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Why is the “root field†premise necessary in this theorem about field extensions?
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In Charles Pinter's "A Book of Abstract Algebra," chapter 31, Theorem 8 (p316-17), he states:
Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $E$, then $K$ is also the root field of some polynomial over $I$.
His short proof is then:
Suppose $K$ is a root field of some polynomial over $E$, and let $K = I(a)$. If $p(x)$ is the minimum polynomial of $a$ over $I$, its coefficients are certainly in $E$ and it has a root $a$ in $K$, so by Theorem 7, all its roots are in $K$. Therefore, $K$ is the root field of $p(x)$ over $I$.
I think "certainly in $E$" is a typo that should read "certainly in $I$," but here's my question:
Why is it required as a premise that $K$ be a root field of some polynomial over $E$?
Since $K$ is a finite extension of $E$ which is a finite extension of $I$, $K$ must also be a finite extension of $I$ (by Ch29 Th2). Therefore it must also be a simple extension (by Ch31 Th2). So we can write $K=I(a)$ for some $a in K$, and be certain a minimum polynomial of $a$ over $I$ exists (since finite and simple extensions are defined to be in terms of algebraic elements). And from here, Pinter's argument (starting at sentence 2) still seems to hold. Am I missing something?
field-theory extension-field
asked Jul 18 at 16:40
WillG
40128
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In Charles Pinter's "A Book of Abstract Algebra," chapter 31, Theorem 8 (p316-17), he states:
Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $E$, then $K$ is also the root field of some polynomial over $I$.
His short proof is then:
Suppose $K$ is a root field of some polynomial over $E$, and let $K = I(a)$. If $p(x)$ is the minimum polynomial of $a$ over $I$, its coefficients are certainly in $E$ and it has a root $a$ in $K$, so by Theorem 7, all its roots are in $K$. Therefore, $K$ is the root field of $p(x)$ over $I$.
I think "certainly in $E$" is a typo that should read "certainly in $I$," but here's my question:
Why is it required as a premise that $K$ be a root field of some polynomial over $E$?
Since $K$ is a finite extension of $E$ which is a finite extension of $I$, $K$ must also be a finite extension of $I$ (by Ch29 Th2). Therefore it must also be a simple extension (by Ch31 Th2). So we can write $K=I(a)$ for some $a in K$, and be certain a minimum polynomial of $a$ over $I$ exists (since finite and simple extensions are defined to be in terms of algebraic elements). And from here, Pinter's argument (starting at sentence 2) still seems to hold. Am I missing something?
field-theory extension-field
asked Jul 18 at 16:40
WillG
40128
For reference, here is Ch31 Th7, which he refers to: Let $K$ be the root field of some polynomial over $F$. For every irreducible polynomial $p(x)$ in $F[x]$, if $p(x)$ has one root in $K$, then $p(x)$ must have all of its roots in $K$.
– WillG
Jul 18 at 16:45
add a comment |Â
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0
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up vote
0
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favorite
In Charles Pinter's "A Book of Abstract Algebra," chapter 31, Theorem 8 (p316-17), he states:
Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $E$, then $K$ is also the root field of some polynomial over $I$.
His short proof is then:
Suppose $K$ is a root field of some polynomial over $E$, and let $K = I(a)$. If $p(x)$ is the minimum polynomial of $a$ over $I$, its coefficients are certainly in $E$ and it has a root $a$ in $K$, so by Theorem 7, all its roots are in $K$. Therefore, $K$ is the root field of $p(x)$ over $I$.
I think "certainly in $E$" is a typo that should read "certainly in $I$," but here's my question:
Why is it required as a premise that $K$ be a root field of some polynomial over $E$?
