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The theory of infinite abelian group of exponent $p$ is model-complete
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This is exercise 2.16.10 in Kenneth's Foundation of Mathematics. How to make use of the hypothesis the group is of exponent $p$ to prove that the theory of infinite abelian group of exponent $p$ is model-complete?
The following lemma is given as a hint in the book:
Let $mathfrakA subseteq mathfrakB$ and if for all $a_1, …, a_n in A$ (universe of $mathfrakA$) and $b in B$, there is an automorphism $phi$ of $mathfrakB$ such that $phi(b) in A$ and $phi(a_i)=a_i$, then $mathfrakA$ is an elementary submodel of $mathfrakB$.
logic first-order-logic model-theory
edited Aug 1 at 3:56
Taroccoesbrocco
3,31941330
asked Jul 31 at 2:25
YuiTo Cheng
545
 |Â
show 3 more comments
up vote
2
down vote
favorite
This is exercise 2.16.10 in Kenneth's Foundation of Mathematics. How to make use of the hypothesis the group is of exponent $p$ to prove that the theory of infinite abelian group of exponent $p$ is model-complete?
The following lemma is given as a hint in the book:
Let $mathfrakA subseteq mathfrakB$ and if for all $a_1, …, a_n in A$ (universe of $mathfrakA$) and $b in B$, there is an automorphism $phi$ of $mathfrakB$ such that $phi(b) in A$ and $phi(a_i)=a_i$, then $mathfrakA$ is an elementary submodel of $mathfrakB$.
logic first-order-logic model-theory
edited Aug 1 at 3:56
Taroccoesbrocco
3,31941330
asked Jul 31 at 2:25
YuiTo Cheng
545
What have you tried? What tools do you know for proving a theory is model complete?
– Alex Kruckman
Jul 31 at 21:01
I am asked to use the automorphism lemma.
– YuiTo Cheng
Aug 1 at 0:13
I recommend adding this information, along with the statement of the lemma, to the question.
– Alex Kruckman
Aug 1 at 0:29
The more context, the better!
– Alex Kruckman
Aug 1 at 1:34
2
I don't know much about model theory, but if the groups is of exponent $p,$ then it is a vector space over the field $mathbbZ/pmathbbZ.$ I don't know if this is of any help.
– Chris Leary
Aug 1 at 2:14
 |Â
show 3 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is exercise 2.16.10 in Kenneth's Foundation of Mathematics. How to make use of the hypothesis the group is of exponent $p$ to prove that the theory of infinite abelian group of exponent $p$ is model-complete?
The following lemma is given as a hint in the book:
Let $mathfrakA subseteq mathfrakB$ and if for all $a_1, …, a_n in A$ (universe of $mathfrakA$) and $b in B$, there is an automorphism $phi$ of $mathfrakB$ such that $phi(b) in A$ and $phi(a_i)=a_i$, then $mathfrakA$ is an elementary submodel of $mathfrakB$.
logic first-order-logic model-theory
edited Aug 1 at 3:56
Taroccoesbrocco
3,31941330
asked Jul 31 at 2:25
YuiTo Cheng
545
This is exercise 2.16.10 in Kenneth's Foundation of Mathematics. How to make use of the hypothesis the group is of exponent $p$ to prove that the theory of infinite abelian group of exponent $p$ is model-complete?
The following lemma is given as a hint in the book:
Let $mathfrakA subseteq mathfrakB$ and if for all $a_1, …, a_n in A$ (universe of $mathfrakA$) and $b in B$, there is an automorphism $phi$ of $mathfrakB$ such that $phi(b) in A$ and $phi(a_i)=a_i$, then $mathfrakA$ is an elementary submodel of $mathfrakB$.
logic first-order-logic model-theory
edited Aug 1 at 3:56
Taroccoesbrocco
3,31941330
asked Jul 31 at 2:25
YuiTo Cheng
545
edited Aug 1 at 3:56
Taroccoesbrocco
3,31941330
edited Aug 1 at 3:56
Taroccoesbrocco
3,31941330
edited Aug 1 at 3:56
Taroccoesbrocco
3,31941330
3,31941330
asked Jul 31 at 2:25
YuiTo Cheng
545
asked Jul 31 at 2:25
YuiTo Cheng
545
asked Jul 31 at 2:25
YuiTo Cheng
545
545
What have you tried? What tools do you know for proving a theory is model complete?
– Alex Kruckman
Jul 31 at 21:01
I am asked to use the automorphism lemma.
– YuiTo Cheng
Aug 1 at 0:13
I recommend adding this information, along with the statement of the lemma, to the question.
