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Determining the ideals of a quotient ring

2

2

$begingroup$

Given an ideal $I = langle x^3 - xrangle subseteq Bbb{R}[x]$, determine the ideals in the quotient ring $Bbb{R}[x]/I$.

I understand that the quotient ring is of the form $k[x_1...x_n]/I$ where I is an ideal in $k[x_1...x_n]$. I also understand that the ideals of the quotient ring are in one-to-one correspondence with the ideals of $k[x_1...x_n]$ containing $I$.

However, I'm really confused on how to actually go about finding the ideals of the quotient ring given a specific ideal. Any hints or suggestions would be appreciated, thank you.

abstract-algebra polynomials ring-theory ideals quotient-spaces

share|cite|improve this question

edited 2 hours ago

Servaes

31.1k342101

asked 6 hours ago

MashaMasha

313

New contributor

Masha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have just one indeterminate. There's a very useful description of all ideals of $mathbb{R}[x]$, because it is a principal ideal domain.
  $endgroup$
  – egreg
  6 hours ago
  

  
  
  
  

add a comment | 

2

2

$begingroup$

Given an ideal $I = langle x^3 - xrangle subseteq Bbb{R}[x]$, determine the ideals in the quotient ring $Bbb{R}[x]/I$.

I understand that the quotient ring is of the form $k[x_1...x_n]/I$ where I is an ideal in $k[x_1...x_n]$. I also understand that the ideals of the quotient ring are in one-to-one correspondence with the ideals of $k[x_1...x_n]$ containing $I$.

However, I'm really confused on how to actually go about finding the ideals of the quotient ring given a specific ideal. Any hints or suggestions would be appreciated, thank you.

abstract-algebra polynomials ring-theory ideals quotient-spaces

share|cite|improve this question

edited 2 hours ago

Servaes

31.1k342101

asked 6 hours ago

MashaMasha

313

New contributor

Masha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have just one indeterminate. There's a very useful description of all ideals of $mathbb{R}[x]$, because it is a principal ideal domain.
  $endgroup$
  – egreg
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Given an ideal $I = langle x^3 - xrangle subseteq Bbb{R}[x]$, determine the ideals in the quotient ring $Bbb{R}[x]/I$.

I understand that the quotient ring is of the form $k[x_1...x_n]/I$ where I is an ideal in $k[x_1...x_n]$. I also understand that the ideals of the quotient ring are in one-to-one correspondence with the ideals of $k[x_1...x_n]$ containing $I$.

However, I'm really confused on how to actually go about finding the ideals of the quotient ring given a specific ideal. Any hints or suggestions would be appreciated, thank you.

abstract-algebra polynomials ring-theory ideals quotient-spaces

share|cite|improve this question

edited 2 hours ago

Servaes

31.1k342101

asked 6 hours ago

MashaMasha

313

New contributor

Masha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 2 hours ago

Servaes

31.1k342101

asked 6 hours ago

MashaMasha

313

New contributor

Masha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 hours ago

Servaes

31.1k342101

asked 6 hours ago

MashaMasha

313

New contributor

Masha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 hours ago

Servaes

31.1k342101

edited 2 hours ago

Servaes

31.1k342101

asked 6 hours ago

MashaMasha

313

New contributor

Masha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 6 hours ago

MashaMasha

313

2 Answers
2

active

oldest

votes

4

$begingroup$

Because $Bbb{R}$ is a field, the polynomial ring $Bbb{R}[x]$ in one indeterminate is a principal ideal domain. That means every ideal is of the form $langle frangle$ for some $finBbb{R}[x]$. Now use the fact that
$$langle x^3-xranglesubsetlangle frangle
qquadiffqquad
x^3-xinlangle frangle
qquadiffqquad
ftext{ divides }x^3-x.$$

share|cite|improve this answer

answered 6 hours ago

ServaesServaes

31.1k342101

$endgroup$

add a comment | 

4

$begingroup$

Note that we can factor $x^3 - x = x(x^2 - 1) = x(x + 1)(x - 1)$. Now, by the Chinese Reimander theorem:

$$mathbb R[X]/langle x^3 - xrangle = mathbb R[X] /langle x(x + 1)(x-1) rangle = mathbb R[X]/langle x rangle timesmathbb R[X]/langle x+1 rangle timesmathbb R[X]/langle x-1 rangle $$

We know that:

1. 
   $mathbb R[X]/langle x rangle simeq mathbb R$, since quotienting by $x$ kills all polynomials of degree 1 or higher. What's left are the reals


2. Similarly, $mathbb R[X]/langle x+1 rangle simeq mathbb R [X]/ langle x-1 rangle simeq mathbb R$.

So, the ring that we have is actually $mathbb R times mathbb R times mathbb R$.