Since $K$ is a finite extension of $E$ which is a finite extension of $I$, $K$ must also be a finite extension of $I$ (by Ch29 Th2). Therefore it must also be a simple extension (by Ch31 Th2). So we can write $K=I(a)$ for some $a in K$, and be certain a minimum polynomial of $a$ over $I$ exists (since finite and simple extensions are defined to be in terms of algebraic elements). And from here, Pinter's argument (starting at sentence 2) still seems to hold. Am I missing something?
field-theory extension-field
asked Jul 18 at 16:40
WillG
40128
In Charles Pinter's "A Book of Abstract Algebra," chapter 31, Theorem 8 (p316-17), he states:
Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $E$, then $K$ is also the root field of some polynomial over $I$.
His short proof is then:
Suppose $K$ is a root field of some polynomial over $E$, and let $K = I(a)$. If $p(x)$ is the minimum polynomial of $a$ over $I$, its coefficients are certainly in $E$ and it has a root $a$ in $K$, so by Theorem 7, all its roots are in $K$. Therefore, $K$ is the root field of $p(x)$ over $I$.
I think "certainly in $E$" is a typo that should read "certainly in $I$," but here's my question:
Why is it required as a premise that $K$ be a root field of some polynomial over $E$?
Since $K$ is a finite extension of $E$ which is a finite extension of $I$, $K$ must also be a finite extension of $I$ (by Ch29 Th2). Therefore it must also be a simple extension (by Ch31 Th2). So we can write $K=I(a)$ for some $a in K$, and be certain a minimum polynomial of $a$ over $I$ exists (since finite and simple extensions are defined to be in terms of algebraic elements). And from here, Pinter's argument (starting at sentence 2) still seems to hold. Am I missing something?
field-theory extension-field
asked Jul 18 at 16:40
WillG
40128
asked Jul 18 at 16:40
WillG
40128
asked Jul 18 at 16:40
WillG
40128
asked Jul 18 at 16:40
WillG
40128
40128
For reference, here is Ch31 Th7, which he refers to: Let $K$ be the root field of some polynomial over $F$. For every irreducible polynomial $p(x)$ in $F[x]$, if $p(x)$ has one root in $K$, then $p(x)$ must have all of its roots in $K$.
– WillG
Jul 18 at 16:45
add a comment |Â
For reference, here is Ch31 Th7, which he refers to: Let $K$ be the root field of some polynomial over $F$. For every irreducible polynomial $p(x)$ in $F[x]$, if $p(x)$ has one root in $K$, then $p(x)$ must have all of its roots in $K$.
– WillG
Jul 18 at 16:45
For reference, here is Ch31 Th7, which he refers to: Let $K$ be the root field of some polynomial over $F$. For every irreducible polynomial $p(x)$ in $F[x]$, if $p(x)$ has one root in $K$, then $p(x)$ must have all of its roots in $K$.
– WillG
Jul 18 at 16:45
For reference, here is Ch31 Th7, which he refers to: Let $K$ be the root field of some polynomial over $F$. For every irreducible polynomial $p(x)$ in $F[x]$, if $p(x)$ has one root in $K$, then $p(x)$ must have all of its roots in $K$.
– WillG
Jul 18 at 16:45
add a comment |Â
2 Answers
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This answer refers to the theorem as written in the 2nd. edition of Pinter's book:
"Root field" seems to be the same of what many of us call "splitting field": $;K;$ is a root field of some polynomial $;f(x)in I[x];$ means $;K=I(a_1,...,a_n);,;a_1,...,a_n;$ all the roots of $;f(x);$ (in some algebraic closure of $;I;$) and any subfield $;Ile Llneq K;$ fulfills that there exists at least one root $;a_n;$ such that $;a_nnotin L;$.
Observe then that $;f(x)in I[x]subset E[x];$ . Either $;E=K;$ and there's nothing to prove, or else $;Elneq K;$ and thus there is some root of $;f(x);$ whic is not in $;E;implies$ $;K;$ is the splitting field (root filed) of $;f(x);$ as polynomial in $;E[x];$ as $;K;$ contains all the roots of $;f;$ ...
answered Jul 18 at 17:41
DonAntonio
173k1484218
Doesn't the condition $K=I(a_1,...,a_n)$ already guarantee that any strict subfield of K lacks at least one of the $a_i$? My understanding was that $I(a_1,...,a_n)$ is the smallest field containing $I$ and $a_1, ..., a_n$, so that certainly if $L subsetneq I(a_1,...,a_n)$ then there exists an $a_i$ such that $a_i notin L$.
