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.

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- 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:

$\displaystyle \phi: $ (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) $\displaystyle \ne 0 $ and b(x) $\displaystyle \ne 0 $

Then $\displaystyle \phi (a(x)b(x)) = \phi (a(x)) \phi (b(x)) = \phi (0) = 0 $ with a(x) $\displaystyle \ne 0 $ and b(x) $\displaystyle \ne 0 $ ...

But ... where to now??

What can one do with a(x) $\displaystyle \ne 0 $ and b(x) $\displaystyle \ne 0 $ ... how to progress ... that is what does this imply about $\displaystyle \phi (a(x)) $ and $\displaystyle \phi (b(x)) $

Maybe we could proceed by arguing that $\displaystyle \phi (a(x)) \phi (b(x)) = 0 $ implies that

$\displaystyle ([a_n] x^n + [a_{n-1} x^{n-1} + ... + a_1 x + a_0) ([b_m] x^m+ [b_{m-1} x^{m-1} + ... + b_1 x + b_0) = 0 $

But this would mean that

$\displaystyle [a_n] [b_m] = 0 [mod 2]$ which means $\displaystyle [a_n] = 0 $ or $\displaystyle [b_m] = 0 $ 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 $\displaystyle \phi $ 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