Suppose $F$ is a field and $K$ is a field extension of F. Let $a,b in K$. show $F[a]b=F[b]a$? [on hold]Field extension question.Intermediate ring of an arbitrary field extension is also a field?Field extension of a finite fieldPolynomial ring and extension fieldA field extension of prime degreeLet $K$ be a field extension of $F$ and let $a in K$. Show that $[F(a):F(a^3)] leq 3$If every intermediate ring of a field extension is a field, then the extension is algebraicField extension and homomorphism between ringsThe subring of a field extension contains the field, is a subfield of the field extension$E$ be an algebraic extension of $F$. Suppose $K$ is an integral domain s.t. $F leq K leq E$. Show that $K$ is a field.
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Suppose $F$ is a field and $K$ is a field extension of F. Let $a,b in K$. show $F[a]b=F[b]a$? [on hold]
Field extension question.Intermediate ring of an arbitrary field extension is also a field?Field extension of a finite fieldPolynomial ring and extension fieldA field extension of prime degreeLet $K$ be a field extension of $F$ and let $a in K$. Show that $[F(a):F(a^3)] leq 3$If every intermediate ring of a field extension is a field, then the extension is algebraicField extension and homomorphism between ringsThe subring of a field extension contains the field, is a subfield of the field extension$E$ be an algebraic extension of $F$. Suppose $K$ is an integral domain s.t. $F leq K leq E$. Show that $K$ is a field.
$begingroup$
Suppose $F$ is a field and $K$ is a field extension of $F$. Let $a,b$ be in $K$. How would one show
$(F[a])b=(F[b])a$
ring-theory
edited Mar 12 at 19:46
Max
881318
asked Mar 12 at 19:38
user520403user520403
1157
$endgroup$
put on hold as off-topic by Servaes, TomGrubb, Shailesh, Lee David Chung Lin, hardmath Mar 13 at 2:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, TomGrubb, Shailesh, Lee David Chung Lin, hardmath
add a comment |
$begingroup$
Suppose $F$ is a field and $K$ is a field extension of $F$. Let $a,b$ be in $K$. How would one show
$(F[a])b=(F[b])a$
ring-theory
edited Mar 12 at 19:46
Max
881318
asked Mar 12 at 19:38
user520403user520403
1157
$endgroup$
put on hold as off-topic by Servaes, TomGrubb, Shailesh, Lee David Chung Lin, hardmath Mar 13 at 2:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, TomGrubb, Shailesh, Lee David Chung Lin, hardmath
$begingroup$
Perhaps the notation means something other than what it conveys to me. I parse $F[a]$ as polynomial expressions in $a$ with coefficients in $F$, and respectively for $F[b]$. Then $(F[a])b$ would be products of those polynomials in $a$ times $b$ (and correspondingly for $(F[b])a$ as polynomials in $b$ times $a$). Without further assumptions we cannot conclude that $(F[a])b=(F[b])a$ with that interpretation.
$endgroup$
– hardmath
Mar 13 at 2:18
$begingroup$
I agree it doesn't quite make sense. Thats why I came here to see if I could find an explanation (its on my homework) Thanks anyways!
$endgroup$
– user520403
Mar 13 at 12:29
$begingroup$
I'm raising an issue here of the notation (via Comment, for clarification). You have the advantage of your Readers, who see only the short problem statement (cudos for including it in the body as well as the title, but more context can be added to the body). You've been taking a class, and if you are unsure of the notation's meaning, there is abundant reference material for you to review. But we Readers must rely on you to be the expert to resolve doubts about the meaning of the notation.
$endgroup$
– hardmath
Mar 13 at 14:16
1
$begingroup$
It seems possible that the notation has been garbled. A sensible problem would ask for proof that $(F[a])[b] = (F[b])[a]$, i.e. that the two notions of rings polynomial jointly in $a,b$ agree.