Since $mathbb R$ is a field, it has only two idels: the zero ideal and the ring itself.

Now, note that:

1. the product of ideals is an ideal of the product ring


2. Every ideal of the product ring is a product of ideals

and we complete the proof, since the set of all ideals will all be:

$$
{0, mathbb R} times {0, mathbb R} times {0, mathbb R}
$$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

Siddharth BhatSiddharth Bhat

3,3471918

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is not correct. $X^{2}-1$ factors as $(X-1)(X+1)$ over $mathbb{R}$. The Chinese Remainder Theorem therefore shows that $mathbb{R}[X]/langle X^{3}-X rangle cong mathbb{R} oplus mathbb{R} oplus mathbb{R}$.
  $endgroup$
  – Alex Wertheim
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh my god, right, I confused $X^2 - 1$ and $X^2 + 1$. Fixing right now!
  $endgroup$
  – Siddharth Bhat
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexWertheim I just fixed it. Could you please double-check to make sure I didn't miss anything?
  $endgroup$
  – Siddharth Bhat
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looks mostly correct to me! (The factorization is $(X^{3}-X) = X(X^{2}-1)$, not $X(1-X^{2})$. But this is a minor point, which doesn't affect the computation of the quotient up to isomorphism.
  $endgroup$
  – Alex Wertheim
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexWertheim Argh, fixed! Thanks for catching that :)
  $endgroup$
  – Siddharth Bhat
  5 hours ago
  

  
  
  
  

 | 
show 3 more comments

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4

$begingroup$

Because $Bbb{R}$ is a field, the polynomial ring $Bbb{R}[x]$ in one indeterminate is a principal ideal domain. That means every ideal is of the form $langle frangle$ for some $finBbb{R}[x]$. Now use the fact that
$$langle x^3-xranglesubsetlangle frangle
qquadiffqquad
x^3-xinlangle frangle
qquadiffqquad
ftext{ divides }x^3-x.$$

share|cite|improve this answer

answered 6 hours ago

ServaesServaes

31.1k342101

$endgroup$

add a comment | 

4

$begingroup$

Because $Bbb{R}$ is a field, the polynomial ring $Bbb{R}[x]$ in one indeterminate is a principal ideal domain. That means every ideal is of the form $langle frangle$ for some $finBbb{R}[x]$. Now use the fact that
$$langle x^3-xranglesubsetlangle frangle
qquadiffqquad
x^3-xinlangle frangle
qquadiffqquad
ftext{ divides }x^3-x.$$

share|cite|improve this answer

answered 6 hours ago

ServaesServaes

31.1k342101

$endgroup$

add a comment | 

$begingroup$

Because $Bbb{R}$ is a field, the polynomial ring $Bbb{R}[x]$ in one indeterminate is a principal ideal domain. That means every ideal is of the form $langle frangle$ for some $finBbb{R}[x]$. Now use the fact that
$$langle x^3-xranglesubsetlangle frangle
qquadiffqquad
x^3-xinlangle frangle
qquadiffqquad
ftext{ divides }x^3-x.$$

share|cite|improve this answer

answered 6 hours ago

ServaesServaes

31.1k342101

$endgroup$

share|cite|improve this answer

answered 6 hours ago

ServaesServaes

31.1k342101

answered 6 hours ago

ServaesServaes

31.1k342101

answered 6 hours ago

ServaesServaes

31.1k342101

4

$begingroup$

Note that we can factor $x^3 - x = x(x^2 - 1) = x(x + 1)(x - 1)$. Now, by the Chinese Reimander theorem:

$$mathbb R[X]/langle x^3 - xrangle = mathbb R[X] /langle x(x + 1)(x-1) rangle = mathbb R[X]/langle x rangle timesmathbb R[X]/langle x+1 rangle timesmathbb R[X]/langle x-1 rangle $$

We know that:

1. 
   $mathbb R[X]/langle x rangle simeq mathbb R$, since quotienting by $x$ kills all polynomials of degree 1 or higher. What's left are the reals


2. Similarly, $mathbb R[X]/langle x+1 rangle simeq mathbb R [X]/ langle x-1 rangle simeq mathbb R$.

So, the ring that we have is actually $mathbb R times mathbb R times mathbb R$.

Since $mathbb R$ is a field, it has only two idels: the zero ideal and the ring itself.

