Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $F[x]$ ?

share|improve this question
    
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

4 Answers 4

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 $E = F(\alpha)$.

share|improve this answer
    
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 $p$.

share|improve this answer
1  
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
1  
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).

share|improve this answer
    
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...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.