R[X] is never a field .... Sharp, Exercise 1.29 .... ....

[Math Amateur]

I am reading R. Y. Sharp's book: "Steps in Commutative Algebra" Cambridge University Press (Second Edition) ... ...

I am focused on Chapter 1: Commutative Rings and Subrings ... ...

I need some help with Exercise 1.29 ...

Exercise 1.29 reads as follows:View attachment 8169I am somewhat unsure about how to go about framing a valid and rigorous proof to demonstrate that $$R[X]$$ is never a field ...

But ... maybe the following is relevant ...

Consider $$a_1 X \in R[X]$$ ...

... then if $$R[X]$$ is a field ... there would be a polynomial $$b_0 + b_1 X + \ ... \ ... \ + b_n X^n$$ such that ...

... $$a_1 X ( b_0 + b_1 X + \ ... \ ... \ + b_n X^n ) = 1$$

That is, we would require

$$a_1 b_0 X + a_1 b_1 X^2 + \ ... \ ... \ + a_1 b_n X^{ n + 1} = 1$$ ... ... ... ... ... (1) ... But ... it is impossible for equation (1) to be satisfied as the term on the RHS has only a term in $$X^0$$ while the LHS only has terms in $$X$$ in powers greater than $$0$$ ...Does the above qualify as a formal and rigorous proof ... if not ... what would constitute a formal and rigorous proof ...Hope someone can help ...

Peter

 

[caffeinemachine]

I am reading R. Y. Sharp's book: "Steps in Commutative Algebra" Cambridge University Press (Second Edition) ... ...

I am focused on Chapter 1: Commutative Rings and Subrings ... ...

I need some help with Exercise 1.29 ...

Exercise 1.29 reads as follows:I am somewhat unsure about how to go about framing a valid and rigorous proof to demonstrate that $$R[X]$$ is never a field ...

But ... maybe the following is relevant ...

Consider $$a_1 X \in R[X]$$ ...

... then if $$R[X]$$ is a field ... there would be a polynomial $$b_0 + b_1 X + \ ... \ ... \ + b_n X^n$$ such that ...

... $$a_1 X ( b_0 + b_1 X + \ ... \ ... \ + b_n X^n ) = 1$$

That is, we would require

$$a_1 b_0 X + a_1 b_1 X^2 + \ ... \ ... \ + a_1 b_n X^{ n + 1} = 1$$ ... ... ... ... ... (1) ... But ... it is impossible for equation (1) to be satisfied as the term on the RHS has only a term in $$X^0$$ while the LHS only has terms in $$X$$ in powers greater than $$0$$ ...Does the above qualify as a formal and rigorous proof ... if not ... what would constitute a formal and rigorous proof ...Hope someone can help ...

Peter

The proof is correct.

 

[Math Amateur]

The proof is correct.

Thanks for confirming the proof caffeinemachine ... appreciate the help ...

Peter