Why isn't an infinite direct product of copies of $Bbb Z$ a free module?











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Why isn't an infinite direct product of copies of $Bbb Z$ a free module?




Actually I was asked to show that it's not projective, but as $Bbb{Z}$ is a PID, so it suffices to show it's not free.



I am stuck here. I saw some questions in SE, but there is no satisfactory answer at all.










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    Have you in mind the (infinite) complete product of Z's?
    – Boris Novikov
    Mar 4 '13 at 14:21

















up vote
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down vote

favorite
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Why isn't an infinite direct product of copies of $Bbb Z$ a free module?




Actually I was asked to show that it's not projective, but as $Bbb{Z}$ is a PID, so it suffices to show it's not free.



I am stuck here. I saw some questions in SE, but there is no satisfactory answer at all.










share|cite|improve this question




















  • 1




    Have you in mind the (infinite) complete product of Z's?
    – Boris Novikov
    Mar 4 '13 at 14:21















up vote
23
down vote

favorite
23









up vote
23
down vote

favorite
23






23






Why isn't an infinite direct product of copies of $Bbb Z$ a free module?




Actually I was asked to show that it's not projective, but as $Bbb{Z}$ is a PID, so it suffices to show it's not free.



I am stuck here. I saw some questions in SE, but there is no satisfactory answer at all.










share|cite|improve this question
















Why isn't an infinite direct product of copies of $Bbb Z$ a free module?




Actually I was asked to show that it's not projective, but as $Bbb{Z}$ is a PID, so it suffices to show it's not free.



I am stuck here. I saw some questions in SE, but there is no satisfactory answer at all.







abstract-algebra modules projective-module






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edited Dec 28 '13 at 9:49







user26857

















asked Mar 4 '13 at 14:08









lee

1,4341023




1,4341023








  • 1




    Have you in mind the (infinite) complete product of Z's?
    – Boris Novikov
    Mar 4 '13 at 14:21
















  • 1




    Have you in mind the (infinite) complete product of Z's?
    – Boris Novikov
    Mar 4 '13 at 14:21










1




1




Have you in mind the (infinite) complete product of Z's?
– Boris Novikov
Mar 4 '13 at 14:21






Have you in mind the (infinite) complete product of Z's?
– Boris Novikov
Mar 4 '13 at 14:21












1 Answer
1






active

oldest

votes

















up vote
24
down vote



accepted










This failure of freeness is a non-trivial result. One way to prove it is to begin with a lemma: If $F$ is a free abelian group and $C$ is a countable subgroup, then the quotient $F/C$ is the direct sum of a countable group and a free group. (I'm omitting "abelian" because I'm lazy and all groups here will be abelian.) [Proof of lemma: Fix a basis $B$ for $F$, let $B_0$ be the countable subset consisting of the basis elements that occur when you expand elements of $C$ in terms of your basis $B$. Then $F$ is the direct sum of $F_0$ freely generated by $B_0$ and $F_1$ freely generated by $B-B_0$. As $Csubseteq F_0$, it follows that $F/C$ is the direct sum of the countable group $F_0/C$ and the free group $F_1$.]



As a corollary, under the hypotheses of the lemma, any divisible subgroup of $F/C$ must be included in the countable summand and must therefore be countable.



Now suppose the direct product $P$ of countably infinitely many copies of $mathbb Z$ were free. The elements of $P$ are the all of the infinite sequences of integers. Let $C$ be the subgroup of $P$ consisting of those sequences that have non-zero entries in only finitely many positions. Then $C$ is countable, so the divisible part of $P/C$ would have to be countable. But this divisible part contains the cosets (in $P/C$) of all the sequences (in $P$) of the form $nmapsto n!cdot a_n$ for arbitrary sequences of integers $(a_n)$. So the divisible part of $P/C$ has the cardinality of the continuum. This contradiction shows that $P$ is not free.



If your question was not only about $P$ but also about products of uncountably many copies of $mathbb Z$, notice that such a product contains a copy of $P$, so you're done if you know that subgroups of free (abelian) groups are free. If you don't know that, just re-run the argument in the preceding paragraph within a copy of $P$ inside your bigger product.



