Intuition about turning a polynomial ring into a field

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Let's consider the polynomial ring $R=F[x]$ over a field $F$. Then by taking the quotient by a principal ideal $I=(f(x))$ generated by an irreducible polynomial $f(x)$, we obtain a field $R'=F[x]/(f(x))$.

It's easy to see that $R'$ is indeed a field. Since the ideals of $R$ which contain $I$ are in bijective correspondence with the ideals of $R'$, we can conclude that $R'$ has only two ideals and is therefore a field (as $I$ is maximal in $R$ since $f(x)$ is irreducible).

I wanted to ask, is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field? I would ideally like some way of demonstrating that the existence of a nonzero polynomial equivalent to zero in $R'$ somehow allows us to describe an algorithm to calculate multiplicative inverses...

abstract-algebra polynomials

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asked Jul 29 at 15:53

Beckham Myers

1185

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  The extended Euclidean algorithm is such an algorithm. When $(f,g)=1$, which for irreducible $f$ happens every time that $g$ is not a multiple of $f$, then there are $a,bin F[x]$ such that $af+bg=1$. Hence the class of $b$ is the inverse of the class of $g$ in $F[x]/(f)$.
  – user578878
  Jul 29 at 16:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is what I was looking for, thanks for your help
  – Beckham Myers
  Jul 29 at 16:51
  

  
  
  
  

add a comment | 

up vote
8
down vote

favorite

1

Let's consider the polynomial ring $R=F[x]$ over a field $F$. Then by taking the quotient by a principal ideal $I=(f(x))$ generated by an irreducible polynomial $f(x)$, we obtain a field $R'=F[x]/(f(x))$.

It's easy to see that $R'$ is indeed a field. Since the ideals of $R$ which contain $I$ are in bijective correspondence with the ideals of $R'$, we can conclude that $R'$ has only two ideals and is therefore a field (as $I$ is maximal in $R$ since $f(x)$ is irreducible).

I wanted to ask, is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field? I would ideally like some way of demonstrating that the existence of a nonzero polynomial equivalent to zero in $R'$ somehow allows us to describe an algorithm to calculate multiplicative inverses...

abstract-algebra polynomials

share|cite|improve this question

asked Jul 29 at 15:53

Beckham Myers

1185

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The extended Euclidean algorithm is such an algorithm. When $(f,g)=1$, which for irreducible $f$ happens every time that $g$ is not a multiple of $f$, then there are $a,bin F[x]$ such that $af+bg=1$. Hence the class of $b$ is the inverse of the class of $g$ in $F[x]/(f)$.
  – user578878
  Jul 29 at 16:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is what I was looking for, thanks for your help
  – Beckham Myers
  Jul 29 at 16:51
  

  
  
  
  

add a comment | 

up vote
8
down vote

favorite

1

up vote
8
down vote

favorite

1

1

Let's consider the polynomial ring $R=F[x]$ over a field $F$. Then by taking the quotient by a principal ideal $I=(f(x))$ generated by an irreducible polynomial $f(x)$, we obtain a field $R'=F[x]/(f(x))$.

It's easy to see that $R'$ is indeed a field. Since the ideals of $R$ which contain $I$ are in bijective correspondence with the ideals of $R'$, we can conclude that $R'$ has only two ideals and is therefore a field (as $I$ is maximal in $R$ since $f(x)$ is irreducible).

I wanted to ask, is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field? I would ideally like some way of demonstrating that the existence of a nonzero polynomial equivalent to zero in $R'$ somehow allows us to describe an algorithm to calculate multiplicative inverses...

abstract-algebra polynomials

share|cite|improve this question

asked Jul 29 at 15:53

Beckham Myers

1185

Let's consider the polynomial ring $R=F[x]$ over a field $F$. Then by taking the quotient by a principal ideal $I=(f(x))$ generated by an irreducible polynomial $f(x)$, we obtain a field $R'=F[x]/(f(x))$.

It's easy to see that $R'$ is indeed a field. Since the ideals of $R$ which contain $I$ are in bijective correspondence with the ideals of $R'$, we can conclude that $R'$ has only two ideals and is therefore a field (as $I$ is maximal in $R$ since $f(x)$ is irreducible).