– Alex Kruckman
Aug 1 at 0:29
The more context, the better!
– Alex Kruckman
Aug 1 at 1:34
2
I don't know much about model theory, but if the groups is of exponent $p,$ then it is a vector space over the field $mathbbZ/pmathbbZ.$ I don't know if this is of any help.
– Chris Leary
Aug 1 at 2:14
 |Â
show 3 more comments
What have you tried? What tools do you know for proving a theory is model complete?
– Alex Kruckman
Jul 31 at 21:01
I am asked to use the automorphism lemma.
– YuiTo Cheng
Aug 1 at 0:13
I recommend adding this information, along with the statement of the lemma, to the question.
– Alex Kruckman
Aug 1 at 0:29
The more context, the better!
– Alex Kruckman
Aug 1 at 1:34
2
I don't know much about model theory, but if the groups is of exponent $p,$ then it is a vector space over the field $mathbbZ/pmathbbZ.$ I don't know if this is of any help.
– Chris Leary
Aug 1 at 2:14
What have you tried? What tools do you know for proving a theory is model complete?
– Alex Kruckman
Jul 31 at 21:01
What have you tried? What tools do you know for proving a theory is model complete?
– Alex Kruckman
Jul 31 at 21:01
I am asked to use the automorphism lemma.
– YuiTo Cheng
Aug 1 at 0:13
I am asked to use the automorphism lemma.
– YuiTo Cheng
Aug 1 at 0:13
I recommend adding this information, along with the statement of the lemma, to the question.
– Alex Kruckman
Aug 1 at 0:29
I recommend adding this information, along with the statement of the lemma, to the question.
– Alex Kruckman
Aug 1 at 0:29
The more context, the better!
– Alex Kruckman
Aug 1 at 1:34
The more context, the better!
– Alex Kruckman
Aug 1 at 1:34
2
2
I don't know much about model theory, but if the groups is of exponent $p,$ then it is a vector space over the field $mathbbZ/pmathbbZ.$ I don't know if this is of any help.
– Chris Leary
Aug 1 at 2:14
I don't know much about model theory, but if the groups is of exponent $p,$ then it is a vector space over the field $mathbbZ/pmathbbZ.$ I don't know if this is of any help.
– Chris Leary
Aug 1 at 2:14
 |Â
show 3 more comments
1 Answer
1
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oldest
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up vote
1
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There are many ways to prove this, but I'll give a proof that follows the hint.
Let $T$ be the theory of infinite abelian groups of exponent $p$. To show $T$ is model complete, we assume $mathfrakAsubseteq mathfrakB$ are models of $T$, and we will show that $mathfrakApreceq mathfrakB$. By the lemma, it suffices to take $a_1,dots,a_nin A$ and $bin B$ and find an automorphism $phi$ of $mathfrakB$ such that $phi(a_i) = a_i$ for all $i$ and $phi(b)in A$. We may assume $bnotin A$, since otherwise the identity automorphism works.
Abelian groups of exponent $p$ are vector spaces over $mathbbF_p$, so we can use some linear algebra. A finite dimensional vector space over $mathbbF_p$ is finite, so $dim(mathfrakA)$ is infinite. Let $V$ be the subspace of $mathfrakA$ spanned by $a_1,dots,a_n$, and let $a_1',dots,a_k'$ be a basis for $V$ (we can take the $a_i'$ to be elements of the tuple $a_1,dots,a_n$, but this isn't necessary). Since $dim(V)leq n$, while $dim(mathfrakA)$ is infinite, $V$ is a proper subset of $A$. Picking some $a^*in Asetminus V$, the set $a_1',dots,a_k',a^*$ is linearly independent. Let $V' = textSpan(a_1',dots,a_k',a^*)$.
Now $V'subseteq A$ and $bnotin A$, so the set $a_1',dots,a_k',a^*,b$ is also linearly independent. Extend this set to a basis for $mathfrakB$. The permutation of the basis obtained by swapping $a^*$ and $b$ and fixing all other basis elements determines an automorphism $phi$ of $mathfrakB$. We have $phi(b) = a^*in A$, and $phi(a_i) = a_i$ for all $i$, since each $a_i$ is a linear combination of the basis elements $a_1',dots,a_k'$, each of which is fixed by $phi$.
answered Aug 1 at 18:11
Alex Kruckman
23k22451
I've actually come up with the same solution.
– YuiTo Cheng
Aug 2 at 0:43
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There are many ways to prove this, but I'll give a proof that follows the hint.
