- #1

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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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- Thread starter mesarmath
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- #1

- 8

- 0

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

- #2

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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 [Broken] 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.

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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- #3

- 8

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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 [Broken] 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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