Let R = Z/4Z

Find all the zero divisors of R[x] ...

... and then, further, find all elements of R[x] that are neither units nor zero divisors.

Printable View

- Dec 25th 2012, 03:31 PMBernhardPolynomial rings - zero divisors
Let R = Z/4Z

Find all the zero divisors of R[x] ...

... and then, further, find all elements of R[x] that are neither units nor zero divisors. - Dec 25th 2012, 10:04 PMDevenoRe: Polynomial rings - zero divisors
hint: what does the homomorphism used in another post suggest?

- Dec 26th 2012, 04:04 PMBernhardRe: Polynomial rings - zero divisors
I presume you mean use the following ring homomorphism:

(Z/4Z) [x] ---> (Z/2Z)[x] [then Z/2Z is a field and the units of Z/2Z[x] are simply the units of Z/2Z ... but zero divisors???)

Then .... ??/

If a(x) is a zero divisor in (Z/4Z)[x] then a(x)b(x) = 0 for some b(x) with a(x) and b(x)

Then with a(x) and b(x) ...

But ... where to now??

What can one do with a(x) and b(x) ... how to progress ... that is what does this imply about and

Maybe we could proceed by arguing that implies that

But this would mean that

which means or or both

This seems to imply that Z/2Z[x] has no zero divisors .... but where to from here ... need some further clarification and help.

Peter - Dec 26th 2012, 10:50 PMDevenoRe: Polynomial rings - zero divisors
of course it has zero divisors, for example:

every polynomial of the form 2(f(x)) is a zero-divisor. - Dec 27th 2012, 02:19 PMBernhardRe: Polynomial rings - zero divisors
Are you referring to Z/4Z which certainly has zero divisors of that form?

I was referring to Z/2Z[x] which (I think) has no zero divisors [2Z is a prime ideal so Z/2Z has no zero divisors and thus Z/2Z[x] has no zero divisors.

Is that correct?

I would like your guidance on the use of the homomorphism Z/4Z[x] ---> Z/2Z[x]

It seems that the kernel of the transformation is those polynomials of the form 2(f(x)) which are the zero divisors of Z/4Z. Can we use this somehow?

Peter