– WillG
Jul 18 at 18:08
@WillG Yes, of course...and that's the proof! Almost trivial.
– DonAntonio
Jul 18 at 19:34
I'm still confused then, as to what exactly goes wrong when we try to assert the same "theorem" but without the root field premise, i.e. "Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. Then $K$ is the root field of some polynomial over $I$." Is this also true? If not, maybe a counterexample would help me understand what goes wrong...
– WillG
Jul 18 at 20:08
By all means: take for example $;I=E=Bbb Q;,;;K:=Bbb Q(sqrt[4]2);$ , and $; K;$ is not a splitting field of any irreducible rational polynomial.
– DonAntonio
Jul 18 at 21:09
Aha. That just blew my mind. I think my confusion was assuming all field extensions of a field $F$ had to be root fields of something over $F$, which clearly is not the case with $mathbbQ(sqrt[4]2)$. But there's one thing I'm still confused about: if $p(x)$ is the minimum polynomial of $a$ over $I$, then yes, its coefficients are in $E$ and it has a root in $K$. But is $p(x)$ necessarily irreducible in $E$? This seems to be required by Theorem 7 (in my comment above).
– WillG
Jul 19 at 18:55
 |Â
show 2 more comments
up vote
0
down vote
accepted
After reviewing this more, I believe the actual theorem from Pinter is wrong! Counterexample: Let $I=mathbbQ$ and $E=K=mathbbQ(sqrt[4]2)$.
I found the 2nd edition of the same book and discovered the theorem is different:
Suppose $Isubseteq Esubseteq K$ where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $I$, then $K$ is also the root field of some polynomial over $E$.
This makes much more sense now.
answered Jul 20 at 5:03
WillG
40128
I honestly didn't even read your whole question. I simply went to Pinter's book and read from there the theorem, which is the one you mention here (2nd. Edition) and whose proof is pretty trivial. Indeed, the 1st. edition has this theorem wrongly written and, of course, it is false as it appears there.
– DonAntonio
Jul 20 at 9:01
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
This answer refers to the theorem as written in the 2nd. edition of Pinter's book:
"Root field" seems to be the same of what many of us call "splitting field": $;K;$ is a root field of some polynomial $;f(x)in I[x];$ means $;K=I(a_1,...,a_n);,;a_1,...,a_n;$ all the roots of $;f(x);$ (in some algebraic closure of $;I;$) and any subfield $;Ile Llneq K;$ fulfills that there exists at least one root $;a_n;$ such that $;a_nnotin L;$.
Observe then that $;f(x)in I[x]subset E[x];$ . Either $;E=K;$ and there's nothing to prove, or else $;Elneq K;$ and thus there is some root of $;f(x);$ whic is not in $;E;implies$ $;K;$ is the splitting field (root filed) of $;f(x);$ as polynomial in $;E[x];$ as $;K;$ contains all the roots of $;f;$ ...
answered Jul 18 at 17:41
DonAntonio
173k1484218
Doesn't the condition $K=I(a_1,...,a_n)$ already guarantee that any strict subfield of K lacks at least one of the $a_i$? My understanding was that $I(a_1,...,a_n)$ is the smallest field containing $I$ and $a_1, ..., a_n$, so that certainly if $L subsetneq I(a_1,...,a_n)$ then there exists an $a_i$ such that $a_i notin L$.
– WillG
Jul 18 at 18:08
@WillG Yes, of course...and that's the proof! Almost trivial.
– DonAntonio
Jul 18 at 19:34
I'm still confused then, as to what exactly goes wrong when we try to assert the same "theorem" but without the root field premise, i.e. "Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. Then $K$ is the root field of some polynomial over $I$." Is this also true? If not, maybe a counterexample would help me understand what goes wrong...
