Fdtxjr

Multi tool use

What was the first language to use conditional keywords?

Denied boarding although I have proper visa and documentation. To whom should I make a complaint?

What does it mean that physics no longer uses mechanical models to describe phenomena?

What would you call this weird metallic apparatus that allows you to lift people?

An adverb for when you're not exaggerating

Why is the AVR GCC compiler using a full `CALL` even though I have set the `-mshort-calls` flag?

Did Deadpool rescue all of the X-Force?

Selecting user stories during sprint planning

How much damage would a cupful of neutron star matter do to the Earth?

How often does castling occur in grandmaster games?

How come Sam didn't become Lord of Horn Hill?

Is it fair for a professor to grade us on the possession of past papers?

Dating a Former Employee

When a candle burns, why does the top of wick glow if bottom of flame is hottest?

What is a fractional matching?

How could we fake a moon landing now?

How do I use the new nonlinear finite element in Mathematica 12 for this equation?

Should I use a zero-interest credit card for a large one-time purchase?

How can I reduce the gap between left and right of cdot with a macro?

Can a new player join a group only when a new campaign starts?

What is this clumpy 20-30cm high yellow-flowered plant?

Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?

How to react to hostile behavior from a senior developer?

Crossing US/Canada Border for less than 24 hours

Find the rank of the $mathbbC[x]$-module $mathbbC^3$ given by a matrix

1

$begingroup$

As the title says, I'm trying to solve a problem which asks me to find the rank of the $mathbbC[x]$-module $N=mathbbC^3$ given by

$$
A=
beginbmatrix
0 & 0 & 1 \
1 & 0 & 0 \
0 & 1 & 0
endbmatrix.
$$

I'm struggling to see how to do this, because I can't seem to think of a basis. The vector $e_1$ clearly spans $N$ since $A(e_1)=e_2,A(e_2)=e_3$ and hence $(ax^2 +bx + c)e_1 = (a, b, c)^t$, but is obviously not linearly independent since $(x^3 - 1)e_1 = (0, 0, 0)^t $ where clearly $x^3-1 neq 0$ in $N$.

Can anyone suggest an alternative basis? Or is there another way to find the rank of $N$ as a $mathbbC[x]$-module?

abstract-algebra matrices ring-theory modules

share|cite|improve this question

edited Mar 27 at 22:36

user26857

39.6k124284

asked Mar 27 at 17:05

nongnerunongneru

83

$endgroup$

add a comment | 

1

$begingroup$

As the title says, I'm trying to solve a problem which asks me to find the rank of the $mathbbC[x]$-module $N=mathbbC^3$ given by

$$
A=
beginbmatrix
0 & 0 & 1 \
1 & 0 & 0 \
0 & 1 & 0
endbmatrix.
$$

I'm struggling to see how to do this, because I can't seem to think of a basis. The vector $e_1$ clearly spans $N$ since $A(e_1)=e_2,A(e_2)=e_3$ and hence $(ax^2 +bx + c)e_1 = (a, b, c)^t$, but is obviously not linearly independent since $(x^3 - 1)e_1 = (0, 0, 0)^t $ where clearly $x^3-1 neq 0$ in $N$.

Can anyone suggest an alternative basis? Or is there another way to find the rank of $N$ as a $mathbbC[x]$-module?

abstract-algebra matrices ring-theory modules

share|cite|improve this question

edited Mar 27 at 22:36

user26857

39.6k124284

asked Mar 27 at 17:05

nongnerunongneru

83

$endgroup$

add a comment | 

$begingroup$

As the title says, I'm trying to solve a problem which asks me to find the rank of the $mathbbC[x]$-module $N=mathbbC^3$ given by

$$
A=
beginbmatrix
0 & 0 & 1 \
1 & 0 & 0 \
0 & 1 & 0
endbmatrix.
$$

I'm struggling to see how to do this, because I can't seem to think of a basis. The vector $e_1$ clearly spans $N$ since $A(e_1)=e_2,A(e_2)=e_3$ and hence $(ax^2 +bx + c)e_1 = (a, b, c)^t$, but is obviously not linearly independent since $(x^3 - 1)e_1 = (0, 0, 0)^t $ where clearly $x^3-1 neq 0$ in $N$.

