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Preservation of multiplicity of roots by $F$-morphisms between field extensions
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I'm trying to study basic facts of field extensions from the model-theory point of view.
Consider the language $L$ of fields, and call $at$ the family of atomic formulae. A morphism of fields is by definition an $at$-morphism of $L$-structures, that is a function $f colon F_1 to F_2$ such that
- for every $phi(vecx) in at$, if $F_1 models phi(veca)$
then $F_2 models phi(f , veca)$.
From the theory of fields, it is known that every morphism of fields is in fact an embedding.
Let $F subseteq E_i = 1, 2$ be two field extensions.
An $F$-morphism is a field morphism $f colon F_1 to F_2$ such that $f(x) = x$ for all $x in F$.
In this case $f$ is an $at(F)$-morphism, where $at(F)$ is the family of atomic formulae with parameters in $F$.
For example, if $alpha in E_1$ is solution to a polynomial equation with coefficients in $F$, then $f(alpha) in E_2$ is solution to the same equation.
It is also well known that
For any set of formulae $Delta$, every $Delta$-embedding is a $existsDelta$-morphism.
Let $alpha in E_1$ be a root of a polynomial $p(x) in F[x]$.
Then we can express that $alpha$ has multiplicity at least $m$
as an $exists,at(F)$-formula as follows:
$$ varphi(alpha) equiv exists , b_0, ldots, b_s , . , p(x) = (x - alpha)^m (b_s x^s + ldots b_0)$$
where the equality of polynomials is replaced by the equality of the corresponding coefficients and $s$ is the appropriate integer.
This implies that if $alpha in E_1$ is a root of $p(x)$ with multiplicity $m$, then $f(alpha) in E_2$ is a root of $p(x)$ with multiplicity at least $m$. The intuition suggests that this multiplicity should be exactly $m$.
Whence we wish to express the property
- $alpha$ has multiplicity less than or equal to $m$
with a formula that is preserved by $F$-morphisms. This seems to be expressible with a $neg exists , at$-formula, or a $forall , at^pm$-formula. But those should not be preserved by $F$-morphisms, as the following example shows:
$$mathbbQ models (forall x , . , x^2 - 2 neq 0), qquad
mathbbR notmodels (forall x , . , x^2 - 2 neq 0).$$
Questions:
- is the intuition of the fact that the exact multiplicity is preserved wrong?
- can this property be expressed as a formula that is preserved?
- also, is there any reference which study field extensions using model theory?
Thank you in advance.
abstract-algebra reference-request field-theory model-theory
asked Jul 14 at 16:49
mononote
1286
add a comment |Â
up vote
1
down vote
favorite
I'm trying to study basic facts of field extensions from the model-theory point of view.
Consider the language $L$ of fields, and call $at$ the family of atomic formulae. A morphism of fields is by definition an $at$-morphism of $L$-structures, that is a function $f colon F_1 to F_2$ such that
- for every $phi(vecx) in at$, if $F_1 models phi(veca)$
then $F_2 models phi(f , veca)$.
From the theory of fields, it is known that every morphism of fields is in fact an embedding.
Let $F subseteq E_i = 1, 2$ be two field extensions.
An $F$-morphism is a field morphism $f colon F_1 to F_2$ such that $f(x) = x$ for all $x in F$.
In this case $f$ is an $at(F)$-morphism, where $at(F)$ is the family of atomic formulae with parameters in $F$.
For example, if $alpha in E_1$ is solution to a polynomial equation with coefficients in $F$, then $f(alpha) in E_2$ is solution to the same equation.
It is also well known that
For any set of formulae $Delta$, every $Delta$-embedding is a $existsDelta$-morphism.
Let $alpha in E_1$ be a root of a polynomial $p(x) in F[x]$.
Then we can express that $alpha$ has multiplicity at least $m$
as an $exists,at(F)$-formula as follows:
$$ varphi(alpha) equiv exists , b_0, ldots, b_s , . , p(x) = (x - alpha)^m (b_s x^s + ldots b_0)$$
where the equality of polynomials is replaced by the equality of the corresponding coefficients and $s$ is the appropriate integer.
