Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

11 E

Expert-verified

Abstract Algebra: An Introduction

Found in: Page 365

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 F has characteristic 0 and K is a splitting field of $f (x) \in F [x]$ . If is the greatest common divisor of $f (x)$ and $f' (x)$ and $h (x) = f (x) / d x \in F [x]$ , prove

(a) $f (x)$ and $h (x)$ have the same roots in K.
(b) $h (x)$ is separable

It is proved that $f (x)$ and have same roots in .
It is proved that is separable.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Describe the condition for separable field

A polynomial $p (x)$ over a given field K is separable if its roots are distinct in an algebraic closure of K. that is the number of roots is equal to the degree of the polynomial.

Prove that f(x) and h(x) have the same roots in K.

Claim: $f (x)$ and $h (x)$ have same roots in $K$ .

Let $c$ be the root of $h (x)$ . This implies that $h (c) = 0$ .

Consider $h (x) = \frac{f (x)}{d (x)}$

Substitute $x = c$ in $h (x)$ .

$h (c) = \frac{f (c)}{d (c)}$

Since $h (c) = 0$ . Therefore, $\frac{f (c)}{d (c)} = 0$ .

That means $f (c) = 0$

Thus, $f (c) = 0$ indicates that $c$ is the root of $f (x)$ .

Therefore, $f (x)$ and $h (x)$ have same roots in $K$ .

Prove that is h(x) separable

Claim: $h (x)$ is separable.

So, just show that $h (x)$ and $h' (x)$ both are relatively prime.

That is $\gcd (h (x), h' (x)) = 1$ .

Consider $h (x) = \frac{f (x)}{d (x)}$ .

Differentiate $h (x)$ with respect to .

$\begin{matrix} h' (x) = \frac{f' (x) d (x) - d' (x) f (x)}{{(d (x))}^{2}} \\ = \frac{f' (x)}{d (x)} - \frac{f (x)}{d (x)} \frac{d' (x)}{d (x)} \end{matrix}$

Suppose on the contrary $h (x)$ and $h' (x)$ both are not relatively prime.

Let is the common root of $h (x)$ and $h' (x)$ , this implies that $h (c) = 0$ and $h' (c) = 0$ .

Since $h (c) = 0$ implies that $f (c) = 0$ . Now,

$h' (c) = \frac{f' (c) d (c) - d' (c) f (c)}{{(d (c))}^{2}}$

So $h' (c) = 0$ if $f' (c) = 0$ . But $f' (c) = 0$ this contradicts the fact that $d (x)$ is the greatest common divisor of $f (x)$ and $f' (x)$ .

Therefore, the assumption is wrong.

Thus, $h (x)$ and $h' (x)$ both are relatively prime.. so, claim follows and $h (x)$ $h (x)$ is separable.

Hence, it is proved that is separable.

Most popular questions for Math Textbooks

Question: For any vector $v \in V$ and any element $a \in F$ , prove that

(a) $O_{F V} = O_{V}$

(b) $a 0_{v} = 0_{v}$

(c) $- (av) = (- a) V = a (- a)$

If $u \in K$ is transcendental over $F$ , prove that $F (u) ≅ F (x)$ , where $F (x)$ is the field of quotients of $F (x)$ .

Question: Let E be the field of all element of K that are algebraic over F. Prove that every element of the set K-E is transcendental.

If $u$ is algebraic over $F$ and is a normal extension of role="math" localid="1658901983256" $F$ prove that $K$ is a splitting field over data-custom-editor="chemistry" $F$ of the minimal polynomial of $u$ .

: Let $a \in F$ . Describe the congruence classes in $F (x)$ modulo the polynomial $(x - a)$ .

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