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Let's say we have the irreducible polynomial $f$ with roots $\alpha_1,\ldots,\alpha_n$. Now let $K$ be its splitting field, in other words $$K=\mathbb Q(\alpha_1,\ldots,\alpha_n).$$

When is it the case that $K=\mathbb Q(\alpha_i)$ for any $i$, i.e. when is the dimension of $K$ over $\mathbb Q$ equal to $n$, the degree of $f$? If this were true,then we would be able to express all the roots by rational combinations of each other. Is there a criterion that would decide whether an extension $\mathbb Q(\alpha)$ of $\mathbb Q$ has this property? Please explain your answers.

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    $\begingroup$ An equivalent condition is that $K$ must have cyclic Galois group of order $\deg(f)$. I'd be surprised if there was a nice, fully general, categorisation of what $\alpha$ or $f$ should look like. For example, it is not at all obvious at first glance that $\mathbb Q(\sqrt{2+\sqrt2})$ should be an example (and note that this answers your second question: no). There are things we can say: $f$ must be solvable by radicals, $\mathbb Q(\alpha)\subset\mathbb Q(e^{2i\pi/N})$ for some $N$ (Kronecker-Weber). It might help if you specified what kind of criterion you are after. $\endgroup$ – Mathmo123 May 4 '16 at 22:18
  • $\begingroup$ (I may have misunderstood what you meant by rational combinations: if you mean that every root is of the form $\frac{g(\alpha)}{h(\alpha)}$ for some rational polynomials $g,h$ then this is true by definition.) $\endgroup$ – Mathmo123 May 4 '16 at 22:34
  • $\begingroup$ @Mathmo123 i meant that when the splitting field can be generated by one element, then all its conjugates lie in Q(alpha).Also, I remember seeing something about a square discriminant having to do with this, but can't really recall. $\endgroup$ – Bogdan Simeonov May 5 '16 at 5:43
  • $\begingroup$ @Mathmo123 also, how does the proof that such an extension must have a cyclic galois group go? $\endgroup$ – Bogdan Simeonov May 5 '16 at 5:48
  • $\begingroup$ The splitting field can always be generated by one element (primitive element theorem, $\mathbb Q$ is perfect), but that element won't necessarily be a root of the same polynomial. Having square discriminant means the Galois group is a subgroup of $A_{\deg(f)}$. This doesn't help much when the degree is large. For your second question, the Galois group must be a transitive subgroup of $S_{\deg(f)}$ of order $\deg(f)$. $\endgroup$ – Mathmo123 May 5 '16 at 8:13
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A sufficient condition is the Galois group $G$'s subgroups are all normal (e.g $G$ is abelian or $G=Q_8$).

Assume a monic polynomial $f(x) \in \mathbb Q[x]$ has Galois group $G$, and its splitting field is L. Choose a root $\alpha$ of $f(x)$ and consider $\mathbb Q(\alpha)$, which must be Galois over $\mathbb Q$ as all subgroups of $G$ are normal. Hence it contains every conjugation of $\alpha$ i.e the roots of $f$. Therefore it is just the splitting field L, so we have $[L:\mathbb Q]=[\mathbb Q(\alpha):\mathbb Q]=\operatorname{deg} f$.

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