When are nonintersecting finite degree field extensions linearly disjoint?
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Let $F$ be a field, and let $K,L$ be finite degree field extensions of $F$ inside a common algebraic closure. Consider the following two properties:
(i) $K$ and $L$ are linearly disjoint over $F$: the natural map $K otimes_F L hookrightarrow KL$ is an injection.
(ii) $K cap L = F$.
It is well known that (i) $implies$ (ii): see e.g. $S$ 13.1 of my field theory notes. This implication ought to be (and usually is) followed up with the comment that (ii) does not imply (i) without some additional hypothesis: for instance take $F = mathbb{Q}$, $K = mathbb{Q}(sqrt[3]{2})$, $L = mathbb{Q}(e^{frac{2 pi i}{3}}sqrt[3]{2})$, or more generally, any two distinct, but conjugate, extensions of prime degree. Thus some normality hypothesis is necessary. What is the weakest such hypothesis?
The following is a standard result: see e.g. $S$ 13.3, loc. cit.
Theorem: If $K/F$ and $L/F$ are both Galois, then (ii) $implies$ (i).
I remember this point coming up in a course I took as a graduate student, and the instructor claimed in passing that it was enough for only one of $K$, $L$ to be Galois. None of the standard field theory texts I own contains a proof of this. But by online searching I found an algebra text of P.M. Cohn which proves something stronger.
Theorem: If at least one of $K,L$ is normal and at least one is separable [it is permissible for the same field to be both normal and separable], then (ii) $implies$ (i).
I wasn't able to freely view the proof, so if someone can pass it along to me I'd be appreciative. Still, I think I know of no counterexamples to the following stronger
Claim: If at least one of $K,L$ is normal, then (ii) $implies$ (i).
Is this in fact true? (I believe I have seen it claimed in certain research papers, e.g. one by Piatetski-Shapiro. But because terminology and running separability hypotheses are not so standardized, I don't take this as conclusive evidence.)
field-theory
asked May 5 '13 at 2:44
Pete L. Clark
79.9k9161311
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up vote
55
down vote
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Let $F$ be a field, and let $K,L$ be finite degree field extensions of $F$ inside a common algebraic closure. Consider the following two properties:
(i) $K$ and $L$ are linearly disjoint over $F$: the natural map $K otimes_F L hookrightarrow KL$ is an injection.
(ii) $K cap L = F$.
It is well known that (i) $implies$ (ii): see e.g. $S$ 13.1 of my field theory notes. This implication ought to be (and usually is) followed up with the comment that (ii) does not imply (i) without some additional hypothesis: for instance take $F = mathbb{Q}$, $K = mathbb{Q}(sqrt[3]{2})$, $L = mathbb{Q}(e^{frac{2 pi i}{3}}sqrt[3]{2})$, or more generally, any two distinct, but conjugate, extensions of prime degree. Thus some normality hypothesis is necessary. What is the weakest such hypothesis?
The following is a standard result: see e.g. $S$ 13.3, loc. cit.
Theorem: If $K/F$ and $L/F$ are both Galois, then (ii) $implies$ (i).
I remember this point coming up in a course I took as a graduate student, and the instructor claimed in passing that it was enough for only one of $K$, $L$ to be Galois. None of the standard field theory texts I own contains a proof of this. But by online searching I found an algebra text of P.M. Cohn which proves something stronger.
Theorem: If at least one of $K,L$ is normal and at least one is separable [it is permissible for the same field to be both normal and separable], then (ii) $implies$ (i).
I wasn't able to freely view the proof, so if someone can pass it along to me I'd be appreciative. Still, I think I know of no counterexamples to the following stronger
Claim: If at least one of $K,L$ is normal, then (ii) $implies$ (i).
Is this in fact true? (I believe I have seen it claimed in certain research papers, e.g. one by Piatetski-Shapiro. But because terminology and running separability hypotheses are not so standardized, I don't take this as conclusive evidence.)
field-theory
asked May 5 '13 at 2:44
Pete L. Clark
79.9k9161311
6
Here are the relevant pages from Cohn's Algebra, Volume 3: page 188/1, page 188/2, page 189/1, page 189/2. Cohn doesn't appear to address your question about counterexamples directly in the body of the section, but perhaps there is something hidden in the exercises which I am unable to recognize because I'm lacking the expertise.
– Martin
May 5 '13 at 11:23
1
@Martin: Thanks! That's very helpful.