This implies that if $alpha in E_1$ is a root of $p(x)$ with multiplicity $m$, then $f(alpha) in E_2$ is a root of $p(x)$ with multiplicity at least $m$. The intuition suggests that this multiplicity should be exactly $m$.
Whence we wish to express the property
- $alpha$ has multiplicity less than or equal to $m$
with a formula that is preserved by $F$-morphisms. This seems to be expressible with a $neg exists , at$-formula, or a $forall , at^pm$-formula. But those should not be preserved by $F$-morphisms, as the following example shows:
$$mathbbQ models (forall x , . , x^2 - 2 neq 0), qquad
mathbbR notmodels (forall x , . , x^2 - 2 neq 0).$$
Questions:
- is the intuition of the fact that the exact multiplicity is preserved wrong?
- can this property be expressed as a formula that is preserved?
- also, is there any reference which study field extensions using model theory?
Thank you in advance.
abstract-algebra reference-request field-theory model-theory
asked Jul 14 at 16:49
mononote
1286
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to study basic facts of field extensions from the model-theory point of view.
Consider the language $L$ of fields, and call $at$ the family of atomic formulae. A morphism of fields is by definition an $at$-morphism of $L$-structures, that is a function $f colon F_1 to F_2$ such that
- for every $phi(vecx) in at$, if $F_1 models phi(veca)$
then $F_2 models phi(f , veca)$.
From the theory of fields, it is known that every morphism of fields is in fact an embedding.
Let $F subseteq E_i = 1, 2$ be two field extensions.
An $F$-morphism is a field morphism $f colon F_1 to F_2$ such that $f(x) = x$ for all $x in F$.
In this case $f$ is an $at(F)$-morphism, where $at(F)$ is the family of atomic formulae with parameters in $F$.
For example, if $alpha in E_1$ is solution to a polynomial equation with coefficients in $F$, then $f(alpha) in E_2$ is solution to the same equation.
It is also well known that
For any set of formulae $Delta$, every $Delta$-embedding is a $existsDelta$-morphism.
Let $alpha in E_1$ be a root of a polynomial $p(x) in F[x]$.
Then we can express that $alpha$ has multiplicity at least $m$
as an $exists,at(F)$-formula as follows:
$$ varphi(alpha) equiv exists , b_0, ldots, b_s , . , p(x) = (x - alpha)^m (b_s x^s + ldots b_0)$$
where the equality of polynomials is replaced by the equality of the corresponding coefficients and $s$ is the appropriate integer.
This implies that if $alpha in E_1$ is a root of $p(x)$ with multiplicity $m$, then $f(alpha) in E_2$ is a root of $p(x)$ with multiplicity at least $m$. The intuition suggests that this multiplicity should be exactly $m$.
Whence we wish to express the property
- $alpha$ has multiplicity less than or equal to $m$
with a formula that is preserved by $F$-morphisms. This seems to be expressible with a $neg exists , at$-formula, or a $forall , at^pm$-formula. But those should not be preserved by $F$-morphisms, as the following example shows:
$$mathbbQ models (forall x , . , x^2 - 2 neq 0), qquad
mathbbR notmodels (forall x , . , x^2 - 2 neq 0).$$
Questions:
- is the intuition of the fact that the exact multiplicity is preserved wrong?
- can this property be expressed as a formula that is preserved?
- also, is there any reference which study field extensions using model theory?
Thank you in advance.
abstract-algebra reference-request field-theory model-theory
asked Jul 14 at 16:49
mononote
1286
I'm trying to study basic facts of field extensions from the model-theory point of view.
Consider the language $L$ of fields, and call $at$ the family of atomic formulae. A morphism of fields is by definition an $at$-morphism of $L$-structures, that is a function $f colon F_1 to F_2$ such that
- for every $phi(vecx) in at$, if $F_1 models phi(veca)$
then $F_2 models phi(f , veca)$.
From the theory of fields, it is known that every morphism of fields is in fact an embedding.