Let $T$ be the theory of infinite abelian groups of exponent $p$. To show $T$ is model complete, we assume $mathfrakAsubseteq mathfrakB$ are models of $T$, and we will show that $mathfrakApreceq mathfrakB$. By the lemma, it suffices to take $a_1,dots,a_nin A$ and $bin B$ and find an automorphism $phi$ of $mathfrakB$ such that $phi(a_i) = a_i$ for all $i$ and $phi(b)in A$. We may assume $bnotin A$, since otherwise the identity automorphism works.
Abelian groups of exponent $p$ are vector spaces over $mathbbF_p$, so we can use some linear algebra. A finite dimensional vector space over $mathbbF_p$ is finite, so $dim(mathfrakA)$ is infinite. Let $V$ be the subspace of $mathfrakA$ spanned by $a_1,dots,a_n$, and let $a_1',dots,a_k'$ be a basis for $V$ (we can take the $a_i'$ to be elements of the tuple $a_1,dots,a_n$, but this isn't necessary). Since $dim(V)leq n$, while $dim(mathfrakA)$ is infinite, $V$ is a proper subset of $A$. Picking some $a^*in Asetminus V$, the set $a_1',dots,a_k',a^*$ is linearly independent. Let $V' = textSpan(a_1',dots,a_k',a^*)$.
Now $V'subseteq A$ and $bnotin A$, so the set $a_1',dots,a_k',a^*,b$ is also linearly independent. Extend this set to a basis for $mathfrakB$. The permutation of the basis obtained by swapping $a^*$ and $b$ and fixing all other basis elements determines an automorphism $phi$ of $mathfrakB$. We have $phi(b) = a^*in A$, and $phi(a_i) = a_i$ for all $i$, since each $a_i$ is a linear combination of the basis elements $a_1',dots,a_k'$, each of which is fixed by $phi$.
answered Aug 1 at 18:11
Alex Kruckman
23k22451
I've actually come up with the same solution.
– YuiTo Cheng
Aug 2 at 0:43
add a comment |Â
up vote
1
down vote
accepted
There are many ways to prove this, but I'll give a proof that follows the hint.
Let $T$ be the theory of infinite abelian groups of exponent $p$. To show $T$ is model complete, we assume $mathfrakAsubseteq mathfrakB$ are models of $T$, and we will show that $mathfrakApreceq mathfrakB$. By the lemma, it suffices to take $a_1,dots,a_nin A$ and $bin B$ and find an automorphism $phi$ of $mathfrakB$ such that $phi(a_i) = a_i$ for all $i$ and $phi(b)in A$. We may assume $bnotin A$, since otherwise the identity automorphism works.
Abelian groups of exponent $p$ are vector spaces over $mathbbF_p$, so we can use some linear algebra. A finite dimensional vector space over $mathbbF_p$ is finite, so $dim(mathfrakA)$ is infinite. Let $V$ be the subspace of $mathfrakA$ spanned by $a_1,dots,a_n$, and let $a_1',dots,a_k'$ be a basis for $V$ (we can take the $a_i'$ to be elements of the tuple $a_1,dots,a_n$, but this isn't necessary). Since $dim(V)leq n$, while $dim(mathfrakA)$ is infinite, $V$ is a proper subset of $A$. Picking some $a^*in Asetminus V$, the set $a_1',dots,a_k',a^*$ is linearly independent. Let $V' = textSpan(a_1',dots,a_k',a^*)$.
Now $V'subseteq A$ and $bnotin A$, so the set $a_1',dots,a_k',a^*,b$ is also linearly independent. Extend this set to a basis for $mathfrakB$. The permutation of the basis obtained by swapping $a^*$ and $b$ and fixing all other basis elements determines an automorphism $phi$ of $mathfrakB$. We have $phi(b) = a^*in A$, and $phi(a_i) = a_i$ for all $i$, since each $a_i$ is a linear combination of the basis elements $a_1',dots,a_k'$, each of which is fixed by $phi$.
answered Aug 1 at 18:11
Alex Kruckman
23k22451
I've actually come up with the same solution.
– YuiTo Cheng
Aug 2 at 0:43
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There are many ways to prove this, but I'll give a proof that follows the hint.
Let $T$ be the theory of infinite abelian groups of exponent $p$. To show $T$ is model complete, we assume $mathfrakAsubseteq mathfrakB$ are models of $T$, and we will show that $mathfrakApreceq mathfrakB$. By the lemma, it suffices to take $a_1,dots,a_nin A$ and $bin B$ and find an automorphism $phi$ of $mathfrakB$ such that $phi(a_i) = a_i$ for all $i$ and $phi(b)in A$. We may assume $bnotin A$, since otherwise the identity automorphism works.
