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Consider a monic, irreducible polynomial $f(x)\in\mathbb Q[x]$ and its splitting field $F$. Let $L$ be a number field ($L$ and $F$ are contained in the complex numbers). Are the following true?

(1): If $f$ remains irreducible in $L[x]$, then $L \cap F =\mathbb Q$. False, via gobucksmath's example.

(2): If $L \cap F =\mathbb Q$, then $f$ remains irreducible in $L$. Equivalent to (3).

(3): If $f$ becomes reducible in $L[x]$, then $L \cap F$ strictly contains $\mathbb Q$. True, by Rene Schipperus's argument.

(4): If $L \cap F$ strictly contains $\mathbb Q$, then $f$ becomes reducible in $L$. False, via gobucksmath's example.

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    Well, I've tried to prove (3) by observing that if f=(x-\alpha_1)...(x-\alpha_n) and if f factors in L, then, using Vieta, all symmetric sums of a subset of {1,2,...,n} must be in L.I tried to prove that they can't be all in Q. – 2017-01-12
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    I think that works because f is irreducible over Q. – 2017-01-12
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    Arent (2) and (3) equivalent ? – 2017-01-12
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    Oh, yes they are! – 2017-01-12

2 Answers 2

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Assume (3). Let $f(x)=g(x)h(x)$ with $g(x), h(x)\in L[x]$ then the coefficients of $g(x)$ are in $L$. But they also in $F$ thus the coefficients are in $\mathbb{Q}$ and so $f(x)$ is not irreducible.

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    Thanks, even though I already mentioned that in a previous comment – 2017-01-12
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    Sorry I didnt see how this was contained in the comment. – 2017-01-12
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    Ok well it seems that you already had all the ingredients.... – 2017-01-12
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I will address (1) and (4): Let $F=\mathbb{Q}(\zeta, \sqrt[3]{2})$, where $\zeta=\zeta_3\not=1$ is the cube root of $1$. $F$ is the splitting field of the polynomial $x^3-2$. Let $L=\mathbb{Q}(\zeta)$. Then we have $\mathbb{Q}\subset L\subset F\subset\mathbb{C}$. $x^3-2$ remains irreducible in $L[X]$, and $L\cap F=L$. This gives us that (1) and (4) are false.