Now, note that:

1. the product of ideals is an ideal of the product ring


2. Every ideal of the product ring is a product of ideals

and we complete the proof, since the set of all ideals will all be:

$$
{0, mathbb R} times {0, mathbb R} times {0, mathbb R}
$$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

Siddharth BhatSiddharth Bhat

3,3471918

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is not correct. $X^{2}-1$ factors as $(X-1)(X+1)$ over $mathbb{R}$. The Chinese Remainder Theorem therefore shows that $mathbb{R}[X]/langle X^{3}-X rangle cong mathbb{R} oplus mathbb{R} oplus mathbb{R}$.
  $endgroup$
  – Alex Wertheim
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh my god, right, I confused $X^2 - 1$ and $X^2 + 1$. Fixing right now!
  $endgroup$
  – Siddharth Bhat
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexWertheim I just fixed it. Could you please double-check to make sure I didn't miss anything?
  $endgroup$
  – Siddharth Bhat
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looks mostly correct to me! (The factorization is $(X^{3}-X) = X(X^{2}-1)$, not $X(1-X^{2})$. But this is a minor point, which doesn't affect the computation of the quotient up to isomorphism.
  $endgroup$
  – Alex Wertheim
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexWertheim Argh, fixed! Thanks for catching that :)
  $endgroup$
  – Siddharth Bhat
  5 hours ago
  

  
  
  
  

 | 
show 3 more comments

4

$begingroup$

Note that we can factor $x^3 - x = x(x^2 - 1) = x(x + 1)(x - 1)$. Now, by the Chinese Reimander theorem:

$$mathbb R[X]/langle x^3 - xrangle = mathbb R[X] /langle x(x + 1)(x-1) rangle = mathbb R[X]/langle x rangle timesmathbb R[X]/langle x+1 rangle timesmathbb R[X]/langle x-1 rangle $$

We know that:

1. 
   $mathbb R[X]/langle x rangle simeq mathbb R$, since quotienting by $x$ kills all polynomials of degree 1 or higher. What's left are the reals


2. Similarly, $mathbb R[X]/langle x+1 rangle simeq mathbb R [X]/ langle x-1 rangle simeq mathbb R$.

So, the ring that we have is actually $mathbb R times mathbb R times mathbb R$.

Since $mathbb R$ is a field, it has only two idels: the zero ideal and the ring itself.

Now, note that:

1. the product of ideals is an ideal of the product ring


2. Every ideal of the product ring is a product of ideals

and we complete the proof, since the set of all ideals will all be:

$$
{0, mathbb R} times {0, mathbb R} times {0, mathbb R}
$$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

Siddharth BhatSiddharth Bhat

3,3471918

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is not correct. $X^{2}-1$ factors as $(X-1)(X+1)$ over $mathbb{R}$. The Chinese Remainder Theorem therefore shows that $mathbb{R}[X]/langle X^{3}-X rangle cong mathbb{R} oplus mathbb{R} oplus mathbb{R}$.
  $endgroup$
  – Alex Wertheim
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh my god, right, I confused $X^2 - 1$ and $X^2 + 1$. Fixing right now!
  $endgroup$
  – Siddharth Bhat
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexWertheim I just fixed it. Could you please double-check to make sure I didn't miss anything?
  $endgroup$
  – Siddharth Bhat
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looks mostly correct to me! (The factorization is $(X^{3}-X) = X(X^{2}-1)$, not $X(1-X^{2})$. But this is a minor point, which doesn't affect the computation of the quotient up to isomorphism.
  $endgroup$
  – Alex Wertheim
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexWertheim Argh, fixed! Thanks for catching that :)
  $endgroup$
  – Siddharth Bhat
  5 hours ago
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Note that we can factor $x^3 - x = x(x^2 - 1) = x(x + 1)(x - 1)$. Now, by the Chinese Reimander theorem:

$$mathbb R[X]/langle x^3 - xrangle = mathbb R[X] /langle x(x + 1)(x-1) rangle = mathbb R[X]/langle x rangle timesmathbb R[X]/langle x+1 rangle timesmathbb R[X]/langle x-1 rangle $$

We know that:

1. 
   $mathbb R[X]/langle x rangle simeq mathbb R$, since quotienting by $x$ kills all polynomials of degree 1 or higher. What's left are the reals


2. Similarly, $mathbb R[X]/langle x+1 rangle simeq mathbb R [X]/ langle x-1 rangle simeq mathbb R$.

So, the ring that we have is actually $mathbb R times mathbb R times mathbb R$.

Since $mathbb R$ is a field, it has only two idels: the zero ideal and the ring itself.

Now, note that:

1. the product of ideals is an ideal of the product ring


2. Every ideal of the product ring is a product of ideals

and we complete the proof, since the set of all ideals will all be:

$$
{0, mathbb R} times {0, mathbb R} times {0, mathbb R}
$$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

Siddharth BhatSiddharth Bhat

3,3471918

$endgroup$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

Siddharth BhatSiddharth Bhat

3,3471918

answered 6 hours ago

Siddharth BhatSiddharth Bhat

3,3471918

answered 6 hours ago

Siddharth BhatSiddharth Bhat

3,3471918