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Possible Duplicate:
Is it true that, as Z-modules, the polynomial ring and the power series ring over integers are dual to each other?Is it true that, as Z-modules, the polynomial ring and the power series ring over integers are dual to each other?

Is there an easy proof? I only found citations but have no access. By the way: If we cross over to the rationals every vector space is free (using Zorn's lemma). But can one construct a basis of "Countable infinite product of the rationals"?

Possible Duplicate:
Is it true that, as Z-modules, the polynomial ring and the power series ring over integers are dual to each other?

Is there an easy proof? I only found citations but have no access. By the way: If we cross over to the rationals every vector space is free (using Zorn's lemma). But can one construct a basis of "Countable infinite product of the rationals"?

Possible Duplicate:
Is it true that, as Z-modules, the polynomial ring and the power series ring over integers are dual to each other?

Is there an easy proof? I only found citations but have no access. By the way: If we cross over to the rationals every vector space is free (using Zorn's lemma). But can one construct a basis of "Countable infinite product of the rationals"?

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Possible Duplicate:
Is it true that, as Z-modules, the polynomial ring and the power series ring over integers are dual to each other?

Is there an easy proof? I only found citations but have no access. By the way: If we cross over to the rationals every vector space is free (using Zorn's lemma). But can one construct a basis of "Countable infinite product of the rationals"?

Is there an easy proof? I only found citations but have no access. By the way: If we cross over to the rationals every vector space is free (using Zorn's lemma). But can one construct a basis of "Countable infinite product of the rationals"?

Possible Duplicate:
Is it true that, as Z-modules, the polynomial ring and the power series ring over integers are dual to each other?

Is there an easy proof? I only found citations but have no access. By the way: If we cross over to the rationals every vector space is free (using Zorn's lemma). But can one construct a basis of "Countable infinite product of the rationals"?

Post Closed as "exact duplicate" by Kevin Buzzard, Andreas Thom, S. Carnahan
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Infinite direct product of the integers not a free module over the integers

Is there an easy proof? I only found citations but have no access. By the way: If we cross over to the rationals every vector space is free (using Zorn's lemma). But can one construct a basis of "Countable infinite product of the rationals"?