– WillG
Jul 18 at 20:08
By all means: take for example $;I=E=Bbb Q;,;;K:=Bbb Q(sqrt[4]2);$ , and $; K;$ is not a splitting field of any irreducible rational polynomial.
– DonAntonio
Jul 18 at 21:09
Aha. That just blew my mind. I think my confusion was assuming all field extensions of a field $F$ had to be root fields of something over $F$, which clearly is not the case with $mathbbQ(sqrt[4]2)$. But there's one thing I'm still confused about: if $p(x)$ is the minimum polynomial of $a$ over $I$, then yes, its coefficients are in $E$ and it has a root in $K$. But is $p(x)$ necessarily irreducible in $E$? This seems to be required by Theorem 7 (in my comment above).
– WillG
Jul 19 at 18:55
 |Â
show 2 more comments
up vote
2
down vote
This answer refers to the theorem as written in the 2nd. edition of Pinter's book:
"Root field" seems to be the same of what many of us call "splitting field": $;K;$ is a root field of some polynomial $;f(x)in I[x];$ means $;K=I(a_1,...,a_n);,;a_1,...,a_n;$ all the roots of $;f(x);$ (in some algebraic closure of $;I;$) and any subfield $;Ile Llneq K;$ fulfills that there exists at least one root $;a_n;$ such that $;a_nnotin L;$.
Observe then that $;f(x)in I[x]subset E[x];$ . Either $;E=K;$ and there's nothing to prove, or else $;Elneq K;$ and thus there is some root of $;f(x);$ whic is not in $;E;implies$ $;K;$ is the splitting field (root filed) of $;f(x);$ as polynomial in $;E[x];$ as $;K;$ contains all the roots of $;f;$ ...
answered Jul 18 at 17:41
DonAntonio
173k1484218
Doesn't the condition $K=I(a_1,...,a_n)$ already guarantee that any strict subfield of K lacks at least one of the $a_i$? My understanding was that $I(a_1,...,a_n)$ is the smallest field containing $I$ and $a_1, ..., a_n$, so that certainly if $L subsetneq I(a_1,...,a_n)$ then there exists an $a_i$ such that $a_i notin L$.
– WillG
Jul 18 at 18:08
@WillG Yes, of course...and that's the proof! Almost trivial.
– DonAntonio
Jul 18 at 19:34
I'm still confused then, as to what exactly goes wrong when we try to assert the same "theorem" but without the root field premise, i.e. "Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. Then $K$ is the root field of some polynomial over $I$." Is this also true? If not, maybe a counterexample would help me understand what goes wrong...
– WillG
Jul 18 at 20:08
By all means: take for example $;I=E=Bbb Q;,;;K:=Bbb Q(sqrt[4]2);$ , and $; K;$ is not a splitting field of any irreducible rational polynomial.
– DonAntonio
Jul 18 at 21:09
Aha. That just blew my mind. I think my confusion was assuming all field extensions of a field $F$ had to be root fields of something over $F$, which clearly is not the case with $mathbbQ(sqrt[4]2)$. But there's one thing I'm still confused about: if $p(x)$ is the minimum polynomial of $a$ over $I$, then yes, its coefficients are in $E$ and it has a root in $K$. But is $p(x)$ necessarily irreducible in $E$? This seems to be required by Theorem 7 (in my comment above).
– WillG
Jul 19 at 18:55
 |Â
show 2 more comments
up vote
2
down vote
up vote
2
down vote
This answer refers to the theorem as written in the 2nd. edition of Pinter's book:
"Root field" seems to be the same of what many of us call "splitting field": $;K;$ is a root field of some polynomial $;f(x)in I[x];$ means $;K=I(a_1,...,a_n);,;a_1,...,a_n;$ all the roots of $;f(x);$ (in some algebraic closure of $;I;$) and any subfield $;Ile Llneq K;$ fulfills that there exists at least one root $;a_n;$ such that $;a_nnotin L;$.