Let $F subseteq E_i = 1, 2$ be two field extensions.
An $F$-morphism is a field morphism $f colon F_1 to F_2$ such that $f(x) = x$ for all $x in F$.
In this case $f$ is an $at(F)$-morphism, where $at(F)$ is the family of atomic formulae with parameters in $F$.
For example, if $alpha in E_1$ is solution to a polynomial equation with coefficients in $F$, then $f(alpha) in E_2$ is solution to the same equation.
It is also well known that
For any set of formulae $Delta$, every $Delta$-embedding is a $existsDelta$-morphism.
Let $alpha in E_1$ be a root of a polynomial $p(x) in F[x]$.
Then we can express that $alpha$ has multiplicity at least $m$
as an $exists,at(F)$-formula as follows:
$$ varphi(alpha) equiv exists , b_0, ldots, b_s , . , p(x) = (x - alpha)^m (b_s x^s + ldots b_0)$$
where the equality of polynomials is replaced by the equality of the corresponding coefficients and $s$ is the appropriate integer.
This implies that if $alpha in E_1$ is a root of $p(x)$ with multiplicity $m$, then $f(alpha) in E_2$ is a root of $p(x)$ with multiplicity at least $m$. The intuition suggests that this multiplicity should be exactly $m$.
Whence we wish to express the property
- $alpha$ has multiplicity less than or equal to $m$
with a formula that is preserved by $F$-morphisms. This seems to be expressible with a $neg exists , at$-formula, or a $forall , at^pm$-formula. But those should not be preserved by $F$-morphisms, as the following example shows:
$$mathbbQ models (forall x , . , x^2 - 2 neq 0), qquad
mathbbR notmodels (forall x , . , x^2 - 2 neq 0).$$
Questions:
- is the intuition of the fact that the exact multiplicity is preserved wrong?
- can this property be expressed as a formula that is preserved?
- also, is there any reference which study field extensions using model theory?
Thank you in advance.
abstract-algebra reference-request field-theory model-theory
asked Jul 14 at 16:49
mononote
1286
asked Jul 14 at 16:49
mononote
1286
asked Jul 14 at 16:49
mononote
1286
asked Jul 14 at 16:49
mononote
1286
1286
add a comment |Â
add a comment |Â
1 Answer
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The property that $a$ is a root of $p(x)$ of multiplicity $m$ can be expressed by the following $L(F)$ formula $varphi(a)$: $$exists b_0,dots,b_s, (p(x) = (x-a)^m(b_sx^s+dots+b_1x + b_0)land b_sa^s+dots+b_1a+b_0neq 0),$$
where, as in your question, equality of polynomials is expressed by the conjunction of atomic formulas expressing equality of their coefficients. This formula says that $p(x)$ can be factored into $(x-a)^m$ times a polynomial which does not have $a$ as a root.
$varphi(a)$ is an existential $L(F)$ formula, and truth of existential formulas is preserved under embeddings. In general, homomorphisms only preserve the truth of positive existential formulas, and $varphi(a)$ is not positive, because of the inequation $(b_sa^s+dots+b_1a+b_0neq 0)$. But as you pointed out, embeddings and homomorphisms coincide for fields. This corresponds to the fact that we can use the Rabinowitsch trick to turn inequations into equations. That is, $varphi(a)$ is equivalent (modulo the theory of fields) to the following positive existential formula:
$$exists b_0,dots,b_s,c, (p(x) = (x-a)^m(b_sx^s+dots+b_1x + b_0)land (b_sa^s+dots+b_1a+b_0)c= 1).$$
answered Jul 14 at 21:58
Alex Kruckman
23.4k22452
Thank you, this was very clear. I wanted to ask again, at the cost of sounding boring, have you ever seen someone using those methods to study field theory from "the basics"? I have seen applications of model theory to prove the existence of algebraic closure, proofs of Nullstellensatz, transfer principles, dimension axiomatizations and studying more specific topics. I find the logical point of view more clear, even if it means spelling out more things. But maybe it is not so interesting/relevant, the usual algebraic arguments are enough, and nobody had reason to report that.