Abelian groups of exponent $p$ are vector spaces over $mathbbF_p$, so we can use some linear algebra. A finite dimensional vector space over $mathbbF_p$ is finite, so $dim(mathfrakA)$ is infinite. Let $V$ be the subspace of $mathfrakA$ spanned by $a_1,dots,a_n$, and let $a_1',dots,a_k'$ be a basis for $V$ (we can take the $a_i'$ to be elements of the tuple $a_1,dots,a_n$, but this isn't necessary). Since $dim(V)leq n$, while $dim(mathfrakA)$ is infinite, $V$ is a proper subset of $A$. Picking some $a^*in Asetminus V$, the set $a_1',dots,a_k',a^*$ is linearly independent. Let $V' = textSpan(a_1',dots,a_k',a^*)$.
Now $V'subseteq A$ and $bnotin A$, so the set $a_1',dots,a_k',a^*,b$ is also linearly independent. Extend this set to a basis for $mathfrakB$. The permutation of the basis obtained by swapping $a^*$ and $b$ and fixing all other basis elements determines an automorphism $phi$ of $mathfrakB$. We have $phi(b) = a^*in A$, and $phi(a_i) = a_i$ for all $i$, since each $a_i$ is a linear combination of the basis elements $a_1',dots,a_k'$, each of which is fixed by $phi$.
answered Aug 1 at 18:11
Alex Kruckman
23k22451
There are many ways to prove this, but I'll give a proof that follows the hint.
Let $T$ be the theory of infinite abelian groups of exponent $p$. To show $T$ is model complete, we assume $mathfrakAsubseteq mathfrakB$ are models of $T$, and we will show that $mathfrakApreceq mathfrakB$. By the lemma, it suffices to take $a_1,dots,a_nin A$ and $bin B$ and find an automorphism $phi$ of $mathfrakB$ such that $phi(a_i) = a_i$ for all $i$ and $phi(b)in A$. We may assume $bnotin A$, since otherwise the identity automorphism works.
Abelian groups of exponent $p$ are vector spaces over $mathbbF_p$, so we can use some linear algebra. A finite dimensional vector space over $mathbbF_p$ is finite, so $dim(mathfrakA)$ is infinite. Let $V$ be the subspace of $mathfrakA$ spanned by $a_1,dots,a_n$, and let $a_1',dots,a_k'$ be a basis for $V$ (we can take the $a_i'$ to be elements of the tuple $a_1,dots,a_n$, but this isn't necessary). Since $dim(V)leq n$, while $dim(mathfrakA)$ is infinite, $V$ is a proper subset of $A$. Picking some $a^*in Asetminus V$, the set $a_1',dots,a_k',a^*$ is linearly independent. Let $V' = textSpan(a_1',dots,a_k',a^*)$.
Now $V'subseteq A$ and $bnotin A$, so the set $a_1',dots,a_k',a^*,b$ is also linearly independent. Extend this set to a basis for $mathfrakB$. The permutation of the basis obtained by swapping $a^*$ and $b$ and fixing all other basis elements determines an automorphism $phi$ of $mathfrakB$. We have $phi(b) = a^*in A$, and $phi(a_i) = a_i$ for all $i$, since each $a_i$ is a linear combination of the basis elements $a_1',dots,a_k'$, each of which is fixed by $phi$.
answered Aug 1 at 18:11
Alex Kruckman
23k22451
answered Aug 1 at 18:11
Alex Kruckman
23k22451
answered Aug 1 at 18:11
Alex Kruckman
23k22451
answered Aug 1 at 18:11
Alex Kruckman
23k22451
23k22451
I've actually come up with the same solution.
– YuiTo Cheng
Aug 2 at 0:43
add a comment |Â
I've actually come up with the same solution.
– YuiTo Cheng
Aug 2 at 0:43
I've actually come up with the same solution.
– YuiTo Cheng
Aug 2 at 0:43
I've actually come up with the same solution.
– YuiTo Cheng
Aug 2 at 0:43
add a comment |Â
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What have you tried? What tools do you know for proving a theory is model complete?
– Alex Kruckman
Jul 31 at 21:01
I am asked to use the automorphism lemma.
– YuiTo Cheng
Aug 1 at 0:13
I recommend adding this information, along with the statement of the lemma, to the question.
– Alex Kruckman
Aug 1 at 0:29
The more context, the better!
– Alex Kruckman
Aug 1 at 1:34
2
I don't know much about model theory, but if the groups is of exponent $p,$ then it is a vector space over the field $mathbbZ/pmathbbZ.$ I don't know if this is of any help.
– Chris Leary
Aug 1 at 2:14