Divisors of Zero in Rings: A[x] and the Impossibility in Polynomial Rings

[Kiwi1]

There are rings such as P_3 in which every element not equal to 0 or 1 is a divisor of zero. Explain why this is not possible in any ring of polynomials A[x], even when A is not an integral domain.

I can't see how to answer this.

If I define A as Z_3 with the usual addition and define multiplication trivially as a*b=0 then I have A[x] where every element is a divisor of zero.

 

[Euge]

Hi Kiwi,

By choosing the trivial action for $\Bbb Z_3$, you would have $1 = 0$ and so $\Bbb Z_3[x]$ would be trivial. You're supposed to have $1\neq 0$.

 

[Kiwi1]

Thanks Euge, I still can't see a solution.

The question does not expressly forbid 1=0 in fact see their example P_3, an 8 element ring where 1=0=empty set.

P_3 also has the interesting property that if p is in P_3 then p^2=0.

Actually p_ip_j=0 iff i=j.

so if a(x)b(x)=0 with a(x)=p_ix+p_j and b(x)=b_nx^n+...+b_0

the constant term must be 0 so b_o=p_j

The term in x must be zero so p_ib_0+b_0b_1=0

I'm lost.

 

[Euge]

Sorry if I wasn't clear. If $A$ is the trivial ring (so $A[x]$) is trivial, then the statement "every element not equal to $0$ or $1$ is a divisor of zero" is vacuously true for $A[x]$.

To form a polynomial ring $A[x]$, you assume $A$ is commutative. Otherwise, you don't have enough information to make $A[x]$ into a ring. You would need an endomorphism $\sigma : A\to A$ and define a multiplication $xa := \sigma(a)x$ (for $a\in A$) to make $A[x]$ into a ring for noncommutative $A$.

Check back to see precisely what they mean by "all rings", as some may consider all rings to be commutative with identity.

 

[Kiwi1]

Thanks, I think I have it.

The introductory paragraphs of my text define a polynomial:

"Let A be a commutative ring with unity, and x an arbitrary symbol ..."

So if A is any such ring and in that ring a,b not zero such that ab=0. Then I can prove that ax+1 is not a divisor of zero.

Assume that ax +1 is a divisor of zero with 'partner' of the form b(x) where:

b(x)=b_nx^n+ ... + b_0

Then

(ax+1)(b_nx^n+ ... + b_0)=0, so b_0=0, and:

(ax+1)(b_nx^{n-1}+ ... + b_1)x=0, so b_1=0

continuting as required we eventually get to:

(ax+1)(b_n)x^n=0, so b_n=0

That is b(x)=0 and ax+1 is not a divisor of zero.