Every morphism of abelian groups vanishing on is identically zero.

Use this fact to prove that is not a free abelian group.

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# Freedom

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2 thoughts on “Freedom”

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Every morphism of abelian groups vanishing on is identically zero.

Use this fact to prove that is not a free abelian group.

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If such module were free, then would be uncountable. But by the statement, every map is determined in a countable set, so such homset is at most countable.

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As for the first part: suppose is a morphism as described in the statement. Since and are coprime for all , for any sequence we can write . Now, we have that . The first summand vanishes, since is zero over . Now, the second summand is equal to , so divides for all , and so . A similar argument shows that as well, and so is identically zero, as we wanted to prove.

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