Mixing
Do extension fields always belong to a bigger field?
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Let $F$ be a field, $E_1$ and $E_2$ are two distinct extension fields of $F$. Is it the case that we can always somehow find a field $G$ that contains both $E_1$ and $E_2$? In other words, could extensions of fields have different 'direction's such that they are incompatible?
Edit: I began to think about this problem while reading a proof. $F$ is a field. $a$ and $b$ are algebraic over $F$. $p(x)$ and $q(x)$ are two polynomials in $F[x]$ of minimum degree that respectively make $a$ and $b$ a zero. The proof claims that there is an extension $K$ of $F$ such that all distinct zeros of $p(x)$ and $q(x)$ lie in $K$. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials?
abstract-algebra field-theory extension-field
edited Sep 3 at 19:15
Watson
14.9k92866
asked Sep 2 at 8:15
qiang heng
17715
 |Â
show 2 more comments
up vote
25
down vote
favorite
Let $F$ be a field, $E_1$ and $E_2$ are two distinct extension fields of $F$. Is it the case that we can always somehow find a field $G$ that contains both $E_1$ and $E_2$? In other words, could extensions of fields have different 'direction's such that they are incompatible?
Edit: I began to think about this problem while reading a proof. $F$ is a field. $a$ and $b$ are algebraic over $F$. $p(x)$ and $q(x)$ are two polynomials in $F[x]$ of minimum degree that respectively make $a$ and $b$ a zero. The proof claims that there is an extension $K$ of $F$ such that all distinct zeros of $p(x)$ and $q(x)$ lie in $K$. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials?
abstract-algebra field-theory extension-field
edited Sep 3 at 19:15
Watson
14.9k92866
asked Sep 2 at 8:15
qiang heng
17715
4
A) This is a good question. The answer is that you can always find a field that contains subfields isomorphic to $E_1$ and $E_2$, but not necessarily as subsets. The term is the compositum of the extension fields. See, for example here. B) Wouldn't the splitting field of the product $p(x)q(x)$ give what you need?
– Jyrki Lahtonen
Sep 2 at 8:33
2
The following kind of problems occur. You could have $F=BbbR$, $E_1=BbbC$, $E_2=BbbR[x]/langle x^2+1rangle$. When we describe $E_1$ and $E_2$ as extensions of $BbbR$ we identify certain subsets within both with $BbbR$ in the well known way. But, those identifications don't immediately identify all the elements of $E_1$ with those of $E_2$. Without such extended identification it stands to reason that in some sense $BbbR$ is the intersection $E_1cap E_2$. Yet, it is easy to show that any extension $Omega$ of $BbbR$ can contain at most a single field isomorphic to either
– Jyrki Lahtonen
Sep 2 at 8:38
2
(cont'd) Explaining that we really need that "up to isomorphism" part in many statements when dealing with field theory.
– Jyrki Lahtonen
Sep 2 at 8:39
3
I agree that this is a good question, because it hints towards a subtlety that is often overlooked. Besides Jyrki's point that one often says or should say "up to isomorphism", another way of dealing with it, at least for algebraic extensions, is that one often fixes one algebraic closure $bar K$ of a given field $K$, and then says that all algebraic extensions $L|K$ are to be taken within that algebraic closure. (And then with abstract arguments convinces oneself that if one had picked a different algebraic closure $bar K '$, one would have got very isomorphic results.)
– Torsten Schoeneberg
Sep 2 at 9:00
1
An example of how difficult these things can become if one is not careful is the question mathoverflow.net/q/82083/27465.
– Torsten Schoeneberg
Sep 2 at 12:32
 |Â
show 2 more comments
up vote
25
down vote
favorite
up vote
25
down vote
favorite
Let $F$ be a field, $E_1$ and $E_2$ are two distinct extension fields of $F$. Is it the case that we can always somehow find a field $G$ that contains both $E_1$ and $E_2$? In other words, could extensions of fields have different 'direction's such that they are incompatible?
