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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 | 

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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