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Abstract Algebra: An Introduction

Found in: Page 94

Book edition 3rd

Author(s) Thomas W Hungerford, David Leep

Pages 608 pages

ISBN 9781111569624

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Short Answer

If F is a field, show that $F [x]$ is not a field.

It is proved that F[x] is not a field.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Polynomial Arithmetic

If any given function R[x] is a ring, then the commutative, associative, and distributive laws hold such that the function f(x)+g(x) exists.

Step 2: Fields:

It is given that F is a field.

Let us assume that F[x] is also a field. Then, $\forall x \in F [x]$ will have an inverse as:

$f (x) = \sum_{i = 0}^{m} a_{i} x^{i}$

Therefore, in this case, we have:

$x f (x) = \sum_{i = 0}^{m} a_{i} x^{i + 1} = 1$

Here, the constant coefficient is zero.

According to the theorem 4.1, it should be 1.

This shows our assumption is invalid.

Hence proved, F[x] is not a field.

Most popular questions for Math Textbooks

Let $f (x), g (x), h (x) \in F [x]$ , with $f (x)$ and $g (x)$ relatively prime. If $f (x) h (x)$ and $g (x) h (x)$ , prove that $f (x) g (x) h (x)$ .

(a) By counting products of the form (x+a) (x+b) show that there are exactly $(p^{2} + p) / 2$ monic polynomials of degree 2 that are not irreducible in $ℤ_{p} [x]$ .

(b) Show that there are exactly $(p^{2} - p) / 2$ monic irreducible polynomials of

degree 2 in $ℤ_{p} [x]$ .

Let $φ : C \to C$ be an isomorphism of rings such that $φ (a) = a$ for each $a \in φ$ . Suppose $r \in C$ is a root of $f (x) \in ℚ [x]$ . Prove that $φ (r)$ is also a root of f (x).

Is the given polynomial irreducible?

(a) $x^{2} - 3 in ℚ [x] ? In ℝ [x] ?$

(b) $x^{2} + x - 2 in ℤ_{3} [x] ? In ℤ_{7} [x] ?$

Show that $\sqrt{p}$ is irrational for every positive prime integer p. [Hint: What are the roots of $x^{2} - p^{2}$ ? Do you prefer this proof to the one in Exercises 30 and 31 of Section 1.37].

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