Edit: I began to think about this problem while reading a proof. $F$ is a field. $a$ and $b$ are algebraic over $F$. $p(x)$ and $q(x)$ are two polynomials in $F[x]$ of minimum degree that respectively make $a$ and $b$ a zero. The proof claims that there is an extension $K$ of $F$ such that all distinct zeros of $p(x)$ and $q(x)$ lie in $K$. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials?
abstract-algebra field-theory extension-field
edited Sep 3 at 19:15
Watson
14.9k92866
asked Sep 2 at 8:15
qiang heng
17715
Let $F$ be a field, $E_1$ and $E_2$ are two distinct extension fields of $F$. Is it the case that we can always somehow find a field $G$ that contains both $E_1$ and $E_2$? In other words, could extensions of fields have different 'direction's such that they are incompatible?
Edit: I began to think about this problem while reading a proof. $F$ is a field. $a$ and $b$ are algebraic over $F$. $p(x)$ and $q(x)$ are two polynomials in $F[x]$ of minimum degree that respectively make $a$ and $b$ a zero. The proof claims that there is an extension $K$ of $F$ such that all distinct zeros of $p(x)$ and $q(x)$ lie in $K$. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials?
abstract-algebra field-theory extension-field
edited Sep 3 at 19:15
Watson
14.9k92866
asked Sep 2 at 8:15
qiang heng
17715
edited Sep 3 at 19:15
Watson
14.9k92866
edited Sep 3 at 19:15
Watson
14.9k92866
edited Sep 3 at 19:15
Watson
14.9k92866
14.9k92866
asked Sep 2 at 8:15
qiang heng
17715
asked Sep 2 at 8:15
qiang heng
17715
asked Sep 2 at 8:15
qiang heng
17715
17715
4
A) This is a good question. The answer is that you can always find a field that contains subfields isomorphic to $E_1$ and $E_2$, but not necessarily as subsets. The term is the compositum of the extension fields. See, for example here. B) Wouldn't the splitting field of the product $p(x)q(x)$ give what you need?
– Jyrki Lahtonen
Sep 2 at 8:33
2
The following kind of problems occur. You could have $F=BbbR$, $E_1=BbbC$, $E_2=BbbR[x]/langle x^2+1rangle$. When we describe $E_1$ and $E_2$ as extensions of $BbbR$ we identify certain subsets within both with $BbbR$ in the well known way. But, those identifications don't immediately identify all the elements of $E_1$ with those of $E_2$. Without such extended identification it stands to reason that in some sense $BbbR$ is the intersection $E_1cap E_2$. Yet, it is easy to show that any extension $Omega$ of $BbbR$ can contain at most a single field isomorphic to either
– Jyrki Lahtonen
Sep 2 at 8:38
2
(cont'd) Explaining that we really need that "up to isomorphism" part in many statements when dealing with field theory.
– Jyrki Lahtonen
Sep 2 at 8:39
3
I agree that this is a good question, because it hints towards a subtlety that is often overlooked. Besides Jyrki's point that one often says or should say "up to isomorphism", another way of dealing with it, at least for algebraic extensions, is that one often fixes one algebraic closure $bar K$ of a given field $K$, and then says that all algebraic extensions $L|K$ are to be taken within that algebraic closure. (And then with abstract arguments convinces oneself that if one had picked a different algebraic closure $bar K '$, one would have got very isomorphic results.)
– Torsten Schoeneberg
Sep 2 at 9:00
1
An example of how difficult these things can become if one is not careful is the question mathoverflow.net/q/82083/27465.
– Torsten Schoeneberg
Sep 2 at 12:32
 |Â
show 2 more comments
4
A) This is a good question. The answer is that you can always find a field that contains subfields isomorphic to $E_1$ and $E_2$, but not necessarily as subsets. The term is the compositum of the extension fields. See, for example here. B) Wouldn't the splitting field of the product $p(x)q(x)$ give what you need?
– Jyrki Lahtonen
Sep 2 at 8:33
2
The following kind of problems occur. You could have $F=BbbR$, $E_1=BbbC$, $E_2=BbbR[x]/langle x^2+1rangle$. When we describe $E_1$ and $E_2$ as extensions of $BbbR$ we identify certain subsets within both with $BbbR$ in the well known way. But, those identifications don't immediately identify all the elements of $E_1$ with those of $E_2$. Without such extended identification it stands to reason that in some sense $BbbR$ is the intersection $E_1cap E_2$. Yet, it is easy to show that any extension $Omega$ of $BbbR$ can contain at most a single field isomorphic to either
– Jyrki Lahtonen
Sep 2 at 8:38
2
(cont'd) Explaining that we really need that "up to isomorphism" part in many statements when dealing with field theory.