Observe then that $;f(x)in I[x]subset E[x];$ . Either $;E=K;$ and there's nothing to prove, or else $;Elneq K;$ and thus there is some root of $;f(x);$ whic is not in $;E;implies$ $;K;$ is the splitting field (root filed) of $;f(x);$ as polynomial in $;E[x];$ as $;K;$ contains all the roots of $;f;$ ...
answered Jul 18 at 17:41
DonAntonio
173k1484218
This answer refers to the theorem as written in the 2nd. edition of Pinter's book:
"Root field" seems to be the same of what many of us call "splitting field": $;K;$ is a root field of some polynomial $;f(x)in I[x];$ means $;K=I(a_1,...,a_n);,;a_1,...,a_n;$ all the roots of $;f(x);$ (in some algebraic closure of $;I;$) and any subfield $;Ile Llneq K;$ fulfills that there exists at least one root $;a_n;$ such that $;a_nnotin L;$.
Observe then that $;f(x)in I[x]subset E[x];$ . Either $;E=K;$ and there's nothing to prove, or else $;Elneq K;$ and thus there is some root of $;f(x);$ whic is not in $;E;implies$ $;K;$ is the splitting field (root filed) of $;f(x);$ as polynomial in $;E[x];$ as $;K;$ contains all the roots of $;f;$ ...
answered Jul 18 at 17:41
DonAntonio
173k1484218
edited Jul 20 at 9:02
answered Jul 18 at 17:41
DonAntonio
173k1484218
answered Jul 18 at 17:41
DonAntonio
173k1484218
answered Jul 18 at 17:41
DonAntonio
173k1484218
173k1484218
Doesn't the condition $K=I(a_1,...,a_n)$ already guarantee that any strict subfield of K lacks at least one of the $a_i$? My understanding was that $I(a_1,...,a_n)$ is the smallest field containing $I$ and $a_1, ..., a_n$, so that certainly if $L subsetneq I(a_1,...,a_n)$ then there exists an $a_i$ such that $a_i notin L$.
– WillG
Jul 18 at 18:08
@WillG Yes, of course...and that's the proof! Almost trivial.
– DonAntonio
Jul 18 at 19:34
I'm still confused then, as to what exactly goes wrong when we try to assert the same "theorem" but without the root field premise, i.e. "Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. Then $K$ is the root field of some polynomial over $I$." Is this also true? If not, maybe a counterexample would help me understand what goes wrong...
– WillG
Jul 18 at 20:08
By all means: take for example $;I=E=Bbb Q;,;;K:=Bbb Q(sqrt[4]2);$ , and $; K;$ is not a splitting field of any irreducible rational polynomial.
– DonAntonio
Jul 18 at 21:09
Aha. That just blew my mind. I think my confusion was assuming all field extensions of a field $F$ had to be root fields of something over $F$, which clearly is not the case with $mathbbQ(sqrt[4]2)$. But there's one thing I'm still confused about: if $p(x)$ is the minimum polynomial of $a$ over $I$, then yes, its coefficients are in $E$ and it has a root in $K$. But is $p(x)$ necessarily irreducible in $E$? This seems to be required by Theorem 7 (in my comment above).
– WillG
Jul 19 at 18:55
 |Â
show 2 more comments
Doesn't the condition $K=I(a_1,...,a_n)$ already guarantee that any strict subfield of K lacks at least one of the $a_i$? My understanding was that $I(a_1,...,a_n)$ is the smallest field containing $I$ and $a_1, ..., a_n$, so that certainly if $L subsetneq I(a_1,...,a_n)$ then there exists an $a_i$ such that $a_i notin L$.
– WillG
Jul 18 at 18:08
@WillG Yes, of course...and that's the proof! Almost trivial.
– DonAntonio
Jul 18 at 19:34
I'm still confused then, as to what exactly goes wrong when we try to assert the same "theorem" but without the root field premise, i.e. "Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. Then $K$ is the root field of some polynomial over $I$." Is this also true? If not, maybe a counterexample would help me understand what goes wrong...
– WillG
Jul 18 at 20:08
By all means: take for example $;I=E=Bbb Q;,;;K:=Bbb Q(sqrt[4]2);$ , and $; K;$ is not a splitting field of any irreducible rational polynomial.