$endgroup$
– hardmath
Mar 13 at 14:18
add a comment |
$begingroup$
Suppose $F$ is a field and $K$ is a field extension of $F$. Let $a,b$ be in $K$. How would one show
$(F[a])b=(F[b])a$
ring-theory
edited Mar 12 at 19:46
Max
881318
asked Mar 12 at 19:38
user520403user520403
1157
$endgroup$
Suppose $F$ is a field and $K$ is a field extension of $F$. Let $a,b$ be in $K$. How would one show
$(F[a])b=(F[b])a$
ring-theory
ring-theory
edited Mar 12 at 19:46
Max
881318
asked Mar 12 at 19:38
user520403user520403
1157
edited Mar 12 at 19:46
Max
881318
asked Mar 12 at 19:38
user520403user520403
1157
edited Mar 12 at 19:46
Max
881318
edited Mar 12 at 19:46
Max
881318
edited Mar 12 at 19:46
Max
881318
881318
asked Mar 12 at 19:38
user520403user520403
1157
asked Mar 12 at 19:38
user520403user520403
1157
asked Mar 12 at 19:38
user520403user520403
1157
1157
put on hold as off-topic by Servaes, TomGrubb, Shailesh, Lee David Chung Lin, hardmath Mar 13 at 2:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, TomGrubb, Shailesh, Lee David Chung Lin, hardmath
put on hold as off-topic by Servaes, TomGrubb, Shailesh, Lee David Chung Lin, hardmath Mar 13 at 2:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, TomGrubb, Shailesh, Lee David Chung Lin, hardmath
$begingroup$
Perhaps the notation means something other than what it conveys to me. I parse $F[a]$ as polynomial expressions in $a$ with coefficients in $F$, and respectively for $F[b]$. Then $(F[a])b$ would be products of those polynomials in $a$ times $b$ (and correspondingly for $(F[b])a$ as polynomials in $b$ times $a$). Without further assumptions we cannot conclude that $(F[a])b=(F[b])a$ with that interpretation.
$endgroup$
– hardmath
Mar 13 at 2:18
$begingroup$
I agree it doesn't quite make sense. Thats why I came here to see if I could find an explanation (its on my homework) Thanks anyways!
$endgroup$
– user520403
Mar 13 at 12:29
$begingroup$
I'm raising an issue here of the notation (via Comment, for clarification). You have the advantage of your Readers, who see only the short problem statement (cudos for including it in the body as well as the title, but more context can be added to the body). You've been taking a class, and if you are unsure of the notation's meaning, there is abundant reference material for you to review. But we Readers must rely on you to be the expert to resolve doubts about the meaning of the notation.
$endgroup$
– hardmath
Mar 13 at 14:16
1
$begingroup$
It seems possible that the notation has been garbled. A sensible problem would ask for proof that $(F[a])[b] = (F[b])[a]$, i.e. that the two notions of rings polynomial jointly in $a,b$ agree.
$endgroup$
– hardmath
Mar 13 at 14:18
add a comment |
$begingroup$
Perhaps the notation means something other than what it conveys to me. I parse $F[a]$ as polynomial expressions in $a$ with coefficients in $F$, and respectively for $F[b]$. Then $(F[a])b$ would be products of those polynomials in $a$ times $b$ (and correspondingly for $(F[b])a$ as polynomials in $b$ times $a$). Without further assumptions we cannot conclude that $(F[a])b=(F[b])a$ with that interpretation.
$endgroup$
– hardmath
Mar 13 at 2:18
$begingroup$
I agree it doesn't quite make sense. Thats why I came here to see if I could find an explanation (its on my homework) Thanks anyways!
$endgroup$
– user520403
Mar 13 at 12:29
$begingroup$
I'm raising an issue here of the notation (via Comment, for clarification). You have the advantage of your Readers, who see only the short problem statement (cudos for including it in the body as well as the title, but more context can be added to the body). You've been taking a class, and if you are unsure of the notation's meaning, there is abundant reference material for you to review. But we Readers must rely on you to be the expert to resolve doubts about the meaning of the notation.
$endgroup$
– hardmath
Mar 13 at 14:16
1
$begingroup$
It seems possible that the notation has been garbled. A sensible problem would ask for proof that $(F[a])[b] = (F[b])[a]$, i.e. that the two notions of rings polynomial jointly in $a,b$ agree.
$endgroup$
– hardmath
Mar 13 at 14:18
$begingroup$
Perhaps the notation means something other than what it conveys to me. I parse $F[a]$ as polynomial expressions in $a$ with coefficients in $F$, and respectively for $F[b]$. Then $(F[a])b$ would be products of those polynomials in $a$ times $b$ (and correspondingly for $(F[b])a$ as polynomials in $b$ times $a$). Without further assumptions we cannot conclude that $(F[a])b=(F[b])a$ with that interpretation.