– Jyrki Lahtonen
Sep 2 at 8:39
3
I agree that this is a good question, because it hints towards a subtlety that is often overlooked. Besides Jyrki's point that one often says or should say "up to isomorphism", another way of dealing with it, at least for algebraic extensions, is that one often fixes one algebraic closure $bar K$ of a given field $K$, and then says that all algebraic extensions $L|K$ are to be taken within that algebraic closure. (And then with abstract arguments convinces oneself that if one had picked a different algebraic closure $bar K '$, one would have got very isomorphic results.)
– Torsten Schoeneberg
Sep 2 at 9:00
1
An example of how difficult these things can become if one is not careful is the question mathoverflow.net/q/82083/27465.
– Torsten Schoeneberg
Sep 2 at 12:32
4
4
A) This is a good question. The answer is that you can always find a field that contains subfields isomorphic to $E_1$ and $E_2$, but not necessarily as subsets. The term is the compositum of the extension fields. See, for example here. B) Wouldn't the splitting field of the product $p(x)q(x)$ give what you need?
– Jyrki Lahtonen
Sep 2 at 8:33
A) This is a good question. The answer is that you can always find a field that contains subfields isomorphic to $E_1$ and $E_2$, but not necessarily as subsets. The term is the compositum of the extension fields. See, for example here. B) Wouldn't the splitting field of the product $p(x)q(x)$ give what you need?
– Jyrki Lahtonen
Sep 2 at 8:33
2
2
The following kind of problems occur. You could have $F=BbbR$, $E_1=BbbC$, $E_2=BbbR[x]/langle x^2+1rangle$. When we describe $E_1$ and $E_2$ as extensions of $BbbR$ we identify certain subsets within both with $BbbR$ in the well known way. But, those identifications don't immediately identify all the elements of $E_1$ with those of $E_2$. Without such extended identification it stands to reason that in some sense $BbbR$ is the intersection $E_1cap E_2$. Yet, it is easy to show that any extension $Omega$ of $BbbR$ can contain at most a single field isomorphic to either
– Jyrki Lahtonen
Sep 2 at 8:38
The following kind of problems occur. You could have $F=BbbR$, $E_1=BbbC$, $E_2=BbbR[x]/langle x^2+1rangle$. When we describe $E_1$ and $E_2$ as extensions of $BbbR$ we identify certain subsets within both with $BbbR$ in the well known way. But, those identifications don't immediately identify all the elements of $E_1$ with those of $E_2$. Without such extended identification it stands to reason that in some sense $BbbR$ is the intersection $E_1cap E_2$. Yet, it is easy to show that any extension $Omega$ of $BbbR$ can contain at most a single field isomorphic to either
– Jyrki Lahtonen
Sep 2 at 8:38
2
2
(cont'd) Explaining that we really need that "up to isomorphism" part in many statements when dealing with field theory.
– Jyrki Lahtonen
Sep 2 at 8:39
(cont'd) Explaining that we really need that "up to isomorphism" part in many statements when dealing with field theory.
– Jyrki Lahtonen
Sep 2 at 8:39
3
3
I agree that this is a good question, because it hints towards a subtlety that is often overlooked. Besides Jyrki's point that one often says or should say "up to isomorphism", another way of dealing with it, at least for algebraic extensions, is that one often fixes one algebraic closure $bar K$ of a given field $K$, and then says that all algebraic extensions $L|K$ are to be taken within that algebraic closure. (And then with abstract arguments convinces oneself that if one had picked a different algebraic closure $bar K '$, one would have got very isomorphic results.)
– Torsten Schoeneberg
Sep 2 at 9:00
I agree that this is a good question, because it hints towards a subtlety that is often overlooked. Besides Jyrki's point that one often says or should say "up to isomorphism", another way of dealing with it, at least for algebraic extensions, is that one often fixes one algebraic closure $bar K$ of a given field $K$, and then says that all algebraic extensions $L|K$ are to be taken within that algebraic closure. (And then with abstract arguments convinces oneself that if one had picked a different algebraic closure $bar K '$, one would have got very isomorphic results.)
