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

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The discussion centers on Exercise 1.29 from R. Y. Sharp's "Steps in Commutative Algebra," which asserts that the polynomial ring $$R[X]$$ is never a field. The user presents a proof attempt involving the polynomial $$a_1 X$$ and demonstrates that if $$R[X]$$ were a field, it would lead to a contradiction in the form of an impossible equation. The proof is confirmed as correct by another participant, reinforcing the conclusion that $$R[X]$$ cannot satisfy the field properties.

PREREQUISITES
  • Understanding of polynomial rings, specifically $$R[X]$$.
  • Familiarity with the definitions of fields and their properties.
  • Knowledge of commutative algebra concepts, particularly rings and subrings.
  • Ability to construct and analyze mathematical proofs.
NEXT STEPS
  • Study the properties of polynomial rings in depth, focusing on their structure and limitations.
  • Learn about the criteria for a set to be classified as a field.
  • Explore additional exercises in R. Y. Sharp's "Steps in Commutative Algebra" to reinforce understanding.
  • Investigate examples of fields and non-fields in algebra to solidify concepts.
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Mathematics students, particularly those studying abstract algebra, educators teaching commutative algebra, and anyone interested in the foundational properties of polynomial rings.

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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
 
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Peter said:
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.
 
caffeinemachine said:
The proof is correct.
Thanks for confirming the proof caffeinemachine ... appreciate the help ...

Peter
 

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