divisor of zero and an invertible element in $\mathbb Z_8[x]$

Question

Find an example of a divisor of zero and an invertible element in $\mathbb Z_8[x]$. (Find nonconstant examples).

So I've read around on this site and through other question and it seems that any unit or zero divisors of $R$ will also be a unit or zero divisors in $R[x]$ thus $x,3x,5x,7x$ or $x^2,3x^2,5x^2,7x^2$ and so on would be divisors of zero.

As for the invertible element, $2x,4x,6x$ would be invertible elements.

Does this seem correct?

EDIT: I do seem to have my units and my zero divisors flipped. So $x,3x,5x,7x$ are units and are invertible elements. $2x,4x,6x$ are zero divisors.

The surest way to convince someone (like a professor) that an element is invertible is to exhibit and check its supposed inverse. — Randall, Nov 06 '17 at 15:43
OK, so $(2x)*4=0$, and if $2x$ is a unit, then inverting on the left you get $4=0$... hmmm, problem. As for your zero divisors, what do you multiply $x$ by to get $0$? — rschwieb, Nov 06 '17 at 15:45
@rschwieb then how is 2 an invertible element in $\mathbb Z_8$? If $2\cdot 4=0$ and 2 is a unit then $4=0$? — K Math, Nov 06 '17 at 15:46
@Kelly R: $4x$ is not a unit. $2\cdot 4x=0$ in the ring. Hence $4x$ is a zero-divisor; and it can't simultaneously be a unit. — paw88789, Nov 06 '17 at 16:04
It appears that you've got your units and zero divisors backwards. — zipirovich, Nov 06 '17 at 16:06

score 2 · Answer 1 · answered Nov 06 '17 at 16:12

thus $x,3x,5x,7x$ or $x^2,3x^2,5x^2,7x^2$ and so on would be divisors of zero.

Well... what would you propose that multiplies by each one and yields zero? Actually none of these are zero divisors... nor are they units.

As for the invertible element, $2x$, $4x$, $6x$ would be invertible elements.

Actually, all of them multiply with $4$ to get zero, so they are all zero divisors, and cannot be units.

So now you have the answer to at least half of your question. The problem of finding a nonconstant unit remains. Here's a hint:

Try to get $(1+ax)(1+bx)=1$. To do this, get $ab=0$ and $a+b=0$ simultanously in the ring.

Could a and b both be 4 or must they be different from one another? — K Math, Nov 06 '17 at 16:15
@KellyR $a=b=4$ is just fine. Nobody said they have to be different. Good job finding that solution! — rschwieb, Nov 06 '17 at 16:15

Dietrich Burde · Answer 2 · 2017-11-06T15:48:26.190

0

For zero-divisors in polynomial rings see this duplicate:

Zero divisors in polynomial rings

edited Nov 06 '17 at 15:48

answered Nov 06 '17 at 15:46

Dietrich Burde

130,978

divisor of zero and an invertible element in $\mathbb Z_8[x]$

2 Answers2