By the way, a theorem of Specker shows that $P$ is not only not free but very far from free. Since $P$ has cardinality $2^{aleph_0}$, if it were free any basis for it would also have cardinality $2^{aleph_0}$, so there would be $2^{2^{aleph_0}}$ homomorphisms from $P$ to $mathbb Z$ (because you could choose the images of the $2^{aleph_0}$ basis elements arbitrarily). Specker showed that there are only countably many homomorphisms $Ptomathbb Z$, namely the $mathbb Z$-linear combinations of the projections.






share|cite|improve this answer





















  • Yes, I also started to think about the possible homomorphisms $PtoBbb Z$ to conclude that they are too few..
    – Berci
    Mar 4 '13 at 14:40






  • 1




    what is a divisible subgroup?
    – lee
    Mar 5 '13 at 15:06










  • An abelian group $G$ is divisible if, for every $gin G$ and every positive integer $n$, there is $hin G$ such that $nh$ (meaning the sum of $n$ copies of $h$) equals $g$.
    – Andreas Blass
    Mar 8 '13 at 22:17











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up vote
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This failure of freeness is a non-trivial result. One way to prove it is to begin with a lemma: If $F$ is a free abelian group and $C$ is a countable subgroup, then the quotient $F/C$ is the direct sum of a countable group and a free group. (I'm omitting "abelian" because I'm lazy and all groups here will be abelian.) [Proof of lemma: Fix a basis $B$ for $F$, let $B_0$ be the countable subset consisting of the basis elements that occur when you expand elements of $C$ in terms of your basis $B$. Then $F$ is the direct sum of $F_0$ freely generated by $B_0$ and $F_1$ freely generated by $B-B_0$. As $Csubseteq F_0$, it follows that $F/C$ is the direct sum of the countable group $F_0/C$ and the free group $F_1$.]



As a corollary, under the hypotheses of the lemma, any divisible subgroup of $F/C$ must be included in the countable summand and must therefore be countable.



Now suppose the direct product $P$ of countably infinitely many copies of $mathbb Z$ were free. The elements of $P$ are the all of the infinite sequences of integers. Let $C$ be the subgroup of $P$ consisting of those sequences that have non-zero entries in only finitely many positions. Then $C$ is countable, so the divisible part of $P/C$ would have to be countable. But this divisible part contains the cosets (in $P/C$) of all the sequences (in $P$) of the form $nmapsto n!cdot a_n$ for arbitrary sequences of integers $(a_n)$. So the divisible part of $P/C$ has the cardinality of the continuum. This contradiction shows that $P$ is not free.



If your question was not only about $P$ but also about products of uncountably many copies of $mathbb Z$, notice that such a product contains a copy of $P$, so you're done if you know that subgroups of free (abelian) groups are free. If you don't know that, just re-run the argument in the preceding paragraph within a copy of $P$ inside your bigger product.



By the way, a theorem of Specker shows that $P$ is not only not free but very far from free. Since $P$ has cardinality $2^{aleph_0}$, if it were free any basis for it would also have cardinality $2^{aleph_0}$, so there would be $2^{2^{aleph_0}}$ homomorphisms from $P$ to $mathbb Z$ (because you could choose the images of the $2^{aleph_0}$ basis elements arbitrarily). Specker showed that there are only countably many homomorphisms $Ptomathbb Z$, namely the $mathbb Z$-linear combinations of the projections.






share|cite|improve this answer





















  • Yes, I also started to think about the possible homomorphisms $PtoBbb Z$ to conclude that they are too few..
    – Berci
    Mar 4 '13 at 14:40






  • 1




    what is a divisible subgroup?
    – lee
    Mar 5 '13 at 15:06










  • An abelian group $G$ is divisible if, for every $gin G$ and every positive integer $n$, there is $hin G$ such that $nh$ (meaning the sum of $n$ copies of $h$) equals $g$.
    – Andreas Blass
    Mar 8 '13 at 22:17















up vote
24
down vote



accepted










This failure of freeness is a non-trivial result. One way to prove it is to begin with a lemma: If $F$ is a free abelian group and $C$ is a countable subgroup, then the quotient $F/C$ is the direct sum of a countable group and a free group. (I'm omitting "abelian" because I'm lazy and all groups here will be abelian.) [Proof of lemma: Fix a basis $B$ for $F$, let $B_0$ be the countable subset consisting of the basis elements that occur when you expand elements of $C$ in terms of your basis $B$. Then $F$ is the direct sum of $F_0$ freely generated by $B_0$ and $F_1$ freely generated by $B-B_0$. As $Csubseteq F_0$, it follows that $F/C$ is the direct sum of the countable group $F_0/C$ and the free group $F_1$.]



