Does every element of an integral domain have an inverse?

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:

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

• 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. – Julien May 12 '13 at 14:45
• @julien oh ok, i interpreted it as a subset. Thanks – Yasin Razlık May 12 '13 at 14:47
• @CutieKrait , do you mean every field is an integral domain? – Dan Rust May 12 '13 at 14:54
• yes‌‌‌‌‌‌‌‌‌‌‌‌. – user59671 May 12 '13 at 14:59
• every finite integral domain is a field and thus has inverse for each element – Bhaskar Vashishth Feb 6 '16 at 4:46

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.