I wanted to ask, is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field? I would ideally like some way of demonstrating that the existence of a nonzero polynomial equivalent to zero in $R'$ somehow allows us to describe an algorithm to calculate multiplicative inverses...

abstract-algebra polynomials

share|cite|improve this question

asked Jul 29 at 15:53

Beckham Myers

1185

share|cite|improve this question

share|cite|improve this question

asked Jul 29 at 15:53

Beckham Myers

1185

asked Jul 29 at 15:53

Beckham Myers

1185

asked Jul 29 at 15:53

Beckham Myers

1185

1185

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The extended Euclidean algorithm is such an algorithm. When $(f,g)=1$, which for irreducible $f$ happens every time that $g$ is not a multiple of $f$, then there are $a,bin F[x]$ such that $af+bg=1$. Hence the class of $b$ is the inverse of the class of $g$ in $F[x]/(f)$.
  – user578878
  Jul 29 at 16:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is what I was looking for, thanks for your help
  – Beckham Myers
  Jul 29 at 16:51
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The extended Euclidean algorithm is such an algorithm. When $(f,g)=1$, which for irreducible $f$ happens every time that $g$ is not a multiple of $f$, then there are $a,bin F[x]$ such that $af+bg=1$. Hence the class of $b$ is the inverse of the class of $g$ in $F[x]/(f)$.
  – user578878
  Jul 29 at 16:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is what I was looking for, thanks for your help
  – Beckham Myers
  Jul 29 at 16:51
  

  
  
  
  

2

2

The extended Euclidean algorithm is such an algorithm. When $(f,g)=1$, which for irreducible $f$ happens every time that $g$ is not a multiple of $f$, then there are $a,bin F[x]$ such that $af+bg=1$. Hence the class of $b$ is the inverse of the class of $g$ in $F[x]/(f)$.
– user578878
Jul 29 at 16:03

The extended Euclidean algorithm is such an algorithm. When $(f,g)=1$, which for irreducible $f$ happens every time that $g$ is not a multiple of $f$, then there are $a,bin F[x]$ such that $af+bg=1$. Hence the class of $b$ is the inverse of the class of $g$ in $F[x]/(f)$.
– user578878
Jul 29 at 16:03

This is what I was looking for, thanks for your help
– Beckham Myers
Jul 29 at 16:51

This is what I was looking for, thanks for your help
– Beckham Myers
Jul 29 at 16:51

add a comment | 

1 Answer
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4
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This is more an answer to your first question is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field?

The main idea, is that for a commutative ring $R$ with unity, and an ideal $J$ of $R$, the following are equivalent :

1. $J$ is a maximal ideal.


2. $R/J$ is a field.

This is a general result on commutative rings with unity. Proving that 1. implies 2. is easy. Take $x notin J$. Then as $J$ is maximal, $(x)+J=R$. Which means that it exists $(a,j) in R times I$ such that $ax+j=1$ or $bar a bar x = bar 1$ in $R/J$ proving that $R/J$ is a field.

Now take the ring $R=F[x]$. It is commutative with unity. But it has more specificities. It is an Euclidean domain as on the polynomials over a field, the polynomial division is an Euclidean division.

And in a Euclidean domain, an irreducible polynomial generates a maximal ideal. Applying the result I mentioned in introduction, you get that $F[x]/(f(x))$ is a field.

Not sure that it is intuitive. However if you can look in details at the proof of the introduction result on maximal ideals, it can be a way to get a kind of intuition on your question.

share|cite|improve this answer

edited Jul 29 at 17:07

answered Jul 29 at 16:46

mathcounterexamples.net

23.3k21651

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up vote
4
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This is more an answer to your first question is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field?

The main idea, is that for a commutative ring $R$ with unity, and an ideal $J$ of $R$, the following are equivalent :

1. $J$ is a maximal ideal.


2. $R/J$ is a field.

This is a general result on commutative rings with unity. Proving that 1. implies 2. is easy. Take $x notin J$. Then as $J$ is maximal, $(x)+J=R$. Which means that it exists $(a,j) in R times I$ such that $ax+j=1$ or $bar a bar x = bar 1$ in $R/J$ proving that $R/J$ is a field.

Now take the ring $R=F[x]$. It is commutative with unity. But it has more specificities. It is an Euclidean domain as on the polynomials over a field, the polynomial division is an Euclidean division.

And in a Euclidean domain, an irreducible polynomial generates a maximal ideal. Applying the result I mentioned in introduction, you get that $F[x]/(f(x))$ is a field.

