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Let $K/F$ be finite Galois field extension, then $K$ is the splitting field of a separable polynomial $p$ over $F$, i.e. $K=F(a_{1},..a_{n})$ where $p=(x-a_{1})...(x-a_{n})$.

My question is: is it true that if $E$ is a subextension of $K/F$ then $E$ is also of the form $E=F(b_{i_1},..b_{i_t})$ where $g=(x-b_{1})...(x-b_{k})$ is in $F[x]$ ?

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I'm confused. Are you saying that $p$ factors in this way in $F[x]$? That's hardly ever true. – Dylan Moreland Jul 20 '12 at 15:11
@DylanMoreland-no. It is like $x^2+1=(x-i)(x+i)$. the factoring is in the extension, but the polynomial is in the base field – Belgi Jul 20 '12 at 15:13
I guess I would have made that more explicit. So your question is: is a subextension of a finite Galois extension a splitting field over $F$? The answer is no, I think. I believe that would imply that any $E/F$ was always Galois, which is not true. – Dylan Moreland Jul 20 '12 at 15:14
@DylanMoreland - I asked if it generated by roots (i.e. some of them), not if it a splitting field so it does not imply the extension is normal – Belgi Jul 20 '12 at 15:20
I don't follow. $g$ is a polynomial in $F[x]$, it splits in $E$ as you've written it, and indeed its roots generate $E$ over $F$. From what you're saying now it sounds like you may want $g = \cdots (x - b_r)$ for some $r \geq k$, and the $b_i$ all lie in $K$? Cheers, – Dylan Moreland Jul 20 '12 at 15:22
up vote 1 down vote accepted

Now that we have the question nailed down: Let $E/F$ be a finite extension. The first question is whether $E$ can be written as $F(b_1, \ldots, b_n)$. This is true. You could, for example, take $\{b_i\}$ to be a basis for $E$ as a vector space over $F$. Each $b_i$ has some minimal polynomial $p_i \in F[x]$ over $F$, and $p_1 \cdots p_r$ seems to do what you want.

Note that the above choice of $\{b_i\}$ was probably not very efficient with regard to $n$. The primitive element theorem says that if $E/F$ is separable then $E = F(\alpha)$.

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Thank you! this is very clear – Belgi Jul 20 '12 at 15:34
Just to be clear $E$ is probably not the splitting field of $p_1\cdots p_r$. – JSchlather Jul 20 '12 at 17:20

The first statement is not even necessarily correct. Take $K = \Bbb{Q}(\sqrt[3]{2},\zeta_3)$ that is the splitting field of $p(x) = x^3 - 2 = (x - \sqrt[3]{2})(x- \sqrt[3]{2}\zeta_3)(x- \sqrt[3]{2}\zeta_3^2) \in \Bbb{F}[x]$ where $F = \Bbb{Q}$. So just because $K$ is the splitting field of a separable polynomial $p$ over $F$, it is true that $K$ is $F$ adjoined a finite number of elements but they need not be roots of the polynomial $p$.

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I don't see why this is a counter example, If I understand correctly then indeed $\mathbb{Q}(\sqrt[3]{2},\zeta_{3})=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2}\zeta_{3},-‌​\sqrt[3]{2},\zeta_{3})$ – Belgi Jul 20 '12 at 15:12
He did not ask if they're roots of the same polynomial. – tomasz Jul 20 '12 at 15:12
I found the question somewhat hard to interpret, so I don't blame Ben for being confused. [I should say that I don't blame Belgi either — this is not the language I would have used to express this problem, but if you're just starting out in the subject then it's completely understandable to write things in such a way.] – Dylan Moreland Jul 20 '12 at 15:17

A finite Galois extension is a finite, separable, normal extension. Any subextension of a separable extension is separable, similarly for finite extensions (mostly obviously). However, it is certainly not true for normality.

Consider arbitrary $F$, $K$ the splitting field of a minimal (separable) polynomial in $F[X]$ with at least two distinct roots. Then $F[\alpha]$, where $\alpha$ is just one of the roots of the polynomial is not normal.

But if you want $E$ to be generated by some roots of some polynomial, then of course this is true. You don't even need $K$ to be Galois, just finite (and hence algebraic). Any finite field extension is algebraic, so for arbitrary $F\subseteq E\subseteq K$ we have $E=F[\alpha_1,\ldots,\alpha_n]$ (this is true for example because $E$ is a subspace of $K$ as a vector field over $F$), an then $\alpha_j$ are algebraic, so if you take for $p$ the product of their respective minimal polynomials over $F$, they will all be roots of $p$.

Furthermore, if $K$ was separable, then if you take only distinct minimal polynomials of $\alpha_j$ (so you won't take $x^2+1$ twice for $\alpha_1=i,\alpha_2=-i$, for instance), $p$ will also be separable (because any $\alpha_j\in K$ is separable).

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There is such a thing as infinite Galois extensions, you know... – Arturo Magidin Jul 20 '12 at 15:34
@ArturoMagidin: Good point. The definition I've been told in an algebra course contained finite, that's where it comes from. Will fix. – tomasz Jul 20 '12 at 15:36

It is true is that for any normal extension $\,A/B\,$ , $\,A\,$ is the splitting field of a familiy of polynomials over $\,B\,$ , so if the extension is finite we can take one polynomial s.t. $\,A\,$ is its splitting field (as there must be a finite number of roots involved and thus we can take the product of a finite number of polynomials. This is stressed in, for example, Lang's "Algebra").

Thus, it is true in our case that $\,K=F(a_1,...,a_n)\,$ , as you wrote.

As for the second question the answer is no: for example, the Galois extension of the polynomial $\,x^4-2\in\Bbb Q[x]\,$ contains the subextension $\,\Bbb Q(\sqrt[4] 2)/\Bbb Q\,$ is not normal...

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