Ttdfjt

Does the 7th major scale note resolve more strongly to the lower tonic (note 1) than the higher tonic (note 8)?

Decoupling cap routing on a 4 layer PCB

29er Road Tire?

What does "Managed by Windows" do in the Power options for network connection?

Out of scope work duties and resignation

Pressure inside an infinite ocean?

Understanding trademark infringements in a world where many dictionary words are trademarks?

Emotional immaturity of comic-book version of superhero Shazam

Why does sound not move through a wall?

Are Finitely generated modules over a ring also finitely generated over a subring containing the identity?

A factorization game

How to adjust tikz picture so it fits to current size of a table cell?

US born but as a child of foreign diplomat

Where can I go to avoid planes overhead?

Adjacent DEM color matching in QGIS

Find the cheapest shipping option based on item weight

How did Valkyrie get a pegasus in "Endgame"?

Building a list of products from the elements in another list

How did the Venus Express detect lightning?

What are the differences between credential stuffing and password spraying?

Identifying characters

What to use instead of cling film to wrap pastry

Where in Bitcoin Core does it do X?

Does it make sense for a function to return a rvalue reference

Isomorphism of fields via the forgetful functor

Why is $textAut(F)$ of the forgetful functor $F$ on $G$-sets isomorphic to $G$?Trying to find a left adjointadjoint of forgetful functor related to localizationAlgebraic theories and forgetful functorWhat categorical property do these forgetful functors have in common?For what $k$ does the forgetful functor $mathsfFld_k to mathsfSet$ have a left adjoint?Fields on uncountable setsExistence of a cofree functor (right adjoint)Legitimate functor metacategory.Trying to find a left adjoint to the forgetful functor from [C,Set] to Set^Ob(C)

2

$begingroup$

The following article https://tinyurl.com/yydxzxe3 says that

"For the case of fields, given a field F and an isomorphism of sets U(F) → S, there is a unique field whose underlying set is S and which is isomorphic to F as a field via the given function"

The U above is the forgetful functor which we know for fields doesn't have a left adjoint but this could still be possible as we need a weaker condition (i.e. all bijections give us isomorphisms not all morphisms ).

But I don't get how. Say $mathbbF_2^n = mathbbF_2[x]/p(x) $ is a fixed finite field and I have a permutation $pi$ on $2^n$ elements. How do I construct $f(x)$ such that $ mathbbF_2[x]/f(x) cong mathbbF_2^n$ via $pi$?

abstract-algebra category-theory finite-fields

share|cite|improve this question

asked Apr 29 at 15:48

Tushant MittalTushant Mittal

752412

$endgroup$

add a comment | 

2

$begingroup$

The following article https://tinyurl.com/yydxzxe3 says that

"For the case of fields, given a field F and an isomorphism of sets U(F) → S, there is a unique field whose underlying set is S and which is isomorphic to F as a field via the given function"

The U above is the forgetful functor which we know for fields doesn't have a left adjoint but this could still be possible as we need a weaker condition (i.e. all bijections give us isomorphisms not all morphisms ).

But I don't get how. Say $mathbbF_2^n = mathbbF_2[x]/p(x) $ is a fixed finite field and I have a permutation $pi$ on $2^n$ elements. How do I construct $f(x)$ such that $ mathbbF_2[x]/f(x) cong mathbbF_2^n$ via $pi$?

abstract-algebra category-theory finite-fields

share|cite|improve this question

asked Apr 29 at 15:48

Tushant MittalTushant Mittal

752412

$endgroup$

add a comment | 

$begingroup$

The following article https://tinyurl.com/yydxzxe3 says that

"For the case of fields, given a field F and an isomorphism of sets U(F) → S, there is a unique field whose underlying set is S and which is isomorphic to F as a field via the given function"

The U above is the forgetful functor which we know for fields doesn't have a left adjoint but this could still be possible as we need a weaker condition (i.e. all bijections give us isomorphisms not all morphisms ).

But I don't get how. Say $mathbbF_2^n = mathbbF_2[x]/p(x) $ is a fixed finite field and I have a permutation $pi$ on $2^n$ elements. How do I construct $f(x)$ such that $ mathbbF_2[x]/f(x) cong mathbbF_2^n$ via $pi$?

abstract-algebra category-theory finite-fields

share|cite|improve this question

asked Apr 29 at 15:48

Tushant MittalTushant Mittal

752412

$endgroup$

share|cite|improve this question

asked Apr 29 at 15:48

Tushant MittalTushant Mittal

752412

share|cite|improve this question

asked Apr 29 at 15:48

Tushant MittalTushant Mittal

752412

asked Apr 29 at 15:48

Tushant MittalTushant Mittal

752412

asked Apr 29 at 15:48

Tushant MittalTushant Mittal

752412

1 Answer
1

active

oldest

votes

7

$begingroup$

The observation here is just that we can "transport structure" along any isomorphism.

Suppose $F$ is a field and $picolon U(F)to S$ is a bijection of sets. We define operations $+$ and $times$ on $S$ in such a way that $(S;+,times)$ is a field and $picolon Fto (S;+,times)$ is a field isomorphism. For any $a,bin S$, define: beginalign* a+b &= pi(pi^-1(a) + pi^-1(b))\ atimes b &= pi(pi^-1(a)times pi^-1(b)).endalign*
Note that on the right hand sides of the above equations, $+$ and $times$ are the field operations in $F$.