– DonAntonio
Jul 18 at 21:09
Aha. That just blew my mind. I think my confusion was assuming all field extensions of a field $F$ had to be root fields of something over $F$, which clearly is not the case with $mathbbQ(sqrt[4]2)$. But there's one thing I'm still confused about: if $p(x)$ is the minimum polynomial of $a$ over $I$, then yes, its coefficients are in $E$ and it has a root in $K$. But is $p(x)$ necessarily irreducible in $E$? This seems to be required by Theorem 7 (in my comment above).
– WillG
Jul 19 at 18:55
Doesn't the condition $K=I(a_1,...,a_n)$ already guarantee that any strict subfield of K lacks at least one of the $a_i$? My understanding was that $I(a_1,...,a_n)$ is the smallest field containing $I$ and $a_1, ..., a_n$, so that certainly if $L subsetneq I(a_1,...,a_n)$ then there exists an $a_i$ such that $a_i notin L$.
– WillG
Jul 18 at 18:08
Doesn't the condition $K=I(a_1,...,a_n)$ already guarantee that any strict subfield of K lacks at least one of the $a_i$? My understanding was that $I(a_1,...,a_n)$ is the smallest field containing $I$ and $a_1, ..., a_n$, so that certainly if $L subsetneq I(a_1,...,a_n)$ then there exists an $a_i$ such that $a_i notin L$.
– WillG
Jul 18 at 18:08
@WillG Yes, of course...and that's the proof! Almost trivial.
– DonAntonio
Jul 18 at 19:34
@WillG Yes, of course...and that's the proof! Almost trivial.
– DonAntonio
Jul 18 at 19:34
I'm still confused then, as to what exactly goes wrong when we try to assert the same "theorem" but without the root field premise, i.e. "Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. Then $K$ is the root field of some polynomial over $I$." Is this also true? If not, maybe a counterexample would help me understand what goes wrong...
– WillG
Jul 18 at 20:08
I'm still confused then, as to what exactly goes wrong when we try to assert the same "theorem" but without the root field premise, i.e. "Suppose $I subseteq E subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. Then $K$ is the root field of some polynomial over $I$." Is this also true? If not, maybe a counterexample would help me understand what goes wrong...
– WillG
Jul 18 at 20:08
By all means: take for example $;I=E=Bbb Q;,;;K:=Bbb Q(sqrt[4]2);$ , and $; K;$ is not a splitting field of any irreducible rational polynomial.
– DonAntonio
Jul 18 at 21:09
By all means: take for example $;I=E=Bbb Q;,;;K:=Bbb Q(sqrt[4]2);$ , and $; K;$ is not a splitting field of any irreducible rational polynomial.
– DonAntonio
Jul 18 at 21:09
Aha. That just blew my mind. I think my confusion was assuming all field extensions of a field $F$ had to be root fields of something over $F$, which clearly is not the case with $mathbbQ(sqrt[4]2)$. But there's one thing I'm still confused about: if $p(x)$ is the minimum polynomial of $a$ over $I$, then yes, its coefficients are in $E$ and it has a root in $K$. But is $p(x)$ necessarily irreducible in $E$? This seems to be required by Theorem 7 (in my comment above).
– WillG
Jul 19 at 18:55
Aha. That just blew my mind. I think my confusion was assuming all field extensions of a field $F$ had to be root fields of something over $F$, which clearly is not the case with $mathbbQ(sqrt[4]2)$. But there's one thing I'm still confused about: if $p(x)$ is the minimum polynomial of $a$ over $I$, then yes, its coefficients are in $E$ and it has a root in $K$. But is $p(x)$ necessarily irreducible in $E$? This seems to be required by Theorem 7 (in my comment above).
– WillG
Jul 19 at 18:55
 |Â
show 2 more comments
up vote
0
down vote
accepted
After reviewing this more, I believe the actual theorem from Pinter is wrong! Counterexample: Let $I=mathbbQ$ and $E=K=mathbbQ(sqrt[4]2)$.
