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Abstract Algebra: An Introduction
Found in: Page 398
Abstract Algebra: An Introduction

Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624

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Short Answer

Assume that F is infinite that v,wK are algebraic over F and that w is the root of a separable polynomial in F[x]. Prove that F(v,w) is a simple extension over F .

F(v,x) is a simple extension over F.

See the step by step solution

Step by Step Solution

Step 1: Definition of minimal polynomial

The monic polynomial p(x) over a field F of least degree such that pα=0 for an algebraic element α over a field F.

Step 2: Showing that F(v,w) is a simple extension over F  

Let K=Fv,w and F is a infinite and pxFx be the minimal polynomial of v and qxFx be the minimal polynomial.of w.

Now let L be the splititing field of px,qx over F and w=w1,w2....wn be the root of qx in Li also wi are distinct.

Similarly v=v1,v2,...vn be the root of the px in Li also vi are distinct. Since F is infinite there exist cF such that

Fcvi-vw-wi for all 1im,1jn 1

Let u=v+cw and wFu and hx=pu-cxFux. Now w is a root of h(x). So,


Suppose some wj is also a root of h(x). Then pu-cwj=0F so that u-cwj 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 rxFux must have degree 1 hence its root w is in F(u). Since v=u-cw with u,wFu and also vFu. Hence K=Fv,wFu 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, Fu=K .

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