I have this doubt. In a unitary and commutative ring $$Z_m = \{[0]_m, [1]_m,\ ...\ ,\ [m - 1]_m\}$$

There are only two "kind" of elements: invertible and zero divisors. Is it true to say that the only regular elements are the invertible elements?



If by regular, you mean not a zero divisor, then yes it is true. This would follow by your observation that there are only two "kinds" of elements in $\mathbb{Z}_m$.

Generally this is not true. For instance in the polynomial ring $\mathbb{R}[x]$, $x$ is neither a zero divisor nor invertible.

  • $\begingroup$ By regular I mean $a \in A: \forall b,c \in A, a*b = a*c \Longrightarrow b=c$ $\endgroup$ – user1365914 Jun 21 '16 at 20:47
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    $\begingroup$ Alright. That is equivalent to not being a zero divisor. $\endgroup$ – Ken Duna Jun 21 '16 at 20:51
  • $\begingroup$ So the "opposite" of zero divisor is being a regular element? I'm sorry if I do make confusion but English is not my primary language and I'm translating all the terms I find in my language from my book to English :) $\endgroup$ – user1365914 Jun 21 '16 at 20:52
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    $\begingroup$ That is correct. Many times people refer to regular elements as "non-zero-divisors". $\endgroup$ – Ken Duna Jun 21 '16 at 20:53
  • $\begingroup$ Today I Learned "regular" elements are called "non zero divisors" in English :). Thank you a lot, have a great day @Ken Duna ! $\endgroup$ – user1365914 Jun 21 '16 at 20:54

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