# Does all splitting fields have characteristic 0?

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?

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

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 $F[X]$.