Is a Submodule of a Cyclic R-Module Also Cyclic if R is a PID?

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SUMMARY

If R is a Principal Ideal Domain (PID), then any submodule of a cyclic R-module is also cyclic. This conclusion is supported by the structure theorem for finitely generated modules over a PID, which states that such modules can be expressed as direct sums of cyclic modules. The argument hinges on the independence of generators within the submodule and the impossibility of having a non-1-generated ideal in a PID, confirming that the submodule must also be cyclic.

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mesarmath
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hi,

i want to show that If R is a PID then a submodule of a cyclic R-module is also cyclic.

do i need to use fundamental theorem for finitely generated R-module over R PID ?

thanks in advance
 
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The module will be a direct sum of cyclic modules (see http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain).The module is cyclic or 1-generated,say by g .Let the submodule contain independent generators (x1,x2,..,xn).If y1 = r1*x1 +...rn*xn = ug
and y2 = s1*x1+...sn*xn = vg were independent , the corresponding ideal (u,v) in the ring will not be 1-generated. This is impossible, as the ring is a PID.
 
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Eynstone said:
The module will be a direct sum of cyclic modules (see http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain).The module is cyclic or 1-generated,say by g .Let the submodule contain independent generators (x1,x2,..,xn).If y1 = r1*x1 +...rn*xn = ug
and y2 = s1*x1+...sn*xn = vg were independent , the corresponding ideal (u,v) in the ring will not be 1-generated. This is impossible, as the ring is a PID.

thanks so much

but your last implication was not so obvious, at least for me :)

but i did it myself

thanks a lot again
 
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