In your example of a permutation $pi$ of the set $U(mathbbF_2^n) = mathbbF_2[x]/p(x)$, this doesn't amount to finding a new quotient of $mathbbF_2[x]$, we just get totally new field structure on the set $mathbbF_2[x]/p(x)$ which has nothing to do with the original one, or with the ring structure on $mathbbF_2[x]$ (e.g. in most cases it will have a different $0$ and a different $1$).

share|cite|improve this answer

edited Apr 29 at 16:50

answered Apr 29 at 16:22

Alex KruckmanAlex Kruckman

29.4k32758

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f)
  $endgroup$
  – Tushant Mittal
  Apr 29 at 16:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  All finite fields are of the above type up to isomorphism. The new field is isomorphic to $mathbbF_2^n$ by construction, so it's isomorphic to the same quotient of $mathbbF_2[x]$, not a new one!
  $endgroup$
  – Alex Kruckman
  Apr 29 at 16:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error!
  $endgroup$
  – Tushant Mittal
  Apr 29 at 16:56
  
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3207070%2fisomorphism-of-fields-via-the-forgetful-functor%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

7

$begingroup$

The observation here is just that we can "transport structure" along any isomorphism.

Suppose $F$ is a field and $picolon U(F)to S$ is a bijection of sets. We define operations $+$ and $times$ on $S$ in such a way that $(S;+,times)$ is a field and $picolon Fto (S;+,times)$ is a field isomorphism. For any $a,bin S$, define: beginalign* a+b &= pi(pi^-1(a) + pi^-1(b))\ atimes b &= pi(pi^-1(a)times pi^-1(b)).endalign*
Note that on the right hand sides of the above equations, $+$ and $times$ are the field operations in $F$.

In your example of a permutation $pi$ of the set $U(mathbbF_2^n) = mathbbF_2[x]/p(x)$, this doesn't amount to finding a new quotient of $mathbbF_2[x]$, we just get totally new field structure on the set $mathbbF_2[x]/p(x)$ which has nothing to do with the original one, or with the ring structure on $mathbbF_2[x]$ (e.g. in most cases it will have a different $0$ and a different $1$).

share|cite|improve this answer

edited Apr 29 at 16:50

answered Apr 29 at 16:22

Alex KruckmanAlex Kruckman

29.4k32758

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f)
  $endgroup$
  – Tushant Mittal
  Apr 29 at 16:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  All finite fields are of the above type up to isomorphism. The new field is isomorphic to $mathbbF_2^n$ by construction, so it's isomorphic to the same quotient of $mathbbF_2[x]$, not a new one!
  $endgroup$
  – Alex Kruckman
  Apr 29 at 16:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error!
  $endgroup$
  – Tushant Mittal
  Apr 29 at 16:56
  
  

  
  
  
  

add a comment | 

7

$begingroup$

The observation here is just that we can "transport structure" along any isomorphism.

Suppose $F$ is a field and $picolon U(F)to S$ is a bijection of sets. We define operations $+$ and $times$ on $S$ in such a way that $(S;+,times)$ is a field and $picolon Fto (S;+,times)$ is a field isomorphism. For any $a,bin S$, define: beginalign* a+b &= pi(pi^-1(a) + pi^-1(b))\ atimes b &= pi(pi^-1(a)times pi^-1(b)).endalign*
Note that on the right hand sides of the above equations, $+$ and $times$ are the field operations in $F$.

In your example of a permutation $pi$ of the set $U(mathbbF_2^n) = mathbbF_2[x]/p(x)$, this doesn't amount to finding a new quotient of $mathbbF_2[x]$, we just get totally new field structure on the set $mathbbF_2[x]/p(x)$ which has nothing to do with the original one, or with the ring structure on $mathbbF_2[x]$ (e.g. in most cases it will have a different $0$ and a different $1$).

share|cite|improve this answer

edited Apr 29 at 16:50

answered Apr 29 at 16:22

Alex KruckmanAlex Kruckman

29.4k32758

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f)
  $endgroup$
  – Tushant Mittal
  Apr 29 at 16:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  All finite fields are of the above type up to isomorphism. The new field is isomorphic to $mathbbF_2^n$ by construction, so it's isomorphic to the same quotient of $mathbbF_2[x]$, not a new one!
  $endgroup$
  – Alex Kruckman
  Apr 29 at 16:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error!
  $endgroup$
  – Tushant Mittal
  Apr 29 at 16:56
  
  

  
  
  
  

add a comment | 

$begingroup$

The observation here is just that we can "transport structure" along any isomorphism.

Suppose $F$ is a field and $picolon U(F)to S$ is a bijection of sets. We define operations $+$ and $times$ on $S$ in such a way that $(S;+,times)$ is a field and $picolon Fto (S;+,times)$ is a field isomorphism. For any $a,bin S$, define: beginalign* a+b &= pi(pi^-1(a) + pi^-1(b))\ atimes b &= pi(pi^-1(a)times pi^-1(b)).endalign*
Note that on the right hand sides of the above equations, $+$ and $times$ are the field operations in $F$.

In your example of a permutation $pi$ of the set $U(mathbbF_2^n) = mathbbF_2[x]/p(x)$, this doesn't amount to finding a new quotient of $mathbbF_2[x]$, we just get totally new field structure on the set $mathbbF_2[x]/p(x)$ which has nothing to do with the original one, or with the ring structure on $mathbbF_2[x]$ (e.g. in most cases it will have a different $0$ and a different $1$).

share|cite|improve this answer

edited Apr 29 at 16:50

answered Apr 29 at 16:22

Alex KruckmanAlex Kruckman

29.4k32758

$endgroup$

share|cite|improve this answer

edited Apr 29 at 16:50

answered Apr 29 at 16:22

Alex KruckmanAlex Kruckman

29.4k32758

answered Apr 29 at 16:22

Alex KruckmanAlex Kruckman

29.4k32758

answered Apr 29 at 16:22

Alex KruckmanAlex Kruckman

29.4k32758