Suppose $F$ is a field and $K$ is a field extension of F. Let $a,b in K$. show $F[a]b=F[b]a$? [closed]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.

Differential and Linear trail propagation in Noekeon

HP P840 HDD RAID 5 many strange drive failures

Is it possible to stack the damage done by the Absorb Elements spell?

Have the tides ever turned twice on any open problem?

Is it insecure to send a password in a `curl` command?

Do native speakers use "ultima" and "proxima" frequently in spoken English?

What does Jesus mean regarding "Raca," and "you fool?" - is he contrasting them?

Matrix using tikz package

Could Sinn Fein swing any Brexit vote in Parliament?

Is honey really a supersaturated solution? Does heating to un-crystalize redissolve it or melt it?

Does .bashrc contain syntax errors?

How to generate binary array whose elements with values 1 are randomly drawn

PTIJ What is the inyan of the Konami code in Uncle Moishy's song?

How to get the n-th line after a grepped one?

Asserting that Atheism and Theism are both faith based positions

Bash - pair each line of file

Geography in 3D perspective

Tikz: place node leftmost of two nodes of different widths

Would it be believable to defy demographics in a story?

Unfrosted light bulb

Why are there no stars visible in cislunar space?

How do hiring committees for research positions view getting "scooped"?

I seem to dance, I am not a dancer. Who am I?

How to define limit operations in general topological spaces? Are nets able to do this?



Suppose $F$ is a field and $K$ is a field extension of F. Let $a,b in K$. show $F[a]b=F[b]a$? [closed]


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.













1












$begingroup$


Suppose $F$ is a field and $K$ is a field extension of $F$. Let $a,b$ be in $K$. How would one show
$(F[a])b=(F[b])a$










share|cite|improve this question











$endgroup$



closed 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
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    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















1












$begingroup$


Suppose $F$ is a field and $K$ is a field extension of $F$. Let $a,b$ be in $K$. How would one show
$(F[a])b=(F[b])a$










share|cite|improve this question











$endgroup$



closed 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
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    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













1












1








1





$begingroup$


Suppose $F$ is a field and $K$ is a field extension of $F$. Let $a,b$ be in $K$. How would one show
$(F[a])b=(F[b])a$










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 12 at 19:46









Max

881318




881318










asked Mar 12 at 19:38









user520403user520403

1157




1157




closed 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
If this question can be reworded to fit the rules in the help center, please edit the question.







closed 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
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $begingroup$
    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










  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

False: Take $K = F(a)$, $a$ transcendental over $F$, $b = 1$.






share|cite|improve this answer









$endgroup$



















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    False: Take $K = F(a)$, $a$ transcendental over $F$, $b = 1$.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      False: Take $K = F(a)$, $a$ transcendental over $F$, $b = 1$.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        False: Take $K = F(a)$, $a$ transcendental over $F$, $b = 1$.






        share|cite|improve this answer









        $endgroup$



        False: Take $K = F(a)$, $a$ transcendental over $F$, $b = 1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 12 at 20:19









        Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

        34.6k42971




        34.6k42971













            Popular posts from this blog

            How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

            random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

            Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye