Subring of an integral domain is an integral domain

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We know that an integral domain is a commutative ring with unity and no zero-divisors. It is obvious that if $R$ is an integral domain and $S$ is a subring of $R$ that $S$ must also be commutative, and if $a,bin S$ and $ab=0$, then $a=0$ or $b=0$. But I'm having trouble proving that $S$ has unity, does the unity of $R$ have to be the unity of $S$?



Also what's a good counterexample for a subring of a field that is not a field.




abstract-algebra ring-theory




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  • A subring WITH unity of an integral domain is a integral domain
    – L.F. Cavenaghi
    Nov 14 '15 at 16:28














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We know that an integral domain is a commutative ring with unity and no zero-divisors. It is obvious that if $R$ is an integral domain and $S$ is a subring of $R$ that $S$ must also be commutative, and if $a,bin S$ and $ab=0$, then $a=0$ or $b=0$. But I'm having trouble proving that $S$ has unity, does the unity of $R$ have to be the unity of $S$?



Also what's a good counterexample for a subring of a field that is not a field.




abstract-algebra ring-theory




share|cite|improve this question



















  • A subring WITH unity of an integral domain is a integral domain
    – L.F. Cavenaghi
    Nov 14 '15 at 16:28












up vote
1
down vote

favorite









up vote
1
down vote

favorite











We know that an integral domain is a commutative ring with unity and no zero-divisors. It is obvious that if $R$ is an integral domain and $S$ is a subring of $R$ that $S$ must also be commutative, and if $a,bin S$ and $ab=0$, then $a=0$ or $b=0$. But I'm having trouble proving that $S$ has unity, does the unity of $R$ have to be the unity of $S$?



Also what's a good counterexample for a subring of a field that is not a field.




abstract-algebra ring-theory




share|cite|improve this question











We know that an integral domain is a commutative ring with unity and no zero-divisors. It is obvious that if $R$ is an integral domain and $S$ is a subring of $R$ that $S$ must also be commutative, and if $a,bin S$ and $ab=0$, then $a=0$ or $b=0$. But I'm having trouble proving that $S$ has unity, does the unity of $R$ have to be the unity of $S$?



Also what's a good counterexample for a subring of a field that is not a field.





abstract-algebra ring-theory





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asked Nov 14 '15 at 16:25









Jihoon

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  • A subring WITH unity of an integral domain is a integral domain
    – L.F. Cavenaghi
    Nov 14 '15 at 16:28
















  • A subring WITH unity of an integral domain is a integral domain
    – L.F. Cavenaghi
    Nov 14 '15 at 16:28















A subring WITH unity of an integral domain is a integral domain
– L.F. Cavenaghi
Nov 14 '15 at 16:28




A subring WITH unity of an integral domain is a integral domain
– L.F. Cavenaghi
Nov 14 '15 at 16:28










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Usually one requires a subring of a unital ring to contain the unit. If you remove this requirement, the result does not hold. For example, $mathbbZ$ is an integral domain, but if we do not require subrings to contain the unit, $2mathbbZ$ is a subring which is not an integral domain.



As for your second question, $mathbbZ$ is a subring of the field $mathbbQ$, but is not itself a field.






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    Usually one requires a subring of a unital ring to contain the unit. If you remove this requirement, the result does not hold. For example, $mathbbZ$ is an integral domain, but if we do not require subrings to contain the unit, $2mathbbZ$ is a subring which is not an integral domain.



    As for your second question, $mathbbZ$ is a subring of the field $mathbbQ$, but is not itself a field.






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      Usually one requires a subring of a unital ring to contain the unit. If you remove this requirement, the result does not hold. For example, $mathbbZ$ is an integral domain, but if we do not require subrings to contain the unit, $2mathbbZ$ is a subring which is not an integral domain.



      As for your second question, $mathbbZ$ is a subring of the field $mathbbQ$, but is not itself a field.






      share|cite|improve this answer























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        up vote
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        Usually one requires a subring of a unital ring to contain the unit. If you remove this requirement, the result does not hold. For example, $mathbbZ$ is an integral domain, but if we do not require subrings to contain the unit, $2mathbbZ$ is a subring which is not an integral domain.



        As for your second question, $mathbbZ$ is a subring of the field $mathbbQ$, but is not itself a field.






        share|cite|improve this answer













        Usually one requires a subring of a unital ring to contain the unit. If you remove this requirement, the result does not hold. For example, $mathbbZ$ is an integral domain, but if we do not require subrings to contain the unit, $2mathbbZ$ is a subring which is not an integral domain.



        As for your second question, $mathbbZ$ is a subring of the field $mathbbQ$, but is not itself a field.







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        answered Nov 14 '15 at 16:29









        Michael Albanese

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