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I am reading a first course in algebra 7th edition written by John B. Fraleigh. I have seen the following two definitions:

1) A field is a commutative ring in which every nonzero element has multiplicative inverse.

2) An integral domain is a commutative ring with unity 1 and containing no zero divisors.

Then i saw a picture in the book that shows fields as subsets of integral domains like in the following picture:

enter image description here

My question is, how do we understand from these two definitions that fields are subsets of integral domains? In the definition of integral domain, i do not see anything saying that every element in the ring should have an inverse, it just says that there must be a multiplicative identity. Am i missing something or is there something missing in the definitions?

Thank you

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  • $\begingroup$ The answer to your title question is no. This (not amazingly useful) picture means that every field is an integral domain. So assume $F$ is a field and show it contains no zero divisor. $\endgroup$ – Julien May 12 '13 at 14:45
  • $\begingroup$ @julien oh ok, i interpreted it as a subset. Thanks $\endgroup$ – Yasin Razlık May 12 '13 at 14:47
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    $\begingroup$ @CutieKrait , do you mean every field is an integral domain? $\endgroup$ – Dan Rust May 12 '13 at 14:54
  • $\begingroup$ yes‌‌‌‌‌‌‌‌‌‌‌‌. $\endgroup$ – user59671 May 12 '13 at 14:59
  • $\begingroup$ every finite integral domain is a field and thus has inverse for each element $\endgroup$ – Bhaskar Vashishth Feb 6 '16 at 4:46
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No, it is not necessarily the case that every element in an integral domain has a multiplicative inverse.

Every field is an integral domain, but not every integral domain is a field. Hence we have that the set of all fields is a proper subset of the set of all integral domains.

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  • $\begingroup$ Thank you, i understand now $\endgroup$ – Yasin Razlık May 12 '13 at 14:53
  • $\begingroup$ @amWhy: that is excellent feedback! :-0 +1 $\endgroup$ – Amzoti May 13 '13 at 0:30
  • $\begingroup$ Yes, that's always rewarding, @Amzoti! $\endgroup$ – Namaste May 13 '13 at 0:31
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As already noticed, every field is an integral domain but the converse doesn't hold (take for example $\mathbb{Z}$). In order to show that a field $F$ is an integral domain, suppose $a,b\in F$ are such that $ab=0$ and assume $a$ is not zero. Then since every non zero element in a field is invertible, one has $$ab=0\implies a^{–1}ab=0\implies b=0. $$ Similarly if $b\neq 0$ one gets that a must be zero and hence $F$ is an integral domain.

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