– Torsten Schoeneberg
Sep 2 at 9:00
1
1
An example of how difficult these things can become if one is not careful is the question mathoverflow.net/q/82083/27465.
– Torsten Schoeneberg
Sep 2 at 12:32
An example of how difficult these things can become if one is not careful is the question mathoverflow.net/q/82083/27465.
– Torsten Schoeneberg
Sep 2 at 12:32
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
20
down vote
accepted
Consider field extensions $E_1/F$ and $E_2/F$. Then the tensor product
$A=E_1otimes_F E_2$ is a commutative ring, not necessarily a field though.
Non-trivial commutative rings have maximal ideals, by a Zorn's lemma argument.
Let $I$ be a maximal ideal of $A$. Then $K=A/I$ is a field. The map $xmapsto
overlinexotimes 1in A/I$ is a ring homomorphism $E_1to K$. As $E_1$
is a field, this is an injective homomorphism, so we can think of $E_1$
being "contained" in $K$. Likewise $E_2$ is "contained" in $K$.
Beware though, the ideal $I$ may not be unique.
answered Sep 2 at 9:25
Lord Shark the Unknown
88.6k955115
1
As an alternative, factor out a prime ideal of $A$, rather than a maximal one, and take the field of fractions of the integral domain you get.
– Lord Shark the Unknown
Sep 2 at 12:25
add a comment |Â
up vote
4
down vote
You could embed $F$ into its algebraic closure $overlineF$ (exists by Zorn's lemma, equivalent to the Axiom of Choice). Then both $E_1$ and $E_2$ are essentially (up to isomorphism) subfield of $overlineF$ and we can take the minimal subfield of $overlineF$ that contains $E_1 cup E_2$. This is assuming algebraic extensions, the most interesting case, I think.
answered Sep 2 at 10:39
Henno Brandsma
92.9k342100
1
However, if we are that careful, one should note that $bar F$ is not unique either, only up to isomorphism fixing $F$.
– Torsten Schoeneberg
Sep 2 at 11:34
3
And the outcome of the process may depend on the choice of subfields isomorphic to $E_1$ and $E_2$. Basically because that choice is not unique unless $E_i/F$ are Galois. Consider $F=BbbQ$, $E_1=F(root3of2)$, $E_2=F[x]/langle x^3-2rangle$. If you choose $E_1$ as the copy of $E_2$ sitting inside $BbbC$ you get $F(E_1cup E_2)=E_1$. OTOH if you choose $F(e^2pi i/3root3of2)$ as the copy of $E_2$, you get the degree six splitting field of $x^3-2$ inside $BbbC$.
– Jyrki Lahtonen
Sep 2 at 12:44
@JyrkiLahtonen thanks for showing me it's more subtle than I thought.
– Henno Brandsma
Sep 2 at 12:46
add a comment |Â
up vote
2
down vote
For your specific example:
Take $F$, then find the splitting field of $p(x)$. Let the splitting field be $G$. Now factorize $q(x)=prod_i=1^ell q_i(x)$ over $G$ into product of irreducible polynomials (Note that $q(x)$ may be irreducible over $F$ but over $G$ it might factorize)then extend $G$ to a splitting field of $q_1$ let it be $G_1$. Now factorize $prod_i=2^ell q_i(x)$ into product of irreducible polynomials over $G_1$ and extend to a splitting field and so on. Since the degree of $q(x)$ is finite, this process will stop. Then you have a field which contains all the zeros of $p(x),q(x)$.
answered Sep 2 at 10:05
Balaji sb
37315
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
20
down vote
accepted
Consider field extensions $E_1/F$ and $E_2/F$. Then the tensor product
$A=E_1otimes_F E_2$ is a commutative ring, not necessarily a field though.
Non-trivial commutative rings have maximal ideals, by a Zorn's lemma argument.
Let $I$ be a maximal ideal of $A$. Then $K=A/I$ is a field. The map $xmapsto
overlinexotimes 1in A/I$ is a ring homomorphism $E_1to K$. As $E_1$
is a field, this is an injective homomorphism, so we can think of $E_1$
being "contained" in $K$. Likewise $E_2$ is "contained" in $K$.
