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IS $\mathbb Z_7$ an example of a commutative ring without zero-divisors that is not an integral domain?

I know it is has no zero divisors because it is prime, and that it is a field, but can someone help me understand integral domain?

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No, because it has unity. The definition of an integral domain is (typically) a commutative ring with $1$ such that there are no zero divisors. In particular, every field is an integral domain. –  user61527 Dec 8 '13 at 2:32
    
So, the only thing that would make something NOT an integral domain is the fact that it has unity? –  sprgrl11 Dec 8 '13 at 2:33
    
Or if there are zero divisors or if it fails to be commutative. –  user61527 Dec 8 '13 at 2:34
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@t-bongers : Why give the answer in the comments ? –  Sergio Parreiras Dec 8 '13 at 2:34
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An integral domain is a commutative ring with $1$ and no zero divisors. Is there a part of this definition you don't understand? –  anon Dec 8 '13 at 2:34

1 Answer 1

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$\Bbb Z_7$ is a field, and every field is an integral domain.

On the other hand, every integral domain $D$ determines its field of fractions where fractions are formally built the same way as one constructs $\Bbb Q$ out of $\Bbb Z$.

We can also say that an integral domains $D$ contain the "integer elements" of its fraction field.
(And, as the example of $\Bbb Z_7$ (or, of any field) shows it is possible that all elements of a field are considered "integer" in this sense.)

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