Assume that F is infinite that are algebraic over F and that w is the root of a separable polynomial in . Prove that is a simple extension over F .
F(v,x) is a simple extension over F.
The monic polynomial p(x) over a field F of least degree such that for an algebraic element over a field F.
Let and F is a infinite and be the minimal polynomial of v and be the minimal polynomial.of w.
Now let L be the splititing field of over F and be the root of in also are distinct.
Similarly be the root of the in also are distinct. Since F is infinite there exist such that
Let and and . Now w is a root of h(x). So,
Suppose some is also a root of h(x). Then so that is one of the root of p(x).
Now this is a contradicts the equation 1. Therefore w is the only common root of q(x) and h(x). Let r(x) be the minimal polynomial of w over F(u) . Then r(x) divides q(x), so that every root of r(x) is a root of q(x). But as r(x) also divides q(x) so all its root are root of h(x) and r(x) has a single root w in L .
Therefore must have degree 1 hence its root w is in F(u). Since with and also . Hence but
Now let F is finite. Then the multiplicative group of nonzero element of K is cyclic. If u is generator of this group then the subfield of F(u) contain 0F and all power of u and hence contain every element of K. So, .
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