Beware though, the ideal $I$ may not be unique.
answered Sep 2 at 9:25
Lord Shark the Unknown
88.6k955115
1
As an alternative, factor out a prime ideal of $A$, rather than a maximal one, and take the field of fractions of the integral domain you get.
– Lord Shark the Unknown
Sep 2 at 12:25
add a comment |Â
up vote
20
down vote
accepted
Consider field extensions $E_1/F$ and $E_2/F$. Then the tensor product
$A=E_1otimes_F E_2$ is a commutative ring, not necessarily a field though.
Non-trivial commutative rings have maximal ideals, by a Zorn's lemma argument.
Let $I$ be a maximal ideal of $A$. Then $K=A/I$ is a field. The map $xmapsto
overlinexotimes 1in A/I$ is a ring homomorphism $E_1to K$. As $E_1$
is a field, this is an injective homomorphism, so we can think of $E_1$
being "contained" in $K$. Likewise $E_2$ is "contained" in $K$.
Beware though, the ideal $I$ may not be unique.
answered Sep 2 at 9:25
Lord Shark the Unknown
88.6k955115
1
As an alternative, factor out a prime ideal of $A$, rather than a maximal one, and take the field of fractions of the integral domain you get.
– Lord Shark the Unknown
Sep 2 at 12:25
add a comment |Â
up vote
20
down vote
accepted
up vote
20
down vote
accepted
Consider field extensions $E_1/F$ and $E_2/F$. Then the tensor product
$A=E_1otimes_F E_2$ is a commutative ring, not necessarily a field though.
Non-trivial commutative rings have maximal ideals, by a Zorn's lemma argument.
Let $I$ be a maximal ideal of $A$. Then $K=A/I$ is a field. The map $xmapsto
overlinexotimes 1in A/I$ is a ring homomorphism $E_1to K$. As $E_1$
is a field, this is an injective homomorphism, so we can think of $E_1$
being "contained" in $K$. Likewise $E_2$ is "contained" in $K$.
Beware though, the ideal $I$ may not be unique.
answered Sep 2 at 9:25
Lord Shark the Unknown
88.6k955115
Consider field extensions $E_1/F$ and $E_2/F$. Then the tensor product
$A=E_1otimes_F E_2$ is a commutative ring, not necessarily a field though.
Non-trivial commutative rings have maximal ideals, by a Zorn's lemma argument.
Let $I$ be a maximal ideal of $A$. Then $K=A/I$ is a field. The map $xmapsto
overlinexotimes 1in A/I$ is a ring homomorphism $E_1to K$. As $E_1$
is a field, this is an injective homomorphism, so we can think of $E_1$
being "contained" in $K$. Likewise $E_2$ is "contained" in $K$.
Beware though, the ideal $I$ may not be unique.
answered Sep 2 at 9:25
Lord Shark the Unknown
88.6k955115
answered Sep 2 at 9:25
Lord Shark the Unknown
88.6k955115
answered Sep 2 at 9:25
Lord Shark the Unknown
88.6k955115
answered Sep 2 at 9:25
Lord Shark the Unknown
88.6k955115
88.6k955115
1
As an alternative, factor out a prime ideal of $A$, rather than a maximal one, and take the field of fractions of the integral domain you get.
– Lord Shark the Unknown
Sep 2 at 12:25
add a comment |Â
1
As an alternative, factor out a prime ideal of $A$, rather than a maximal one, and take the field of fractions of the integral domain you get.
– Lord Shark the Unknown
Sep 2 at 12:25
1
1
As an alternative, factor out a prime ideal of $A$, rather than a maximal one, and take the field of fractions of the integral domain you get.
– Lord Shark the Unknown
Sep 2 at 12:25
As an alternative, factor out a prime ideal of $A$, rather than a maximal one, and take the field of fractions of the integral domain you get.
– Lord Shark the Unknown
Sep 2 at 12:25
add a comment |Â
up vote
4
down vote
You could embed $F$ into its algebraic closure $overlineF$ (exists by Zorn's lemma, equivalent to the Axiom of Choice). Then both $E_1$ and $E_2$ are essentially (up to isomorphism) subfield of $overlineF$ and we can take the minimal subfield of $overlineF$ that contains $E_1 cup E_2$. This is assuming algebraic extensions, the most interesting case, I think.
answered Sep 2 at 10:39
Henno Brandsma
92.9k342100
1
However, if we are that careful, one should note that $bar F$ is not unique either, only up to isomorphism fixing $F$.
