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Prove or disprove: a commutative unital ring that is not an integral domain may be contained in a field.

Looking at the definitions, I thought that an integral domain was a ring with unity.

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No -- an integral domain is a ring (with unity) without zero divisors. –  Henning Makholm Feb 13 '12 at 0:44

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up vote 4 down vote accepted

An integral domain is more than a ring with unity. An integral domain must be commutative and not have divisors of zero. A divisor of zero of a ring $R$ is an element $r\in R\setminus\{0\}$ for which there exists another element $s\in R\setminus\{0\}$ such that $rs=0$.

Try proving that a field cannot have divisors of zero. This will lead you to the conclusion that a commutative unital ring that has divisors of zero cannot be contained in a field as a subring. It may of course be contained as a subset.

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Thank you - I realized my mistake was in writing the definition down incorrectly. –  user24700 Feb 13 '12 at 0:57
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I see some glaring errors here. The last paragraph sounds totally erroneous to me! I'll downvote and I'll retract that once the corrections are made. $(\mathbb Z, +, .) \subset (\mathbb R, +, .)$. What is probably meant is about integral domain. I am not sure about the last sentence either! –  user21436 Feb 13 '12 at 22:19
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@KannappanSampath You're right of course. I'm correcting the post, thanks. For the last sentence, take $\{1,2,3,4,5,6\}\subset\mathbb R$ and biject this set with $\mathbb Z_6$ to obtain a ring structure via transport of structure. Of course, this is a very strange thing to do. –  user23211 Feb 13 '12 at 23:20
    
Agreed about the last statement. Please ping me here or at the Chat so I can retract the $ \downarrow$ vote :) –  user21436 Feb 13 '12 at 23:27

An integral domain is a special type of commutative, unital ring - that is, a commutative, unital ring is only an integral domain if it satisfies a certain extra property.

Specifically, a commutative, unital ring $R$ is called an integral domain when the following statement is true: $$\text{For all }a,b\in R,\; ab=0\implies a=0\,\text{ or }\,b=0.$$ Therefore, a commutative, unital ring $R$ is not an integral domain when the negation of the above statement is true: $$\text{There exist }a,b\in R\text{ such that }ab=0,\text{ but }a\neq0\text{ and }b\neq0.$$ Suppose that a non-integral-domain $R$ were a subring of a field $F$. What do you know about nonzero elements of a field? Do you see how can you apply that to reach a contradiction?

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