2
$\begingroup$

Does all splitting field have characteristic 0?

This may be a bad question, but I am wondering because the author of a book I am reading summarises a lot of properties when he is about to start with Galois theory, one of the properties is:

If E is a finite extension of F and is a separable splitting field over F, then $|G(E/F)|=\{E:F\}=[E:F]$.

However in the relevant chapter I can not see this proved, the closest thing I find is:

We have completed our aim, which was to show that field of characteristic 0 and finite fields have only separable finite extensions, that is, these fields are perfect. For finite extensions E of such fields F, we then have $[E:F]=\{E:F\}$.

Does the first sentence in some way follow from the second? It would if I could prove that all splitting fields are are either finite or have characteristic 0?

$\endgroup$
  • $\begingroup$ And the book you are reading is...? $\endgroup$ – whacka May 13 '15 at 5:05
  • $\begingroup$ A first course in Abstract Algebrah, John B. Fraleigh, seventh edition $\endgroup$ – user119615 May 13 '15 at 5:06
0
$\begingroup$

A field $E$ over $F$ is a splitting field if it is the smallest field extension of $F$ containing all the roots of some polynomial $p \in F[X]$ (defined regardless of the characteristic of $F$). Let $\alpha_1, \ldots, \alpha_n$ be the roots of $p$ in an algebraic closure. Then the splitting field of $p$ over $F$ is $E=F(\alpha_1, \ldots, \alpha_n)$ (smallest field containing $\alpha_i$). This is a finite extension of $F$ since each $\alpha_i$ is algebraic over $F$ (and any extension of $F$). You can form a (finite) tower of finite extensions: $F \subset F(\alpha_1) \subset F(\alpha_1,\alpha_2) \subset \ldots \subset E$.

Compare this with a normal extension, which may be infinite. One characterization of a normal extension $E$ over $F$ is that $E$ is the splitting field for a (possibly infinite) family of polynomials in $F[X]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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