– Torsten Schoeneberg
Sep 2 at 11:34
3
And the outcome of the process may depend on the choice of subfields isomorphic to $E_1$ and $E_2$. Basically because that choice is not unique unless $E_i/F$ are Galois. Consider $F=BbbQ$, $E_1=F(root3of2)$, $E_2=F[x]/langle x^3-2rangle$. If you choose $E_1$ as the copy of $E_2$ sitting inside $BbbC$ you get $F(E_1cup E_2)=E_1$. OTOH if you choose $F(e^2pi i/3root3of2)$ as the copy of $E_2$, you get the degree six splitting field of $x^3-2$ inside $BbbC$.
– Jyrki Lahtonen
Sep 2 at 12:44
@JyrkiLahtonen thanks for showing me it's more subtle than I thought.
– Henno Brandsma
Sep 2 at 12:46
add a comment |Â
up vote
4
down vote
You could embed $F$ into its algebraic closure $overlineF$ (exists by Zorn's lemma, equivalent to the Axiom of Choice). Then both $E_1$ and $E_2$ are essentially (up to isomorphism) subfield of $overlineF$ and we can take the minimal subfield of $overlineF$ that contains $E_1 cup E_2$. This is assuming algebraic extensions, the most interesting case, I think.
answered Sep 2 at 10:39
Henno Brandsma
92.9k342100
1
However, if we are that careful, one should note that $bar F$ is not unique either, only up to isomorphism fixing $F$.
– Torsten Schoeneberg
Sep 2 at 11:34
3
And the outcome of the process may depend on the choice of subfields isomorphic to $E_1$ and $E_2$. Basically because that choice is not unique unless $E_i/F$ are Galois. Consider $F=BbbQ$, $E_1=F(root3of2)$, $E_2=F[x]/langle x^3-2rangle$. If you choose $E_1$ as the copy of $E_2$ sitting inside $BbbC$ you get $F(E_1cup E_2)=E_1$. OTOH if you choose $F(e^2pi i/3root3of2)$ as the copy of $E_2$, you get the degree six splitting field of $x^3-2$ inside $BbbC$.
– Jyrki Lahtonen
Sep 2 at 12:44
@JyrkiLahtonen thanks for showing me it's more subtle than I thought.
– Henno Brandsma
Sep 2 at 12:46
add a comment |Â
up vote
4
down vote
up vote
4
down vote
You could embed $F$ into its algebraic closure $overlineF$ (exists by Zorn's lemma, equivalent to the Axiom of Choice). Then both $E_1$ and $E_2$ are essentially (up to isomorphism) subfield of $overlineF$ and we can take the minimal subfield of $overlineF$ that contains $E_1 cup E_2$. This is assuming algebraic extensions, the most interesting case, I think.
answered Sep 2 at 10:39
Henno Brandsma
92.9k342100
You could embed $F$ into its algebraic closure $overlineF$ (exists by Zorn's lemma, equivalent to the Axiom of Choice). Then both $E_1$ and $E_2$ are essentially (up to isomorphism) subfield of $overlineF$ and we can take the minimal subfield of $overlineF$ that contains $E_1 cup E_2$. This is assuming algebraic extensions, the most interesting case, I think.
answered Sep 2 at 10:39
Henno Brandsma
92.9k342100
answered Sep 2 at 10:39
Henno Brandsma
92.9k342100
answered Sep 2 at 10:39
Henno Brandsma
92.9k342100
answered Sep 2 at 10:39
Henno Brandsma
92.9k342100
92.9k342100
1
However, if we are that careful, one should note that $bar F$ is not unique either, only up to isomorphism fixing $F$.