– Pete L. Clark
May 5 '13 at 11:27
add a comment |
up vote
55
down vote
favorite
up vote
55
down vote
favorite
Let $F$ be a field, and let $K,L$ be finite degree field extensions of $F$ inside a common algebraic closure. Consider the following two properties:
(i) $K$ and $L$ are linearly disjoint over $F$: the natural map $K otimes_F L hookrightarrow KL$ is an injection.
(ii) $K cap L = F$.
It is well known that (i) $implies$ (ii): see e.g. $S$ 13.1 of my field theory notes. This implication ought to be (and usually is) followed up with the comment that (ii) does not imply (i) without some additional hypothesis: for instance take $F = mathbb{Q}$, $K = mathbb{Q}(sqrt[3]{2})$, $L = mathbb{Q}(e^{frac{2 pi i}{3}}sqrt[3]{2})$, or more generally, any two distinct, but conjugate, extensions of prime degree. Thus some normality hypothesis is necessary. What is the weakest such hypothesis?
The following is a standard result: see e.g. $S$ 13.3, loc. cit.
Theorem: If $K/F$ and $L/F$ are both Galois, then (ii) $implies$ (i).
I remember this point coming up in a course I took as a graduate student, and the instructor claimed in passing that it was enough for only one of $K$, $L$ to be Galois. None of the standard field theory texts I own contains a proof of this. But by online searching I found an algebra text of P.M. Cohn which proves something stronger.
Theorem: If at least one of $K,L$ is normal and at least one is separable [it is permissible for the same field to be both normal and separable], then (ii) $implies$ (i).
I wasn't able to freely view the proof, so if someone can pass it along to me I'd be appreciative. Still, I think I know of no counterexamples to the following stronger
Claim: If at least one of $K,L$ is normal, then (ii) $implies$ (i).
Is this in fact true? (I believe I have seen it claimed in certain research papers, e.g. one by Piatetski-Shapiro. But because terminology and running separability hypotheses are not so standardized, I don't take this as conclusive evidence.)
field-theory
asked May 5 '13 at 2:44
Pete L. Clark
79.9k9161311
Let $F$ be a field, and let $K,L$ be finite degree field extensions of $F$ inside a common algebraic closure. Consider the following two properties:
(i) $K$ and $L$ are linearly disjoint over $F$: the natural map $K otimes_F L hookrightarrow KL$ is an injection.
(ii) $K cap L = F$.
It is well known that (i) $implies$ (ii): see e.g. $S$ 13.1 of my field theory notes. This implication ought to be (and usually is) followed up with the comment that (ii) does not imply (i) without some additional hypothesis: for instance take $F = mathbb{Q}$, $K = mathbb{Q}(sqrt[3]{2})$, $L = mathbb{Q}(e^{frac{2 pi i}{3}}sqrt[3]{2})$, or more generally, any two distinct, but conjugate, extensions of prime degree. Thus some normality hypothesis is necessary. What is the weakest such hypothesis?
The following is a standard result: see e.g. $S$ 13.3, loc. cit.
Theorem: If $K/F$ and $L/F$ are both Galois, then (ii) $implies$ (i).
I remember this point coming up in a course I took as a graduate student, and the instructor claimed in passing that it was enough for only one of $K$, $L$ to be Galois. None of the standard field theory texts I own contains a proof of this. But by online searching I found an algebra text of P.M. Cohn which proves something stronger.
Theorem: If at least one of $K,L$ is normal and at least one is separable [it is permissible for the same field to be both normal and separable], then (ii) $implies$ (i).
I wasn't able to freely view the proof, so if someone can pass it along to me I'd be appreciative. Still, I think I know of no counterexamples to the following stronger
Claim: If at least one of $K,L$ is normal, then (ii) $implies$ (i).
Is this in fact true? (I believe I have seen it claimed in certain research papers, e.g. one by Piatetski-Shapiro. But because terminology and running separability hypotheses are not so standardized, I don't take this as conclusive evidence.)
field-theory
field-theory
asked May 5 '13 at 2:44
Pete L. Clark
79.9k9161311
asked May 5 '13 at 2:44
Pete L. Clark
79.9k9161311
edited May 5 '13 at 10:47
asked May 5 '13 at 2:44
Pete L. Clark
79.9k9161311
asked May 5 '13 at 2:44
Pete L. Clark
79.9k9161311
asked May 5 '13 at 2:44
Pete L. Clark
79.9k9161311
79.9k9161311
6
Here are the relevant pages from Cohn's Algebra, Volume 3: page 188/1, page 188/2, page 189/1, page 189/2. Cohn doesn't appear to address your question about counterexamples directly in the body of the section, but perhaps there is something hidden in the exercises which I am unable to recognize because I'm lacking the expertise.
