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 \(\displaystyle p(x) = x \text{ or } x^2 \text{ or } x^3 \text{ etc. } \) would require 'polynomials' like \(\displaystyle x^{-1} \text{ or } x^{-2} \text{ or } x^{-3} \text{ etc. } \) 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

 

[blue_raver22]

Sure, I'd be happy to help you get started on this problem. First, let's review the definitions of a commutative ring and a field.

A commutative ring is a set R with two operations, addition and multiplication, that satisfy the following properties:

1. Addition and multiplication are both associative.
2. Addition and multiplication are both commutative.
3. The ring has an additive identity element, denoted by 0.
4. The ring has a multiplicative identity element, denoted by 1.
5. For any elements a, b, c in the ring, a(b+c) = ab + ac (left distributive property).

A field is a commutative ring with the additional property that every nonzero element has a multiplicative inverse. This means that for every nonzero element a in the field, there exists an element b such that ab = 1.

Now, let's consider the polynomial ring R[x], which consists of all polynomials with coefficients in R. We can define addition and multiplication of polynomials in the same way as for numbers, using the distributive property. For example, (3x^2 + 2x + 1) + (x^2 + 3) = 4x^2 + 2x + 4.

To show that R[x] is not a field, we need to show that there exists a polynomial that does not have a multiplicative inverse. Let's consider the polynomial x+1. If we multiply this polynomial by any other polynomial, the result will always have a nonzero constant term (since x times any other polynomial will have a nonzero constant term, and adding 1 will not change this). Therefore, there is no polynomial in R[x] that can be multiplied by x+1 to give the constant polynomial 1, which is the multiplicative identity element. This means that x+1 does not have a multiplicative inverse in R[x], and therefore R[x] is not a field.

I hope this helps you get started on the problem. Good luck!