– Torsten Schoeneberg
Sep 2 at 11:34
3
And the outcome of the process may depend on the choice of subfields isomorphic to $E_1$ and $E_2$. Basically because that choice is not unique unless $E_i/F$ are Galois. Consider $F=BbbQ$, $E_1=F(root3of2)$, $E_2=F[x]/langle x^3-2rangle$. If you choose $E_1$ as the copy of $E_2$ sitting inside $BbbC$ you get $F(E_1cup E_2)=E_1$. OTOH if you choose $F(e^2pi i/3root3of2)$ as the copy of $E_2$, you get the degree six splitting field of $x^3-2$ inside $BbbC$.
– Jyrki Lahtonen
Sep 2 at 12:44
@JyrkiLahtonen thanks for showing me it's more subtle than I thought.
– Henno Brandsma
Sep 2 at 12:46
add a comment |Â
1
However, if we are that careful, one should note that $bar F$ is not unique either, only up to isomorphism fixing $F$.
– Torsten Schoeneberg
Sep 2 at 11:34
3
And the outcome of the process may depend on the choice of subfields isomorphic to $E_1$ and $E_2$. Basically because that choice is not unique unless $E_i/F$ are Galois. Consider $F=BbbQ$, $E_1=F(root3of2)$, $E_2=F[x]/langle x^3-2rangle$. If you choose $E_1$ as the copy of $E_2$ sitting inside $BbbC$ you get $F(E_1cup E_2)=E_1$. OTOH if you choose $F(e^2pi i/3root3of2)$ as the copy of $E_2$, you get the degree six splitting field of $x^3-2$ inside $BbbC$.
– Jyrki Lahtonen
Sep 2 at 12:44
@JyrkiLahtonen thanks for showing me it's more subtle than I thought.
– Henno Brandsma
Sep 2 at 12:46
1
1
However, if we are that careful, one should note that $bar F$ is not unique either, only up to isomorphism fixing $F$.
– Torsten Schoeneberg
Sep 2 at 11:34
However, if we are that careful, one should note that $bar F$ is not unique either, only up to isomorphism fixing $F$.
– Torsten Schoeneberg
Sep 2 at 11:34
3
3
And the outcome of the process may depend on the choice of subfields isomorphic to $E_1$ and $E_2$. Basically because that choice is not unique unless $E_i/F$ are Galois. Consider $F=BbbQ$, $E_1=F(root3of2)$, $E_2=F[x]/langle x^3-2rangle$. If you choose $E_1$ as the copy of $E_2$ sitting inside $BbbC$ you get $F(E_1cup E_2)=E_1$. OTOH if you choose $F(e^2pi i/3root3of2)$ as the copy of $E_2$, you get the degree six splitting field of $x^3-2$ inside $BbbC$.
– Jyrki Lahtonen
Sep 2 at 12:44
And the outcome of the process may depend on the choice of subfields isomorphic to $E_1$ and $E_2$. Basically because that choice is not unique unless $E_i/F$ are Galois. Consider $F=BbbQ$, $E_1=F(root3of2)$, $E_2=F[x]/langle x^3-2rangle$. If you choose $E_1$ as the copy of $E_2$ sitting inside $BbbC$ you get $F(E_1cup E_2)=E_1$. OTOH if you choose $F(e^2pi i/3root3of2)$ as the copy of $E_2$, you get the degree six splitting field of $x^3-2$ inside $BbbC$.
– Jyrki Lahtonen
Sep 2 at 12:44
@JyrkiLahtonen thanks for showing me it's more subtle than I thought.
– Henno Brandsma
Sep 2 at 12:46
@JyrkiLahtonen thanks for showing me it's more subtle than I thought.
– Henno Brandsma
Sep 2 at 12:46
add a comment |Â
up vote
2
down vote
For your specific example:
Take $F$, then find the splitting field of $p(x)$. Let the splitting field be $G$. Now factorize $q(x)=prod_i=1^ell q_i(x)$ over $G$ into product of irreducible polynomials (Note that $q(x)$ may be irreducible over $F$ but over $G$ it might factorize)then extend $G$ to a splitting field of $q_1$ let it be $G_1$. Now factorize $prod_i=2^ell q_i(x)$ into product of irreducible polynomials over $G_1$ and extend to a splitting field and so on. Since the degree of $q(x)$ is finite, this process will stop. Then you have a field which contains all the zeros of $p(x),q(x)$.