I found the 2nd edition of the same book and discovered the theorem is different:
Suppose $Isubseteq Esubseteq K$ where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $I$, then $K$ is also the root field of some polynomial over $E$.
This makes much more sense now.
answered Jul 20 at 5:03
WillG
40128
I honestly didn't even read your whole question. I simply went to Pinter's book and read from there the theorem, which is the one you mention here (2nd. Edition) and whose proof is pretty trivial. Indeed, the 1st. edition has this theorem wrongly written and, of course, it is false as it appears there.
– DonAntonio
Jul 20 at 9:01
add a comment |Â
up vote
0
down vote
accepted
After reviewing this more, I believe the actual theorem from Pinter is wrong! Counterexample: Let $I=mathbbQ$ and $E=K=mathbbQ(sqrt[4]2)$.
I found the 2nd edition of the same book and discovered the theorem is different:
Suppose $Isubseteq Esubseteq K$ where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $I$, then $K$ is also the root field of some polynomial over $E$.
This makes much more sense now.
answered Jul 20 at 5:03
WillG
40128
I honestly didn't even read your whole question. I simply went to Pinter's book and read from there the theorem, which is the one you mention here (2nd. Edition) and whose proof is pretty trivial. Indeed, the 1st. edition has this theorem wrongly written and, of course, it is false as it appears there.
– DonAntonio
Jul 20 at 9:01
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After reviewing this more, I believe the actual theorem from Pinter is wrong! Counterexample: Let $I=mathbbQ$ and $E=K=mathbbQ(sqrt[4]2)$.
I found the 2nd edition of the same book and discovered the theorem is different:
Suppose $Isubseteq Esubseteq K$ where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $I$, then $K$ is also the root field of some polynomial over $E$.
This makes much more sense now.
answered Jul 20 at 5:03
WillG
40128
After reviewing this more, I believe the actual theorem from Pinter is wrong! Counterexample: Let $I=mathbbQ$ and $E=K=mathbbQ(sqrt[4]2)$.
I found the 2nd edition of the same book and discovered the theorem is different:
Suppose $Isubseteq Esubseteq K$ where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $I$, then $K$ is also the root field of some polynomial over $E$.
This makes much more sense now.
answered Jul 20 at 5:03
WillG
40128
answered Jul 20 at 5:03
WillG
40128
answered Jul 20 at 5:03
WillG
40128
answered Jul 20 at 5:03
WillG
40128
40128
I honestly didn't even read your whole question. I simply went to Pinter's book and read from there the theorem, which is the one you mention here (2nd. Edition) and whose proof is pretty trivial. Indeed, the 1st. edition has this theorem wrongly written and, of course, it is false as it appears there.
– DonAntonio
Jul 20 at 9:01
add a comment |Â
I honestly didn't even read your whole question. I simply went to Pinter's book and read from there the theorem, which is the one you mention here (2nd. Edition) and whose proof is pretty trivial. Indeed, the 1st. edition has this theorem wrongly written and, of course, it is false as it appears there.
– DonAntonio
Jul 20 at 9:01
I honestly didn't even read your whole question. I simply went to Pinter's book and read from there the theorem, which is the one you mention here (2nd. Edition) and whose proof is pretty trivial. Indeed, the 1st. edition has this theorem wrongly written and, of course, it is false as it appears there.
– DonAntonio
Jul 20 at 9:01
I honestly didn't even read your whole question. I simply went to Pinter's book and read from there the theorem, which is the one you mention here (2nd. Edition) and whose proof is pretty trivial. Indeed, the 1st. edition has this theorem wrongly written and, of course, it is false as it appears there.
– DonAntonio
Jul 20 at 9:01
add a comment |Â
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For reference, here is Ch31 Th7, which he refers to: Let $K$ be the root field of some polynomial over $F$. For every irreducible polynomial $p(x)$ in $F[x]$, if $p(x)$ has one root in $K$, then $p(x)$ must have all of its roots in $K$.
– WillG
Jul 18 at 16:45