Can anyone suggest an alternative basis? Or is there another way to find the rank of $N$ as a $mathbbC[x]$-module?

abstract-algebra matrices ring-theory modules

share|cite|improve this question

edited Mar 27 at 22:36

user26857

39.6k124284

asked Mar 27 at 17:05

nongnerunongneru

83

$endgroup$

share|cite|improve this question

edited Mar 27 at 22:36

user26857

39.6k124284

asked Mar 27 at 17:05

nongnerunongneru

83

share|cite|improve this question

edited Mar 27 at 22:36

user26857

39.6k124284

asked Mar 27 at 17:05

nongnerunongneru

83

edited Mar 27 at 22:36

user26857

39.6k124284

edited Mar 27 at 22:36

user26857

39.6k124284

asked Mar 27 at 17:05

nongnerunongneru

83

asked Mar 27 at 17:05

nongnerunongneru

83

1 Answer
1

active

oldest

votes

0

$begingroup$

The problem you have has nothing to do with $e_1$. For any $vinmathbb C^3$, you will have $(x^3-1)cdot v=0$. So $N$ cannot have a basis in the setup you are looking at.

What one usually does is to consider, instead of $mathbb C[x]$, the quotient $mathbb C[x]/(x^3-1)$. With this new ring of coefficients, the set $e_1$ will be a basis. In fact, any one-element set will be a basis, and so your module is free of rank one.

share|cite|improve this answer

answered Mar 27 at 19:41

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or, if by rank the question means the dimension of $N otimes_mathbbC[x] mathbbC(x)$ as a $mathbbC(x)$-vector space - then the rank would be 0.
  $endgroup$
  – Daniel Schepler
  Mar 27 at 19:56
  

  
  
  
  

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%2f3164784%2ffind-the-rank-of-the-mathbbcx-module-mathbbc3-given-by-a-matrix%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

The problem you have has nothing to do with $e_1$. For any $vinmathbb C^3$, you will have $(x^3-1)cdot v=0$. So $N$ cannot have a basis in the setup you are looking at.

What one usually does is to consider, instead of $mathbb C[x]$, the quotient $mathbb C[x]/(x^3-1)$. With this new ring of coefficients, the set $e_1$ will be a basis. In fact, any one-element set will be a basis, and so your module is free of rank one.

share|cite|improve this answer

answered Mar 27 at 19:41

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or, if by rank the question means the dimension of $N otimes_mathbbC[x] mathbbC(x)$ as a $mathbbC(x)$-vector space - then the rank would be 0.
  $endgroup$
  – Daniel Schepler
  Mar 27 at 19:56
  

  
  
  
  

add a comment | 

0

$begingroup$

The problem you have has nothing to do with $e_1$. For any $vinmathbb C^3$, you will have $(x^3-1)cdot v=0$. So $N$ cannot have a basis in the setup you are looking at.

What one usually does is to consider, instead of $mathbb C[x]$, the quotient $mathbb C[x]/(x^3-1)$. With this new ring of coefficients, the set $e_1$ will be a basis. In fact, any one-element set will be a basis, and so your module is free of rank one.

share|cite|improve this answer

answered Mar 27 at 19:41

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or, if by rank the question means the dimension of $N otimes_mathbbC[x] mathbbC(x)$ as a $mathbbC(x)$-vector space - then the rank would be 0.
  $endgroup$
  – Daniel Schepler
  Mar 27 at 19:56
  

  
  
  
  

add a comment | 

$begingroup$

The problem you have has nothing to do with $e_1$. For any $vinmathbb C^3$, you will have $(x^3-1)cdot v=0$. So $N$ cannot have a basis in the setup you are looking at.

What one usually does is to consider, instead of $mathbb C[x]$, the quotient $mathbb C[x]/(x^3-1)$. With this new ring of coefficients, the set $e_1$ will be a basis. In fact, any one-element set will be a basis, and so your module is free of rank one.

share|cite|improve this answer

answered Mar 27 at 19:41

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

share|cite|improve this answer

answered Mar 27 at 19:41

Martin ArgeramiMartin Argerami

130k1184185

answered Mar 27 at 19:41

Martin ArgeramiMartin Argerami

130k1184185

answered Mar 27 at 19:41

Martin ArgeramiMartin Argerami

130k1184185