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

[Math Amateur]

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:

Homework Equations

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

The Attempt at a Solution

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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

 

[andrewkirk]

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.

 

[Math Amateur]

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

 

[Math Amateur]

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

 

[andrewkirk]

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.

 

[Math Amateur]

Thanks again Andrew ...

Really value your help and assistance ...

Peter