Mixing
Conditions under which a union of free abelian groups is free
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Let $(A_nu)_nu<mu$ be an increasing sequence of free abelian groups such that for any limit ordinal $lambda<mu$, $A_lambda=bigcup_nu<lambdaA_nu$. Let $A=bigcup_nu<muA_nu$. Let $kappa$ be the cofinality of $mu$.
Assume that $A_sigma/A_nu+1$ is free whenever $nu<sigma<mu$.
Let $p_sigma:sigma<kappa$ be a closed unbounded subset of $mu$ that is strictly increasing. Assume that for every $sigma<kappa$ such that $p_sigma$ is a limit ordinal, $A_p_sigma+1/A_p_sigma$ is free.
A paper says it is "evident" that $A$ is free and that $A/A_p_sigma$ is free for all $sigma<kappa$.
It is not evident to me. Why are these statements true?
Source: the proof of Lemma 2.1 in Paul Eklof, "On the Existence of $kappa$-Free Abelian Groups," Proceedings of the American Mathematical Society 47 (1975), 65-72.
free-abelian-group
asked Jul 14 at 21:49
Tri
1646
add a comment |Â
up vote
0
down vote
favorite
Let $(A_nu)_nu<mu$ be an increasing sequence of free abelian groups such that for any limit ordinal $lambda<mu$, $A_lambda=bigcup_nu<lambdaA_nu$. Let $A=bigcup_nu<muA_nu$. Let $kappa$ be the cofinality of $mu$.
Assume that $A_sigma/A_nu+1$ is free whenever $nu<sigma<mu$.
Let $p_sigma:sigma<kappa$ be a closed unbounded subset of $mu$ that is strictly increasing. Assume that for every $sigma<kappa$ such that $p_sigma$ is a limit ordinal, $A_p_sigma+1/A_p_sigma$ is free.
A paper says it is "evident" that $A$ is free and that $A/A_p_sigma$ is free for all $sigma<kappa$.
It is not evident to me. Why are these statements true?
Source: the proof of Lemma 2.1 in Paul Eklof, "On the Existence of $kappa$-Free Abelian Groups," Proceedings of the American Mathematical Society 47 (1975), 65-72.
free-abelian-group
asked Jul 14 at 21:49
Tri
1646
I'm not sure if this is what you're looking for, but in Rotman's Homological Algebra, on Page 103, just after Example 3.7, he proves that if $P=cup_alpha P_alpha$, and if each $P_alpha$ is a direct summand of each $P_alpha+1$, and $P_alpha=cup_beta<alpha P_beta$ for every limit ordinal $alpha$, then $Pcongoplus_alpha P_alpha+1/P_alpha$.
– Ashwin Trisal
Jul 15 at 2:52
Thank you. That was the key.
– Tri
Jul 15 at 5:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(A_nu)_nu<mu$ be an increasing sequence of free abelian groups such that for any limit ordinal $lambda<mu$, $A_lambda=bigcup_nu<lambdaA_nu$. Let $A=bigcup_nu<muA_nu$. Let $kappa$ be the cofinality of $mu$.
Assume that $A_sigma/A_nu+1$ is free whenever $nu<sigma<mu$.
Let $p_sigma:sigma<kappa$ be a closed unbounded subset of $mu$ that is strictly increasing. Assume that for every $sigma<kappa$ such that $p_sigma$ is a limit ordinal, $A_p_sigma+1/A_p_sigma$ is free.
A paper says it is "evident" that $A$ is free and that $A/A_p_sigma$ is free for all $sigma<kappa$.
It is not evident to me. Why are these statements true?
Source: the proof of Lemma 2.1 in Paul Eklof, "On the Existence of $kappa$-Free Abelian Groups," Proceedings of the American Mathematical Society 47 (1975), 65-72.
free-abelian-group
asked Jul 14 at 21:49
Tri
1646
Let $(A_nu)_nu<mu$ be an increasing sequence of free abelian groups such that for any limit ordinal $lambda<mu$, $A_lambda=bigcup_nu<lambdaA_nu$. Let $A=bigcup_nu<muA_nu$. Let $kappa$ be the cofinality of $mu$.