– mononote
Jul 14 at 23:17
1
@mononote Hmm... I'm not aware of a development of the basics of field theory that uses the language of model theory. In any development of algebra, one could use this language, but my sense is that it would really be a cosmetic choice, i.e. the proofs would just be translated, with the main ideas remaining the same. So people typically stick with the traditional presentation. That said, it's probably a good exercise in both model theory and field theory, and it could be enlightening in some places, to try translating as much as possible into model-theoretic language.
– Alex Kruckman
Jul 15 at 0:14
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
The property that $a$ is a root of $p(x)$ of multiplicity $m$ can be expressed by the following $L(F)$ formula $varphi(a)$: $$exists b_0,dots,b_s, (p(x) = (x-a)^m(b_sx^s+dots+b_1x + b_0)land b_sa^s+dots+b_1a+b_0neq 0),$$
where, as in your question, equality of polynomials is expressed by the conjunction of atomic formulas expressing equality of their coefficients. This formula says that $p(x)$ can be factored into $(x-a)^m$ times a polynomial which does not have $a$ as a root.
$varphi(a)$ is an existential $L(F)$ formula, and truth of existential formulas is preserved under embeddings. In general, homomorphisms only preserve the truth of positive existential formulas, and $varphi(a)$ is not positive, because of the inequation $(b_sa^s+dots+b_1a+b_0neq 0)$. But as you pointed out, embeddings and homomorphisms coincide for fields. This corresponds to the fact that we can use the Rabinowitsch trick to turn inequations into equations. That is, $varphi(a)$ is equivalent (modulo the theory of fields) to the following positive existential formula:
$$exists b_0,dots,b_s,c, (p(x) = (x-a)^m(b_sx^s+dots+b_1x + b_0)land (b_sa^s+dots+b_1a+b_0)c= 1).$$
answered Jul 14 at 21:58
Alex Kruckman
23.4k22452
Thank you, this was very clear. I wanted to ask again, at the cost of sounding boring, have you ever seen someone using those methods to study field theory from "the basics"? I have seen applications of model theory to prove the existence of algebraic closure, proofs of Nullstellensatz, transfer principles, dimension axiomatizations and studying more specific topics. I find the logical point of view more clear, even if it means spelling out more things. But maybe it is not so interesting/relevant, the usual algebraic arguments are enough, and nobody had reason to report that.
– mononote
Jul 14 at 23:17
1
@mononote Hmm... I'm not aware of a development of the basics of field theory that uses the language of model theory. In any development of algebra, one could use this language, but my sense is that it would really be a cosmetic choice, i.e. the proofs would just be translated, with the main ideas remaining the same. So people typically stick with the traditional presentation. That said, it's probably a good exercise in both model theory and field theory, and it could be enlightening in some places, to try translating as much as possible into model-theoretic language.
– Alex Kruckman
Jul 15 at 0:14
add a comment |Â
up vote
1
down vote
accepted
The property that $a$ is a root of $p(x)$ of multiplicity $m$ can be expressed by the following $L(F)$ formula $varphi(a)$: $$exists b_0,dots,b_s, (p(x) = (x-a)^m(b_sx^s+dots+b_1x + b_0)land b_sa^s+dots+b_1a+b_0neq 0),$$
where, as in your question, equality of polynomials is expressed by the conjunction of atomic formulas expressing equality of their coefficients. This formula says that $p(x)$ can be factored into $(x-a)^m$ times a polynomial which does not have $a$ as a root.