$endgroup$
– hardmath
Mar 13 at 2:18
$begingroup$
Perhaps the notation means something other than what it conveys to me. I parse $F[a]$ as polynomial expressions in $a$ with coefficients in $F$, and respectively for $F[b]$. Then $(F[a])b$ would be products of those polynomials in $a$ times $b$ (and correspondingly for $(F[b])a$ as polynomials in $b$ times $a$). Without further assumptions we cannot conclude that $(F[a])b=(F[b])a$ with that interpretation.
$endgroup$
– hardmath
Mar 13 at 2:18
$begingroup$
I agree it doesn't quite make sense. Thats why I came here to see if I could find an explanation (its on my homework) Thanks anyways!
$endgroup$
– user520403
Mar 13 at 12:29
$begingroup$
I agree it doesn't quite make sense. Thats why I came here to see if I could find an explanation (its on my homework) Thanks anyways!
$endgroup$
– user520403
Mar 13 at 12:29
$begingroup$
I'm raising an issue here of the notation (via Comment, for clarification). You have the advantage of your Readers, who see only the short problem statement (cudos for including it in the body as well as the title, but more context can be added to the body). You've been taking a class, and if you are unsure of the notation's meaning, there is abundant reference material for you to review. But we Readers must rely on you to be the expert to resolve doubts about the meaning of the notation.
$endgroup$
– hardmath
Mar 13 at 14:16
$begingroup$
I'm raising an issue here of the notation (via Comment, for clarification). You have the advantage of your Readers, who see only the short problem statement (cudos for including it in the body as well as the title, but more context can be added to the body). You've been taking a class, and if you are unsure of the notation's meaning, there is abundant reference material for you to review. But we Readers must rely on you to be the expert to resolve doubts about the meaning of the notation.
$endgroup$
– hardmath
Mar 13 at 14:16
1
1
$begingroup$
It seems possible that the notation has been garbled. A sensible problem would ask for proof that $(F[a])[b] = (F[b])[a]$, i.e. that the two notions of rings polynomial jointly in $a,b$ agree.
$endgroup$
– hardmath
Mar 13 at 14:18
$begingroup$
It seems possible that the notation has been garbled. A sensible problem would ask for proof that $(F[a])[b] = (F[b])[a]$, i.e. that the two notions of rings polynomial jointly in $a,b$ agree.
$endgroup$
– hardmath
Mar 13 at 14:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
False: Take $K = F(a)$, $a$ transcendental over $F$, $b = 1$.
answered Mar 12 at 20:19
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
False: Take $K = F(a)$, $a$ transcendental over $F$, $b = 1$.
answered Mar 12 at 20:19
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
add a comment |
$begingroup$
False: Take $K = F(a)$, $a$ transcendental over $F$, $b = 1$.
answered Mar 12 at 20:19
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
add a comment |
$begingroup$
False: Take $K = F(a)$, $a$ transcendental over $F$, $b = 1$.
answered Mar 12 at 20:19
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
False: Take $K = F(a)$, $a$ transcendental over $F$, $b = 1$.
answered Mar 12 at 20:19
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
answered Mar 12 at 20:19
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
answered Mar 12 at 20:19
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
answered Mar 12 at 20:19
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
34.6k42971
add a comment |
add a comment |
$begingroup$
Perhaps the notation means something other than what it conveys to me. I parse $F[a]$ as polynomial expressions in $a$ with coefficients in $F$, and respectively for $F[b]$. Then $(F[a])b$ would be products of those polynomials in $a$ times $b$ (and correspondingly for $(F[b])a$ as polynomials in $b$ times $a$). Without further assumptions we cannot conclude that $(F[a])b=(F[b])a$ with that interpretation.
$endgroup$
– hardmath
Mar 13 at 2:18
$begingroup$
I agree it doesn't quite make sense. Thats why I came here to see if I could find an explanation (its on my homework) Thanks anyways!
$endgroup$
– user520403
Mar 13 at 12:29
$begingroup$
I'm raising an issue here of the notation (via Comment, for clarification). You have the advantage of your Readers, who see only the short problem statement (cudos for including it in the body as well as the title, but more context can be added to the body). You've been taking a class, and if you are unsure of the notation's meaning, there is abundant reference material for you to review. But we Readers must rely on you to be the expert to resolve doubts about the meaning of the notation.
$endgroup$
– hardmath
Mar 13 at 14:16
1
$begingroup$
It seems possible that the notation has been garbled. A sensible problem would ask for proof that $(F[a])[b] = (F[b])[a]$, i.e. that the two notions of rings polynomial jointly in $a,b$ agree.
$endgroup$
– hardmath
Mar 13 at 14:18