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Expert-verified Found in: Page 94 ### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624 # If F is a field, show that $F\left[x\right]$ is not a field.

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

See the step by step solution

## 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, $\forall x\in F\left[x\right]$ will have an inverse as:

$f\left(x\right)=\sum _{i=0}^{m}{a}_{i}{x}^{i}$

Therefore, in this case, we have:

$xf\left(x\right)=\sum _{i=0}^{m}{a}_{i}{x}^{i+1}=1$

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. ### Want to see more solutions like these? 