answered Sep 2 at 10:05
Balaji sb
37315
add a comment |Â
up vote
2
down vote
For your specific example:
Take $F$, then find the splitting field of $p(x)$. Let the splitting field be $G$. Now factorize $q(x)=prod_i=1^ell q_i(x)$ over $G$ into product of irreducible polynomials (Note that $q(x)$ may be irreducible over $F$ but over $G$ it might factorize)then extend $G$ to a splitting field of $q_1$ let it be $G_1$. Now factorize $prod_i=2^ell q_i(x)$ into product of irreducible polynomials over $G_1$ and extend to a splitting field and so on. Since the degree of $q(x)$ is finite, this process will stop. Then you have a field which contains all the zeros of $p(x),q(x)$.
answered Sep 2 at 10:05
Balaji sb
37315
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For your specific example:
Take $F$, then find the splitting field of $p(x)$. Let the splitting field be $G$. Now factorize $q(x)=prod_i=1^ell q_i(x)$ over $G$ into product of irreducible polynomials (Note that $q(x)$ may be irreducible over $F$ but over $G$ it might factorize)then extend $G$ to a splitting field of $q_1$ let it be $G_1$. Now factorize $prod_i=2^ell q_i(x)$ into product of irreducible polynomials over $G_1$ and extend to a splitting field and so on. Since the degree of $q(x)$ is finite, this process will stop. Then you have a field which contains all the zeros of $p(x),q(x)$.
answered Sep 2 at 10:05
Balaji sb
37315
For your specific example:
Take $F$, then find the splitting field of $p(x)$. Let the splitting field be $G$. Now factorize $q(x)=prod_i=1^ell q_i(x)$ over $G$ into product of irreducible polynomials (Note that $q(x)$ may be irreducible over $F$ but over $G$ it might factorize)then extend $G$ to a splitting field of $q_1$ let it be $G_1$. Now factorize $prod_i=2^ell q_i(x)$ into product of irreducible polynomials over $G_1$ and extend to a splitting field and so on. Since the degree of $q(x)$ is finite, this process will stop. Then you have a field which contains all the zeros of $p(x),q(x)$.
answered Sep 2 at 10:05
Balaji sb
37315
edited Sep 2 at 10:10
answered Sep 2 at 10:05
Balaji sb
37315
answered Sep 2 at 10:05
Balaji sb
37315
answered Sep 2 at 10:05
Balaji sb
37315
37315
add a comment |Â
add a comment |Â
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4
A) This is a good question. The answer is that you can always find a field that contains subfields isomorphic to $E_1$ and $E_2$, but not necessarily as subsets. The term is the compositum of the extension fields. See, for example here. B) Wouldn't the splitting field of the product $p(x)q(x)$ give what you need?
– Jyrki Lahtonen
Sep 2 at 8:33
2
The following kind of problems occur. You could have $F=BbbR$, $E_1=BbbC$, $E_2=BbbR[x]/langle x^2+1rangle$. When we describe $E_1$ and $E_2$ as extensions of $BbbR$ we identify certain subsets within both with $BbbR$ in the well known way. But, those identifications don't immediately identify all the elements of $E_1$ with those of $E_2$. Without such extended identification it stands to reason that in some sense $BbbR$ is the intersection $E_1cap E_2$. Yet, it is easy to show that any extension $Omega$ of $BbbR$ can contain at most a single field isomorphic to either
– Jyrki Lahtonen
Sep 2 at 8:38
2
(cont'd) Explaining that we really need that "up to isomorphism" part in many statements when dealing with field theory.
– Jyrki Lahtonen
Sep 2 at 8:39
3
I agree that this is a good question, because it hints towards a subtlety that is often overlooked. Besides Jyrki's point that one often says or should say "up to isomorphism", another way of dealing with it, at least for algebraic extensions, is that one often fixes one algebraic closure $bar K$ of a given field $K$, and then says that all algebraic extensions $L|K$ are to be taken within that algebraic closure. (And then with abstract arguments convinces oneself that if one had picked a different algebraic closure $bar K '$, one would have got very isomorphic results.)
– Torsten Schoeneberg
Sep 2 at 9:00
1
An example of how difficult these things can become if one is not careful is the question mathoverflow.net/q/82083/27465.
– Torsten Schoeneberg
Sep 2 at 12:32