As a corollary, under the hypotheses of the lemma, any divisible subgroup of $F/C$ must be included in the countable summand and must therefore be countable.



Now suppose the direct product $P$ of countably infinitely many copies of $mathbb Z$ were free. The elements of $P$ are the all of the infinite sequences of integers. Let $C$ be the subgroup of $P$ consisting of those sequences that have non-zero entries in only finitely many positions. Then $C$ is countable, so the divisible part of $P/C$ would have to be countable. But this divisible part contains the cosets (in $P/C$) of all the sequences (in $P$) of the form $nmapsto n!cdot a_n$ for arbitrary sequences of integers $(a_n)$. So the divisible part of $P/C$ has the cardinality of the continuum. This contradiction shows that $P$ is not free.



If your question was not only about $P$ but also about products of uncountably many copies of $mathbb Z$, notice that such a product contains a copy of $P$, so you're done if you know that subgroups of free (abelian) groups are free. If you don't know that, just re-run the argument in the preceding paragraph within a copy of $P$ inside your bigger product.



By the way, a theorem of Specker shows that $P$ is not only not free but very far from free. Since $P$ has cardinality $2^{aleph_0}$, if it were free any basis for it would also have cardinality $2^{aleph_0}$, so there would be $2^{2^{aleph_0}}$ homomorphisms from $P$ to $mathbb Z$ (because you could choose the images of the $2^{aleph_0}$ basis elements arbitrarily). Specker showed that there are only countably many homomorphisms $Ptomathbb Z$, namely the $mathbb Z$-linear combinations of the projections.






share|cite|improve this answer





















  • Yes, I also started to think about the possible homomorphisms $PtoBbb Z$ to conclude that they are too few..
    – Berci
    Mar 4 '13 at 14:40






  • 1




    what is a divisible subgroup?
    – lee
    Mar 5 '13 at 15:06










  • An abelian group $G$ is divisible if, for every $gin G$ and every positive integer $n$, there is $hin G$ such that $nh$ (meaning the sum of $n$ copies of $h$) equals $g$.
    – Andreas Blass
    Mar 8 '13 at 22:17













up vote
24
down vote



accepted







up vote
24
down vote



accepted






This failure of freeness is a non-trivial result. One way to prove it is to begin with a lemma: If $F$ is a free abelian group and $C$ is a countable subgroup, then the quotient $F/C$ is the direct sum of a countable group and a free group. (I'm omitting "abelian" because I'm lazy and all groups here will be abelian.) [Proof of lemma: Fix a basis $B$ for $F$, let $B_0$ be the countable subset consisting of the basis elements that occur when you expand elements of $C$ in terms of your basis $B$. Then $F$ is the direct sum of $F_0$ freely generated by $B_0$ and $F_1$ freely generated by $B-B_0$. As $Csubseteq F_0$, it follows that $F/C$ is the direct sum of the countable group $F_0/C$ and the free group $F_1$.]



As a corollary, under the hypotheses of the lemma, any divisible subgroup of $F/C$ must be included in the countable summand and must therefore be countable.



Now suppose the direct product $P$ of countably infinitely many copies of $mathbb Z$ were free. The elements of $P$ are the all of the infinite sequences of integers. Let $C$ be the subgroup of $P$ consisting of those sequences that have non-zero entries in only finitely many positions. Then $C$ is countable, so the divisible part of $P/C$ would have to be countable. But this divisible part contains the cosets (in $P/C$) of all the sequences (in $P$) of the form $nmapsto n!cdot a_n$ for arbitrary sequences of integers $(a_n)$. So the divisible part of $P/C$ has the cardinality of the continuum. This contradiction shows that $P$ is not free.



If your question was not only about $P$ but also about products of uncountably many copies of $mathbb Z$, notice that such a product contains a copy of $P$, so you're done if you know that subgroups of free (abelian) groups are free. If you don't know that, just re-run the argument in the preceding paragraph within a copy of $P$ inside your bigger product.



