I am aware about these algebraic structures which are always contained in each other $Rings\supset Commutative\ rings\ with\ unity\supset Integral\ Domains\supset fields$

There have been many questions in my mind about the existence of units and zero divisors Is it possible that in an ID we have elements which are not units? Is it possible to have a commutative ring with unity which have elements

  1. that are both units and zero divisors
  2. that are units but not zero divisors
  3. that are not units but zero divisors
  4. that are not both

how can we prove or negate each statement


1 Answer 1


A very interesting result about domains and units, which you should try proving yourself, is that every finite integral domain is a field. Hence if we are looking for a domain with non-unit elements, we should look at infinite domains. One example is $\mathbb{Z}$, which is an ID but the only units are $1$ and $-1$.

  1. A unit cannot be a zero divisor: let $a, a^{-1}, b \in R$, such that $a.b=0, a \neq 0$. Then, multiplying by $a^{-1}$ on the left, we see that $a^{-1}.a.b=0 \implies b=0$, hence $a$ is not a zero-divisor. Note that we didn't need the fact that multiplication is commutative.
  2. Every non-zero element of a field is a unit but not a zero-divisor.
  3. In $\mathbb{Z}/ \mathbb{6Z}$, which is not a domain, $\bar{2}.\bar{3}=\bar{0}$ but neither are units in $\mathbb{Z}/ \mathbb{6Z}$
  4. Again, we take a look at infinite IDs, like $\mathbb{Z}$ where there are no zero divisors and no units apart from $1$ and $-1$.

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