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

  • #1
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Homework Statement



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:

Sharp - Exercise 1.29 ... .png



Homework Equations



Sharp's definitions, notation and remarks regarding R[X] are as follows:

Sharp - 1 - Defn of R[X] and R[[X]] ... PART 1.png

Sharp - 2 - Defn of R[X] and R[[X]] ... PART 2 ... .png



The Attempt at a Solution


[/B]
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
 

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  • #2
andrewkirk
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The proof that it cannot be a field is quite short. We observe that, for nonzero ##r\in R##, the polynomial ##X+r## cannot have a multiplicative inverse. Say there is an inverse that is a ##k##-degree polynomial, which we write as ##\sum_{j=0}^k c_j X^j## where ##c_k\neq 0##. Now the product of that with ##r+X## gives a polynomial for which the coefficient of ##X^{k+1}## is ##c_k##. But the product must equal 1, so we must have either
  • ##k=-1##, which is not allowed, as the degree of a polynomial must be non-negative, or
  • ##c_k=0##, which contradicts our supposition that it is nonzero.
Since either way we get a contradiction, we must reject the assumption that ##X+r## has a multiplicative inverse.

The difference of this from your proof is that in my polynomial without an inverse the coefficient of X is 1, whereas in yours is ##a_1##. Using a non-unit coefficient leaves open the possibility that the product of that with another ring element may be zero, unless we assume R is an integral domain, which we should not have to assume.

For the integral domain part, assume R is not an integral domain, so there are two nonzero elements r, s whose product is zero. Then use those to make two nonzero polynomials whose product is zero.
 
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  • #3
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The proof that it cannot be a field is quite short. We observe that, for nonzero ##r\in R##, the polynomial ##X+r## cannot have a multiplicative inverse. Say there is an inverse that is a ##k##-degree polynomial, which we write as ##\sum_{j=0}^k c_j X^j## where ##c_k\neq 0##. Now the product of that with ##r+X## gives a polynomial for which the coefficient of ##X^{k+1}## is ##c_k##. But the product must equal 1, so we must have either
  • ##k=-1##, which is not allowed, as the degree of a polynomial must be non-negative, or
  • ##c_k=0##, which contradicts our supposition that it is nonzero.
Since either way we get a contradiction, we must reject the assumption that ##X+r## has a multiplicative inverse.

The difference of this from your proof is that in my polynomial without an inverse the coefficient of X is 1, whereas in yours is ##a_1##. Using a non-unit coefficient leaves open the possibility that the product of that with another ring element may be zero, unless we assume R is an integral domain, which we should not have to assume.

For the integral domain part, assume R is not an integral domain, so there are two nonzero elements r, s whose product is zero. Then use those to make two nonzero polynomials whose product is zero.

Thanks Andrew ... most helpful ...

Peter
 
  • #4
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Just a thought Andrew ... my equation (1) could not be true even if a number (or even all) of the terms ##a_1 b_0, a_1 b_1, \ ... \ ... \ , a_1 b_n## were equal to zero ... so surely even if ##R## was not an integral domain, my equation (1) would not have a solution ...

Am I missing something ... ?

Peter
 
  • #5
andrewkirk
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Actually your proof is fine. I didn't read it until after I wrote mine down, as I assumed based on your first line that you didn't think much of it. I think you are entitled to be more confident in your work.

Your proof uses the fact that the product has no nonzero terms of order 0 and so cannot be 1, whereas mine uses the fact that the product has a nonzero term of order greater than 0 and hence cannot be 1. So they're more different than I first thought. Neither relies on R being an integral domain.
 
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  • #6
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Thanks again Andrew ...

Really value your help and assistance ...

Peter
 

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