Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

14E

Expert-verified

Abstract Algebra: An Introduction

Found in: Page 398

Book edition 3rd

Author(s) Thomas W Hungerford, David Leep

Pages 608 pages

ISBN 9781111569624

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later

Short Answer

Assume that F is infinite that $v, w \in K$ 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

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of minimal polynomial

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

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

Let $K = F (v, w)$ and F is a infinite and $p (x) \in F [x]$ be the minimal polynomial of v and $q (x) \in F [x]$ be the minimal polynomial.of w.

Now let L be the splititing field of $p (x), q (x)$ over F and $w = w_{1}, w_{2} . . . . w_{n}$ be the root of $q (x)$ in $L_{i}$ also $w_{i}$ are distinct.

Similarly $v = v_{1}, v_{2}, . . . v_{n}$ be the root of the $p (x)$ in $L_{i}$ also $v_{i}$ are distinct. Since F is infinite there exist $c \in F$ such that

$F c \neq \frac{v_{i} - v}{w - w_{i}}$ for all $1 \leq i \leq m, 1 \leq j \leq n (1)$

Let $u = v + c w$ and $w \in F (u)$ and $h (x) = p (u - c x) \in F (u) [x]$ . Now w is a root of h(x). So,

$\begin{array}{rcl} h (w) & = & p (u - cw) \\ = & p_{v} \\ = & 0_{F} \end{array}$

Suppose some $w_{j}$ is also a root of h(x). Then $p (u - c w_{j}) = 0_{F}$ so that $u - c w_{j}$ is one of the root of p(x).

Let,

$\begin{array}{rcl} u - {cw}_{j} & = & v_{j} \\ u & = & v_{j} + {cw}_{j} \end{array}$

Therefore,

$v + cw - {cw}_{j} = v_{j} c = \frac{v_{j} - v}{w - w_{j}}$

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 $r (x) \in F [u] (x)$ must have degree 1 hence its root w is in F(u). Since $v = u - c w$ with $u, w \in F (u)$ and also $v \in F (u)$ . Hence $K = F (v, w) \subseteq F (u)$ but

$u = v + c w \in K F (u) \subseteq K F (u) = K$

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 0_F and all power of u and hence contain every element of K. So, $F (u) = K$ .

Most popular questions for Math Textbooks

Find the minimal polynomial of the given element over Q .

(a) $\sqrt{1 + \sqrt{5}}$

(b) $\sqrt{3 i} + \sqrt{2}$

Question: Let $f : A \to B$ be a function. Prove that the following relation $A : u ~ v$ is an equivalence relation of if and only if $f (u) = f (v)$ .

Find the minimal polynomial of $\sqrt{2} + \overset{⏜}{i}$ over $Q$ and over R

Let F,K and L be field such that $F ⊑ K ⊑ L$ . If $[L : F]$ is finite then prove that $[L : K]$ and $[K : F]$ are also finite and both are $\leq [L : F]$ .

let ${V_{1}, . . . . . . . V_{n}}$ be as basic of $V$ over $F$ let $c_{1}, . . . . . c_{n}$ be nonzero element of $F$ , then prove that ${c_{1} v_{1}, c_{2}, . . . . c_{n} v_{n}}$ is also a basic of $V$ over $F$ .

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free