Embedding of torsion free module

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If R is any commutative ring with 1 and M be any finitely generated torsion free R-module. Can we embed M into free R-module of finite rank?



As we know that it can be embedded if R is a domain. Since we can then take the tensor product of M with K(quotient field of R) and find suitable basis so that we can embed. But I don’t know whether it is true for any commutative ring or not?




abstract-algebra commutative-algebra algebraic-number-theory




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  • What definition are you using for torsion-freeness for a non-domain?
    – Mohan
    2 days ago










  • m is said to be torsion if rm = 0 for some non zero r.
    – Sunny Rathore
    2 days ago










  • got it..it means that R must be domain.
    – Sunny Rathore
    2 days ago














up vote
0
down vote

favorite












If R is any commutative ring with 1 and M be any finitely generated torsion free R-module. Can we embed M into free R-module of finite rank?



As we know that it can be embedded if R is a domain. Since we can then take the tensor product of M with K(quotient field of R) and find suitable basis so that we can embed. But I don’t know whether it is true for any commutative ring or not?




abstract-algebra commutative-algebra algebraic-number-theory




share|cite|improve this question



















  • What definition are you using for torsion-freeness for a non-domain?
    – Mohan
    2 days ago










  • m is said to be torsion if rm = 0 for some non zero r.
    – Sunny Rathore
    2 days ago










  • got it..it means that R must be domain.
    – Sunny Rathore
    2 days ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











If R is any commutative ring with 1 and M be any finitely generated torsion free R-module. Can we embed M into free R-module of finite rank?



As we know that it can be embedded if R is a domain. Since we can then take the tensor product of M with K(quotient field of R) and find suitable basis so that we can embed. But I don’t know whether it is true for any commutative ring or not?




abstract-algebra commutative-algebra algebraic-number-theory




share|cite|improve this question











If R is any commutative ring with 1 and M be any finitely generated torsion free R-module. Can we embed M into free R-module of finite rank?



As we know that it can be embedded if R is a domain. Since we can then take the tensor product of M with K(quotient field of R) and find suitable basis so that we can embed. But I don’t know whether it is true for any commutative ring or not?





abstract-algebra commutative-algebra algebraic-number-theory





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked 2 days ago









Sunny Rathore

49327




49327











  • What definition are you using for torsion-freeness for a non-domain?
    – Mohan
    2 days ago










  • m is said to be torsion if rm = 0 for some non zero r.
    – Sunny Rathore
    2 days ago










  • got it..it means that R must be domain.
    – Sunny Rathore
    2 days ago
















  • What definition are you using for torsion-freeness for a non-domain?
    – Mohan
    2 days ago










  • m is said to be torsion if rm = 0 for some non zero r.
    – Sunny Rathore
    2 days ago










  • got it..it means that R must be domain.
    – Sunny Rathore
    2 days ago















What definition are you using for torsion-freeness for a non-domain?
– Mohan
2 days ago




What definition are you using for torsion-freeness for a non-domain?
– Mohan
2 days ago












m is said to be torsion if rm = 0 for some non zero r.
– Sunny Rathore
2 days ago




m is said to be torsion if rm = 0 for some non zero r.
– Sunny Rathore
2 days ago












got it..it means that R must be domain.
– Sunny Rathore
2 days ago




got it..it means that R must be domain.
– Sunny Rathore
2 days ago















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