# Polynomial rings - zero divisors

• December 25th 2012, 03:31 PM
Bernhard
Polynomial 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.
• December 25th 2012, 10:04 PM
Deveno
Re: Polynomial rings - zero divisors
hint: what does the homomorphism used in another post suggest?
• December 26th 2012, 04:04 PM
Bernhard
Re: Polynomial rings - zero divisors
I presume you mean use the following ring homomorphism:

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

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

But ... where to now??

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

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

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

But this would mean that

$[a_n] [b_m] = 0 [mod 2]$ which means $[a_n] = 0$ or $[b_m] = 0$ 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
• December 26th 2012, 10:50 PM
Deveno
Re: Polynomial rings - zero divisors
of course it has zero divisors, for example:

every polynomial of the form 2(f(x)) is a zero-divisor.
• December 27th 2012, 02:19 PM
Bernhard
Re: 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 $\phi$ 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