Not sure that it is intuitive. However if you can look in details at the proof of the introduction result on maximal ideals, it can be a way to get a kind of intuition on your question.

share|cite|improve this answer

edited Jul 29 at 17:07

answered Jul 29 at 16:46

mathcounterexamples.net

23.3k21651

add a comment | 

up vote
4
down vote

This is more an answer to your first question is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field?

The main idea, is that for a commutative ring $R$ with unity, and an ideal $J$ of $R$, the following are equivalent :

1. $J$ is a maximal ideal.


2. $R/J$ is a field.

This is a general result on commutative rings with unity. Proving that 1. implies 2. is easy. Take $x notin J$. Then as $J$ is maximal, $(x)+J=R$. Which means that it exists $(a,j) in R times I$ such that $ax+j=1$ or $bar a bar x = bar 1$ in $R/J$ proving that $R/J$ is a field.

Now take the ring $R=F[x]$. It is commutative with unity. But it has more specificities. It is an Euclidean domain as on the polynomials over a field, the polynomial division is an Euclidean division.

And in a Euclidean domain, an irreducible polynomial generates a maximal ideal. Applying the result I mentioned in introduction, you get that $F[x]/(f(x))$ is a field.

Not sure that it is intuitive. However if you can look in details at the proof of the introduction result on maximal ideals, it can be a way to get a kind of intuition on your question.

share|cite|improve this answer

edited Jul 29 at 17:07

answered Jul 29 at 16:46

mathcounterexamples.net

23.3k21651

add a comment | 

up vote
4
down vote

up vote
4
down vote

This is more an answer to your first question is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field?

The main idea, is that for a commutative ring $R$ with unity, and an ideal $J$ of $R$, the following are equivalent :

1. $J$ is a maximal ideal.


2. $R/J$ is a field.

This is a general result on commutative rings with unity. Proving that 1. implies 2. is easy. Take $x notin J$. Then as $J$ is maximal, $(x)+J=R$. Which means that it exists $(a,j) in R times I$ such that $ax+j=1$ or $bar a bar x = bar 1$ in $R/J$ proving that $R/J$ is a field.

Now take the ring $R=F[x]$. It is commutative with unity. But it has more specificities. It is an Euclidean domain as on the polynomials over a field, the polynomial division is an Euclidean division.

And in a Euclidean domain, an irreducible polynomial generates a maximal ideal. Applying the result I mentioned in introduction, you get that $F[x]/(f(x))$ is a field.

Not sure that it is intuitive. However if you can look in details at the proof of the introduction result on maximal ideals, it can be a way to get a kind of intuition on your question.

share|cite|improve this answer

edited Jul 29 at 17:07

answered Jul 29 at 16:46

mathcounterexamples.net

23.3k21651

This is more an answer to your first question is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field?

The main idea, is that for a commutative ring $R$ with unity, and an ideal $J$ of $R$, the following are equivalent :

1. $J$ is a maximal ideal.


2. $R/J$ is a field.

This is a general result on commutative rings with unity. Proving that 1. implies 2. is easy. Take $x notin J$. Then as $J$ is maximal, $(x)+J=R$. Which means that it exists $(a,j) in R times I$ such that $ax+j=1$ or $bar a bar x = bar 1$ in $R/J$ proving that $R/J$ is a field.

Now take the ring $R=F[x]$. It is commutative with unity. But it has more specificities. It is an Euclidean domain as on the polynomials over a field, the polynomial division is an Euclidean division.

And in a Euclidean domain, an irreducible polynomial generates a maximal ideal. Applying the result I mentioned in introduction, you get that $F[x]/(f(x))$ is a field.

Not sure that it is intuitive. However if you can look in details at the proof of the introduction result on maximal ideals, it can be a way to get a kind of intuition on your question.

share|cite|improve this answer

edited Jul 29 at 17:07

answered Jul 29 at 16:46

mathcounterexamples.net

23.3k21651

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 17:07

edited Jul 29 at 17:07

edited Jul 29 at 17:07

answered Jul 29 at 16:46

mathcounterexamples.net

23.3k21651

answered Jul 29 at 16:46

mathcounterexamples.net

23.3k21651

answered Jul 29 at 16:46

mathcounterexamples.net

23.3k21651

23.3k21651

add a comment | 

add a comment | 

 

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