By the way, a theorem of Specker shows that $P$ is not only not free but very far from free. Since $P$ has cardinality $2^{aleph_0}$, if it were free any basis for it would also have cardinality $2^{aleph_0}$, so there would be $2^{2^{aleph_0}}$ homomorphisms from $P$ to $mathbb Z$ (because you could choose the images of the $2^{aleph_0}$ basis elements arbitrarily). Specker showed that there are only countably many homomorphisms $Ptomathbb Z$, namely the $mathbb Z$-linear combinations of the projections.






share|cite|improve this answer












This failure of freeness is a non-trivial result. One way to prove it is to begin with a lemma: If $F$ is a free abelian group and $C$ is a countable subgroup, then the quotient $F/C$ is the direct sum of a countable group and a free group. (I'm omitting "abelian" because I'm lazy and all groups here will be abelian.) [Proof of lemma: Fix a basis $B$ for $F$, let $B_0$ be the countable subset consisting of the basis elements that occur when you expand elements of $C$ in terms of your basis $B$. Then $F$ is the direct sum of $F_0$ freely generated by $B_0$ and $F_1$ freely generated by $B-B_0$. As $Csubseteq F_0$, it follows that $F/C$ is the direct sum of the countable group $F_0/C$ and the free group $F_1$.]



As a corollary, under the hypotheses of the lemma, any divisible subgroup of $F/C$ must be included in the countable summand and must therefore be countable.



Now suppose the direct product $P$ of countably infinitely many copies of $mathbb Z$ were free. The elements of $P$ are the all of the infinite sequences of integers. Let $C$ be the subgroup of $P$ consisting of those sequences that have non-zero entries in only finitely many positions. Then $C$ is countable, so the divisible part of $P/C$ would have to be countable. But this divisible part contains the cosets (in $P/C$) of all the sequences (in $P$) of the form $nmapsto n!cdot a_n$ for arbitrary sequences of integers $(a_n)$. So the divisible part of $P/C$ has the cardinality of the continuum. This contradiction shows that $P$ is not free.



If your question was not only about $P$ but also about products of uncountably many copies of $mathbb Z$, notice that such a product contains a copy of $P$, so you're done if you know that subgroups of free (abelian) groups are free. If you don't know that, just re-run the argument in the preceding paragraph within a copy of $P$ inside your bigger product.



By the way, a theorem of Specker shows that $P$ is not only not free but very far from free. Since $P$ has cardinality $2^{aleph_0}$, if it were free any basis for it would also have cardinality $2^{aleph_0}$, so there would be $2^{2^{aleph_0}}$ homomorphisms from $P$ to $mathbb Z$ (because you could choose the images of the $2^{aleph_0}$ basis elements arbitrarily). Specker showed that there are only countably many homomorphisms $Ptomathbb Z$, namely the $mathbb Z$-linear combinations of the projections.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 4 '13 at 14:33









Andreas Blass

48.9k350106




48.9k350106












  • Yes, I also started to think about the possible homomorphisms $PtoBbb Z$ to conclude that they are too few..
    – Berci
    Mar 4 '13 at 14:40






  • 1




    what is a divisible subgroup?
    – lee
    Mar 5 '13 at 15:06










  • An abelian group $G$ is divisible if, for every $gin G$ and every positive integer $n$, there is $hin G$ such that $nh$ (meaning the sum of $n$ copies of $h$) equals $g$.
    – Andreas Blass
    Mar 8 '13 at 22:17


















  • Yes, I also started to think about the possible homomorphisms $PtoBbb Z$ to conclude that they are too few..
    – Berci
    Mar 4 '13 at 14:40






  • 1




    what is a divisible subgroup?
    – lee
    Mar 5 '13 at 15:06










  • An abelian group $G$ is divisible if, for every $gin G$ and every positive integer $n$, there is $hin G$ such that $nh$ (meaning the sum of $n$ copies of $h$) equals $g$.
    – Andreas Blass
    Mar 8 '13 at 22:17
















Yes, I also started to think about the possible homomorphisms $PtoBbb Z$ to conclude that they are too few..
– Berci
Mar 4 '13 at 14:40




Yes, I also started to think about the possible homomorphisms $PtoBbb Z$ to conclude that they are too few..
– Berci
Mar 4 '13 at 14:40




1




1




what is a divisible subgroup?
– lee
Mar 5 '13 at 15:06




what is a divisible subgroup?
– lee
Mar 5 '13 at 15:06












An abelian group $G$ is divisible if, for every $gin G$ and every positive integer $n$, there is $hin G$ such that $nh$ (meaning the sum of $n$ copies of $h$) equals $g$.
– Andreas Blass
Mar 8 '13 at 22:17




An abelian group $G$ is divisible if, for every $gin G$ and every positive integer $n$, there is $hin G$ such that $nh$ (meaning the sum of $n$ copies of $h$) equals $g$.
– Andreas Blass
Mar 8 '13 at 22:17


















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