– Martin
May 5 '13 at 11:23
1
@Martin: Thanks! That's very helpful.
– Pete L. Clark
May 5 '13 at 11:27
add a comment |
6
Here are the relevant pages from Cohn's Algebra, Volume 3: page 188/1, page 188/2, page 189/1, page 189/2. Cohn doesn't appear to address your question about counterexamples directly in the body of the section, but perhaps there is something hidden in the exercises which I am unable to recognize because I'm lacking the expertise.
– Martin
May 5 '13 at 11:23
1
@Martin: Thanks! That's very helpful.
– Pete L. Clark
May 5 '13 at 11:27
6
6
Here are the relevant pages from Cohn's Algebra, Volume 3: page 188/1, page 188/2, page 189/1, page 189/2. Cohn doesn't appear to address your question about counterexamples directly in the body of the section, but perhaps there is something hidden in the exercises which I am unable to recognize because I'm lacking the expertise.
– Martin
May 5 '13 at 11:23
Here are the relevant pages from Cohn's Algebra, Volume 3: page 188/1, page 188/2, page 189/1, page 189/2. Cohn doesn't appear to address your question about counterexamples directly in the body of the section, but perhaps there is something hidden in the exercises which I am unable to recognize because I'm lacking the expertise.
– Martin
May 5 '13 at 11:23
1
1
@Martin: Thanks! That's very helpful.
– Pete L. Clark
May 5 '13 at 11:27
@Martin: Thanks! That's very helpful.
– Pete L. Clark
May 5 '13 at 11:27
add a comment |
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There are counterexamples to the Claim. Recall that an extension of a field $F$ of characteristic $p>0$ is separable if and only if it is linearly disjoint from $F^{1/p^infty}$ over $F$. Note that $F^{1/p^infty}/F$ is a normal extension. There is an insparable algebraic extension $K/F$ which does not intersect $F^{1/p^infty} smallsetminus F$ (see Does an inseparable extension have a purely inseparable element?). This gives a counterexample.
answered Nov 22 at 14:25
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There are counterexamples to the Claim. Recall that an extension of a field $F$ of characteristic $p>0$ is separable if and only if it is linearly disjoint from $F^{1/p^infty}$ over $F$. Note that $F^{1/p^infty}/F$ is a normal extension. There is an insparable algebraic extension $K/F$ which does not intersect $F^{1/p^infty} smallsetminus F$ (see Does an inseparable extension have a purely inseparable element?). This gives a counterexample.
answered Nov 22 at 14:25
unigermalto
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up vote
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down vote
There are counterexamples to the Claim. Recall that an extension of a field $F$ of characteristic $p>0$ is separable if and only if it is linearly disjoint from $F^{1/p^infty}$ over $F$. Note that $F^{1/p^infty}/F$ is a normal extension. There is an insparable algebraic extension $K/F$ which does not intersect $F^{1/p^infty} smallsetminus F$ (see Does an inseparable extension have a purely inseparable element?). This gives a counterexample.
answered Nov 22 at 14:25
unigermalto
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up vote
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up vote
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There are counterexamples to the Claim. Recall that an extension of a field $F$ of characteristic $p>0$ is separable if and only if it is linearly disjoint from $F^{1/p^infty}$ over $F$. Note that $F^{1/p^infty}/F$ is a normal extension. There is an insparable algebraic extension $K/F$ which does not intersect $F^{1/p^infty} smallsetminus F$ (see Does an inseparable extension have a purely inseparable element?). This gives a counterexample.
answered Nov 22 at 14:25
unigermalto
134
There are counterexamples to the Claim. Recall that an extension of a field $F$ of characteristic $p>0$ is separable if and only if it is linearly disjoint from $F^{1/p^infty}$ over $F$. Note that $F^{1/p^infty}/F$ is a normal extension. There is an insparable algebraic extension $K/F$ which does not intersect $F^{1/p^infty} smallsetminus F$ (see Does an inseparable extension have a purely inseparable element?). This gives a counterexample.
answered Nov 22 at 14:25
unigermalto
134
answered Nov 22 at 14:25
unigermalto
134
answered Nov 22 at 14:25
unigermalto
134
answered Nov 22 at 14:25
unigermalto
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Here are the relevant pages from Cohn's Algebra, Volume 3: page 188/1, page 188/2, page 189/1, page 189/2. Cohn doesn't appear to address your question about counterexamples directly in the body of the section, but perhaps there is something hidden in the exercises which I am unable to recognize because I'm lacking the expertise.
– Martin
May 5 '13 at 11:23
1
@Martin: Thanks! That's very helpful.
– Pete L. Clark
May 5 '13 at 11:27