$varphi(a)$ is an existential $L(F)$ formula, and truth of existential formulas is preserved under embeddings. In general, homomorphisms only preserve the truth of positive existential formulas, and $varphi(a)$ is not positive, because of the inequation $(b_sa^s+dots+b_1a+b_0neq 0)$. But as you pointed out, embeddings and homomorphisms coincide for fields. This corresponds to the fact that we can use the Rabinowitsch trick to turn inequations into equations. That is, $varphi(a)$ is equivalent (modulo the theory of fields) to the following positive existential formula:
$$exists b_0,dots,b_s,c, (p(x) = (x-a)^m(b_sx^s+dots+b_1x + b_0)land (b_sa^s+dots+b_1a+b_0)c= 1).$$
answered Jul 14 at 21:58
Alex Kruckman
23.4k22452
Thank you, this was very clear. I wanted to ask again, at the cost of sounding boring, have you ever seen someone using those methods to study field theory from "the basics"? I have seen applications of model theory to prove the existence of algebraic closure, proofs of Nullstellensatz, transfer principles, dimension axiomatizations and studying more specific topics. I find the logical point of view more clear, even if it means spelling out more things. But maybe it is not so interesting/relevant, the usual algebraic arguments are enough, and nobody had reason to report that.
– mononote
Jul 14 at 23:17
1
@mononote Hmm... I'm not aware of a development of the basics of field theory that uses the language of model theory. In any development of algebra, one could use this language, but my sense is that it would really be a cosmetic choice, i.e. the proofs would just be translated, with the main ideas remaining the same. So people typically stick with the traditional presentation. That said, it's probably a good exercise in both model theory and field theory, and it could be enlightening in some places, to try translating as much as possible into model-theoretic language.
– Alex Kruckman
Jul 15 at 0:14
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The property that $a$ is a root of $p(x)$ of multiplicity $m$ can be expressed by the following $L(F)$ formula $varphi(a)$: $$exists b_0,dots,b_s, (p(x) = (x-a)^m(b_sx^s+dots+b_1x + b_0)land b_sa^s+dots+b_1a+b_0neq 0),$$
where, as in your question, equality of polynomials is expressed by the conjunction of atomic formulas expressing equality of their coefficients. This formula says that $p(x)$ can be factored into $(x-a)^m$ times a polynomial which does not have $a$ as a root.
$varphi(a)$ is an existential $L(F)$ formula, and truth of existential formulas is preserved under embeddings. In general, homomorphisms only preserve the truth of positive existential formulas, and $varphi(a)$ is not positive, because of the inequation $(b_sa^s+dots+b_1a+b_0neq 0)$. But as you pointed out, embeddings and homomorphisms coincide for fields. This corresponds to the fact that we can use the Rabinowitsch trick to turn inequations into equations. That is, $varphi(a)$ is equivalent (modulo the theory of fields) to the following positive existential formula:
$$exists b_0,dots,b_s,c, (p(x) = (x-a)^m(b_sx^s+dots+b_1x + b_0)land (b_sa^s+dots+b_1a+b_0)c= 1).$$
answered Jul 14 at 21:58
Alex Kruckman
23.4k22452
The property that $a$ is a root of $p(x)$ of multiplicity $m$ can be expressed by the following $L(F)$ formula $varphi(a)$: $$exists b_0,dots,b_s, (p(x) = (x-a)^m(b_sx^s+dots+b_1x + b_0)land b_sa^s+dots+b_1a+b_0neq 0),$$
where, as in your question, equality of polynomials is expressed by the conjunction of atomic formulas expressing equality of their coefficients. This formula says that $p(x)$ can be factored into $(x-a)^m$ times a polynomial which does not have $a$ as a root.
$varphi(a)$ is an existential $L(F)$ formula, and truth of existential formulas is preserved under embeddings. In general, homomorphisms only preserve the truth of positive existential formulas, and $varphi(a)$ is not positive, because of the inequation $(b_sa^s+dots+b_1a+b_0neq 0)$. But as you pointed out, embeddings and homomorphisms coincide for fields. This corresponds to the fact that we can use the Rabinowitsch trick to turn inequations into equations. That is, $varphi(a)$ is equivalent (modulo the theory of fields) to the following positive existential formula:
$$exists b_0,dots,b_s,c, (p(x) = (x-a)^m(b_sx^s+dots+b_1x + b_0)land (b_sa^s+dots+b_1a+b_0)c= 1).$$
answered Jul 14 at 21:58
Alex Kruckman
23.4k22452
answered Jul 14 at 21:58
Alex Kruckman
23.4k22452
answered Jul 14 at 21:58
Alex Kruckman
23.4k22452
answered Jul 14 at 21:58
Alex Kruckman
23.4k22452
23.4k22452
Thank you, this was very clear. I wanted to ask again, at the cost of sounding boring, have you ever seen someone using those methods to study field theory from "the basics"? I have seen applications of model theory to prove the existence of algebraic closure, proofs of Nullstellensatz, transfer principles, dimension axiomatizations and studying more specific topics. I find the logical point of view more clear, even if it means spelling out more things. But maybe it is not so interesting/relevant, the usual algebraic arguments are enough, and nobody had reason to report that.
