothesis $K={\bf Q}$; this argument ought to work more generally.

EDIT: More generally, if $L$ and $M$ are splitting fields of irreducible cubics over $K$ and are not equal then their intersection can't be of degree 3 over $K$; it can only be $K$ or the quadratic extension given by adjoining the square root of the discriminant (assuming the discriminant is not a square in $K$). That quadratic case can happen, as non-isomorphic cubic fields can be generated by polynommials with the same discriminant. Over the rationals, there are two non-isomorphic cubics with discriminant -1836, also three with discriminant 22356.

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    Dear Gerry,If $f=X^3+X^2-2X-1$, then $disc(f)=49\gt0$ but $[L_f:\mathbb Q]=3$ and $L_f$ thus has no quadratic subfield at all.2012-05-23
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    I can modify $f$ and $g$, so that they have no quadratic terms. Especially, define $f(x):=x^3-3x+\alpha$, $g(x):=x^3-3x+\beta$. What can we conclude now?2012-05-23
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    The polynomial $g=X^3-3X+1$ has discriminant $disc(g)=81$ so that here also $[L_g:\mathbb Q]=3$ and $L_g$ has no quadratic subfield at all.2012-05-23
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    @Georges, of course, you're right; if the discriminant is a square, we're dealing with a cyclic cubic extension --- but this doesn't invalidate the argument. That cubic extension can't be the intersection we're looking for, because it would have to equal the real cubic subfield of $L_g$, which still leads to a contradiction. But it was sloppy of me to leave out this case. Lisa, if my argument shows the intersection is $K$, then that's already the strongest conclusion you can ask for. Special forms for $f,g$ can't make it any stronger.2012-05-23
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    Dear Gerry, indeed your original argument was valid: I was nitpicking! I agree that the intersection of $L_f$ and $L_g$ is $\mathbb Q$ but this is *a priori* weaker than saying that $L_f$ and $L_g$ are linearly disjoint (think of the fields generated over $\mathbb Q$ by the real cubic root of $2$ and an unreal one : they are *not* linearly disjoint) . Are the fields $L_f$ and $L_g$ actually linearly disjoint?2012-05-23
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    @Georges, they are if Lisa is right about linear disjointness of splitting fields.2012-05-23
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    Dear Gerry, yes, you and Lisa are right. Indeed, linear disjointness even obtains as soon as one of the extensions is finite and Galois, the other being finite but arbitrary . As usual, the more general statement is easier, and I let myself be distracted by the particulars of the question !2012-05-23