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I'm confused on what units and non-trivial divisors of zero are when it comes to rings. For example, say I have this finite ring: R=GF(2)[x] mod x^3 + 1 = 0.

Now I know the elements are 0, 1, x, x + 1, x^2, x^2 + 1, x^2 + x, and x^2 + x + 1. Aren't those all non-trivial divisors of zero besides 1? And for the units, I read a unit of a ring is one of those elements, we'll say 'e', such that there exists the inverse of 'e' where e * e^-1 = 1. Do I multiply each of these elements by its inverse to find the units?

Also, does it matter if we can't obtain a field or not to find these two things? I know the example above doesn't give a field, but something like x^3 + x + 1, which has the same elements, does give a field.

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    $\begingroup$ The two rings $\Bbb F_2[x]/(x^3+1)$ and $\Bbb F_2[x]/(x^3+x+1$ do not have the same elements, because their elements are equivalence classes with respect to different equivalence relations. $\endgroup$ – Lubin Sep 21 '15 at 3:07
  • $\begingroup$ Hmm, what elements would be different in the x^3 + x + 1?? $\endgroup$ – pfinferno Sep 21 '15 at 3:40
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    $\begingroup$ They’re all different. Zero in the one is the set of polynomials divisible by $x^3+1$, while zero in the other is the set of polynomials divisible by $x^3+x+1$. You will be begging for trouble if you think that the sets are the same just because they are represented by the same polynomials of degree less than $3$. $\endgroup$ – Lubin Sep 21 '15 at 3:48
  • $\begingroup$ Oh no I didn't think the sets were the same, but that, like you said, they are represented by the same polynomials. Like when I set up an addition/multiplication table, I have the same polynomials written down. $\endgroup$ – pfinferno Sep 21 '15 at 3:53
  • $\begingroup$ Right you are, then. Keep it up. $\endgroup$ – Lubin Sep 21 '15 at 3:56
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A unit in a ring is an element which has an inverse (i.e., an $a$ such that there is $b$ such that $a \cdot b = 1$). A zero divisor is an element $a \ne 0$ such that there is a $b \ne 0$ with $a \cdot b = 0$ (here I use the names $1$ for the multiplicative identity and $0$ for the additive one).

The elements of your ring are rather few, you can work out the full multiplication table and check for the above. Bonus is that your ring is commutative, as is easy to prove.

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  • $\begingroup$ Oh I have the addition and multiplication tables written out already. So I'm looking for the row/column intersection where there's a 1 for the units? $\endgroup$ – pfinferno Sep 21 '15 at 1:15
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    $\begingroup$ The units are the elements whose row contains a $1$; the zero-divisors are the elements whose row contains a $0$ (other than in the column headed by zero, of course). $\endgroup$ – Lubin Sep 21 '15 at 3:04
  • $\begingroup$ Okay that makes sense now. Thank you. $\endgroup$ – pfinferno Sep 21 '15 at 3:38
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In a field, all elements other than $0$ are units, and there are no nontrivial zero-divisors. In a finite ring (with 1), any non-unit is a zero-divisor. That's not true in general for infinite rings.

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  • $\begingroup$ Wait, so if all my units besides 0 are the elements, then there's no non-trivial zero divisors? $\endgroup$ – pfinferno Sep 21 '15 at 1:15
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    $\begingroup$ Well, since your polynomial $x^3+1$ is not irreducible, there are going to be zero-divisors in the factor ring. $\endgroup$ – Lubin Sep 21 '15 at 3:02
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    $\begingroup$ Do you mean "if all elements other than $0$ are units, ..."? Since a unit can't be a zero-divisor, that would indeed imply there are no non-trivial zero-divisors. But that's not the case here: as @Lubin mentioned, $x^3 + 1$ is not irreducible, and a factor of $x^3+1$ is a zero-divisor in your ring. $\endgroup$ – Robert Israel Sep 21 '15 at 3:57

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