– mononote
Jul 14 at 23:17
1
@mononote Hmm... I'm not aware of a development of the basics of field theory that uses the language of model theory. In any development of algebra, one could use this language, but my sense is that it would really be a cosmetic choice, i.e. the proofs would just be translated, with the main ideas remaining the same. So people typically stick with the traditional presentation. That said, it's probably a good exercise in both model theory and field theory, and it could be enlightening in some places, to try translating as much as possible into model-theoretic language.
– Alex Kruckman
Jul 15 at 0:14
add a comment |Â
Thank you, this was very clear. I wanted to ask again, at the cost of sounding boring, have you ever seen someone using those methods to study field theory from "the basics"? I have seen applications of model theory to prove the existence of algebraic closure, proofs of Nullstellensatz, transfer principles, dimension axiomatizations and studying more specific topics. I find the logical point of view more clear, even if it means spelling out more things. But maybe it is not so interesting/relevant, the usual algebraic arguments are enough, and nobody had reason to report that.
– mononote
Jul 14 at 23:17
1
@mononote Hmm... I'm not aware of a development of the basics of field theory that uses the language of model theory. In any development of algebra, one could use this language, but my sense is that it would really be a cosmetic choice, i.e. the proofs would just be translated, with the main ideas remaining the same. So people typically stick with the traditional presentation. That said, it's probably a good exercise in both model theory and field theory, and it could be enlightening in some places, to try translating as much as possible into model-theoretic language.
– Alex Kruckman
Jul 15 at 0:14
Thank you, this was very clear. I wanted to ask again, at the cost of sounding boring, have you ever seen someone using those methods to study field theory from "the basics"? I have seen applications of model theory to prove the existence of algebraic closure, proofs of Nullstellensatz, transfer principles, dimension axiomatizations and studying more specific topics. I find the logical point of view more clear, even if it means spelling out more things. But maybe it is not so interesting/relevant, the usual algebraic arguments are enough, and nobody had reason to report that.
– mononote
Jul 14 at 23:17
Thank you, this was very clear. I wanted to ask again, at the cost of sounding boring, have you ever seen someone using those methods to study field theory from "the basics"? I have seen applications of model theory to prove the existence of algebraic closure, proofs of Nullstellensatz, transfer principles, dimension axiomatizations and studying more specific topics. I find the logical point of view more clear, even if it means spelling out more things. But maybe it is not so interesting/relevant, the usual algebraic arguments are enough, and nobody had reason to report that.
– mononote
Jul 14 at 23:17
1
1
@mononote Hmm... I'm not aware of a development of the basics of field theory that uses the language of model theory. In any development of algebra, one could use this language, but my sense is that it would really be a cosmetic choice, i.e. the proofs would just be translated, with the main ideas remaining the same. So people typically stick with the traditional presentation. That said, it's probably a good exercise in both model theory and field theory, and it could be enlightening in some places, to try translating as much as possible into model-theoretic language.
– Alex Kruckman
Jul 15 at 0:14
@mononote Hmm... I'm not aware of a development of the basics of field theory that uses the language of model theory. In any development of algebra, one could use this language, but my sense is that it would really be a cosmetic choice, i.e. the proofs would just be translated, with the main ideas remaining the same. So people typically stick with the traditional presentation. That said, it's probably a good exercise in both model theory and field theory, and it could be enlightening in some places, to try translating as much as possible into model-theoretic language.
– Alex Kruckman
Jul 15 at 0:14
add a comment |Â
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