Assume that $A_sigma/A_nu+1$ is free whenever $nu<sigma<mu$.
Let $p_sigma:sigma<kappa$ be a closed unbounded subset of $mu$ that is strictly increasing. Assume that for every $sigma<kappa$ such that $p_sigma$ is a limit ordinal, $A_p_sigma+1/A_p_sigma$ is free.
A paper says it is "evident" that $A$ is free and that $A/A_p_sigma$ is free for all $sigma<kappa$.
It is not evident to me. Why are these statements true?
Source: the proof of Lemma 2.1 in Paul Eklof, "On the Existence of $kappa$-Free Abelian Groups," Proceedings of the American Mathematical Society 47 (1975), 65-72.
free-abelian-group
asked Jul 14 at 21:49
Tri
1646
asked Jul 14 at 21:49
Tri
1646
asked Jul 14 at 21:49
Tri
1646
asked Jul 14 at 21:49
Tri
1646
1646
I'm not sure if this is what you're looking for, but in Rotman's Homological Algebra, on Page 103, just after Example 3.7, he proves that if $P=cup_alpha P_alpha$, and if each $P_alpha$ is a direct summand of each $P_alpha+1$, and $P_alpha=cup_beta<alpha P_beta$ for every limit ordinal $alpha$, then $Pcongoplus_alpha P_alpha+1/P_alpha$.
– Ashwin Trisal
Jul 15 at 2:52
Thank you. That was the key.
– Tri
Jul 15 at 5:42
add a comment |Â
I'm not sure if this is what you're looking for, but in Rotman's Homological Algebra, on Page 103, just after Example 3.7, he proves that if $P=cup_alpha P_alpha$, and if each $P_alpha$ is a direct summand of each $P_alpha+1$, and $P_alpha=cup_beta<alpha P_beta$ for every limit ordinal $alpha$, then $Pcongoplus_alpha P_alpha+1/P_alpha$.
– Ashwin Trisal
Jul 15 at 2:52
Thank you. That was the key.
– Tri
Jul 15 at 5:42
I'm not sure if this is what you're looking for, but in Rotman's Homological Algebra, on Page 103, just after Example 3.7, he proves that if $P=cup_alpha P_alpha$, and if each $P_alpha$ is a direct summand of each $P_alpha+1$, and $P_alpha=cup_beta<alpha P_beta$ for every limit ordinal $alpha$, then $Pcongoplus_alpha P_alpha+1/P_alpha$.
– Ashwin Trisal
Jul 15 at 2:52
I'm not sure if this is what you're looking for, but in Rotman's Homological Algebra, on Page 103, just after Example 3.7, he proves that if $P=cup_alpha P_alpha$, and if each $P_alpha$ is a direct summand of each $P_alpha+1$, and $P_alpha=cup_beta<alpha P_beta$ for every limit ordinal $alpha$, then $Pcongoplus_alpha P_alpha+1/P_alpha$.
– Ashwin Trisal
Jul 15 at 2:52
Thank you. That was the key.
– Tri
Jul 15 at 5:42
Thank you. That was the key.
– Tri
Jul 15 at 5:42
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2851987%2fconditions-under-which-a-union-of-free-abelian-groups-is-free%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
I'm not sure if this is what you're looking for, but in Rotman's Homological Algebra, on Page 103, just after Example 3.7, he proves that if $P=cup_alpha P_alpha$, and if each $P_alpha$ is a direct summand of each $P_alpha+1$, and $P_alpha=cup_beta<alpha P_beta$ for every limit ordinal $alpha$, then $Pcongoplus_alpha P_alpha+1/P_alpha$.
– Ashwin Trisal
Jul 15 at 2:52
Thank you. That was the key.
– Tri
Jul 15 at 5:42