MHB R commutative ring then R[x] is never a field

[Math Amateur]

I am reading Joseph Rotman's book Advanced Modern Algebra.

I need help with Problem 2.20 on page 94.

Problem 2.20 reads as follows:

2.20. Prove that if R is a commutative ring then R[x] is never a field.

Could someone please help me get started on this problem.

Peter

***EDIT*** Presumably one shows there are polynomials that do not have inverses??/

 

[Math Amateur]

I am reading Joseph Rotman's book Advanced Modern Algebra.

I need help with Problem 2.20 on age 94.

Problem 2.20 reads as follows:

2.20. Prove that if R is a commutative ring then R[x] is never a field.

Could someone please help me get started on this problem.

Peter

***EDIT*** Presumably one shows there are polynomials that do not have inverses??/

I have been reflecting on my own post ... and two points come to mind regarding Rotman's Problem 2.20.

Firstly since R is a commutative ring we cannot guarantee that the constant polynomials have inverses ... so on this ground alone R[x] can not be a field ... ... unless of course, R is a field in which case the constant polynomials will have inverses ... but then ...

Secondly, polynomials such as $$ p(x) = x \text{ or } x^2 \text{ or } x^3 \text{ etc. } $$ would require 'polynomials' like $$ x^{-1} \text{ or } x^{-2} \text{ or } x^{-3} \text{ etc. } $$ and these do not come under the definition of polynomials ...

Can someone please confirm that the above analysis presents an answer to Rotman's problem 2.2)? I would appreciate any critique of the "proof".

Peter

 

[Deveno]

Claim: $x$ has no inverse in $R[x]$, if $R \neq \{0\}$.

Proof (by contradiction):

Suppose it did, so we have some polynomial:

$a_0 + a_1x + \cdots + a_nx^n \in R[x]$ with:

$x(a_0 + a_1x + \cdots + a_nx^n) = 1$, that is:

$0 + a_0x + a_1x^2 + \cdots + a_nx^{n+1} = 1 + 0x + 0x^2 + \cdots + 0x^{n+1}$.

Equating coefficients (which we can do since $R$ is commutative, and thus we can view $R[x]$ as an $R$-module with BASIS: $\{1,x,x^2,\dots\}$-recall that commutative rings have the invariant basis property), we have:

$a_0 = a_1 = \dots = a_n = 0$,
$0 = 1$,

which means that $R[x]$ is a TRIVIAL ring, contradiction.

(and this means, yes, you're right!).

 

[Math Amateur]

Claim: $x$ has no inverse in $R[x]$, if $R \neq \{0\}$.

Proof (by contradiction):

Suppose it did, so we have some polynomial:

$a_0 + a_1x + \cdots + a_nx^n \in R[x]$ with:

$x(a_0 + a_1x + \cdots + a_nx^n) = 1$, that is:

$0 + a_0x + a_1x^2 + \cdots + a_nx^{n+1} = 1 + 0x + 0x^2 + \cdots + 0x^{n+1}$.

Equating coefficients (which we can do since $R$ is commutative, and thus we can view $R[x]$ as an $R$-module with BASIS: $\{1,x,x^2,\dots\}$-recall that commutative rings have the invariant basis property), we have:

$a_0 = a_1 = \dots = a_n = 0$,
$0 = 1$,

which means that $R[x]$ is a TRIVIAL ring, contradiction.

(and this means, yes, you're right!).

Thanks Deveno ... Definitely needed help with the formal proof ... Appreciate the help

Peter