Galois Theory!

**dabien** · November 21st 2009, 09:44 AM

Set F=F2(t) where t is a variable, let @ be a root of the equation x^2+x+t=0, set B=f(@) and set b=@^(1/2) and E=F(b). then B is separable over F and E is purely inseparable over B. show that there is no intermediate field B' such that E is separable over B' and B' purely inseparable over F.

Q2: Suppose that F is a field of characteristic p and that @ is a separable element in some extension. Show that then F(@^p)=F(@)

Any insights would be great!

**aliceinwonderland** · November 22nd 2009, 12:10 AM

Originally Posted by dabien

Set F=F2(t) where t is a variable, let @ be a root of the equation x^2+x+t=0, set B=f(@) (F@?)and set b=@^(1/2) and E=F(b). then B is separable over F and E is purely inseparable over B.

The deriative of $p(x)=x^2+x+t$ over $\mathbb{F}_2(t)$ is 1, which implies that the deriative has no root at all. Thus $p(x)=x^2+x+t$ is separable. It follows that $B=F(\alpha)$ is separable over F, where $\alpha$ is a root of p(x). Let $k(x)=x^2-\alpha \text{ } (=x^2+\alpha) \in B[x]$ . We see that $k(x)$ is irreducible in $B[x]$ . However, $k(x)$ is factored as $k(x)={(x - \sqrt{\alpha})}^2$ in $E[x]$ . Thus E is purely inseparable over B.

show that there is no intermediate field B' such that E is separable over B' and B' purely inseparable over F.

[EDIT]

Q2: Suppose that F is a field of characteristic p and that @ is a separable element in some extension. Show that then F(@^p)=F(@)
Any insights would be great!

Observe that $\mathbb{F}^p=\mathbb{F}$ , where $\mathbb{F} = F(\alpha)$ and p is a prime number.

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