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Is the definition of integral extension, why we use monic polynomial?

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Is the definition of integral extension, why we use monic polynomial?

Why do we use this definition of “algebraic integer”?Simple integral extension questionWhy mentioning monic is important for gcd and lcm?Polynomial roots in the ring extensionQuestion on integral extension and minimal polynomialNon-uniqueness of minimal polynomial for integral extensionsAbout the definition of an integral element in commutative ringsAn irreducible polynomial of degree n over field is irreducible over extension of degree m if m and n are coprime

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4

$begingroup$

This is the definition of integral over $R$:

Let $S$ be an extension ring of $R$ and $sin S$. If there exists a monic polynomial $f(X)in R[X]$ such that $s$ is a root of $f$ (i.e., $f(s)=0$), then $s$ is said to be integral over $R$.

My question is: why do we use monic polynomial!?

abstract-algebra

share|cite|improve this question

edited 8 hours ago

egreg

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asked 8 hours ago

Sanu Adhikary Sanu Adhikary

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See Why do we use this definition of “algebraic integer”?
  $endgroup$
  – Bill Dubuque
  4 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

This is the definition of integral over $R$:

Let $S$ be an extension ring of $R$ and $sin S$. If there exists a monic polynomial $f(X)in R[X]$ such that $s$ is a root of $f$ (i.e., $f(s)=0$), then $s$ is said to be integral over $R$.

My question is: why do we use monic polynomial!?

abstract-algebra

share|cite|improve this question

edited 8 hours ago

egreg

192k1414 gold badges9191 silver badges217217 bronze badges

asked 8 hours ago

Sanu Adhikary Sanu Adhikary

3933 bronze badges

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See Why do we use this definition of “algebraic integer”?
  $endgroup$
  – Bill Dubuque
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This is the definition of integral over $R$:

Let $S$ be an extension ring of $R$ and $sin S$. If there exists a monic polynomial $f(X)in R[X]$ such that $s$ is a root of $f$ (i.e., $f(s)=0$), then $s$ is said to be integral over $R$.

My question is: why do we use monic polynomial!?

abstract-algebra

share|cite|improve this question

edited 8 hours ago

egreg

192k1414 gold badges9191 silver badges217217 bronze badges

asked 8 hours ago

Sanu Adhikary Sanu Adhikary

3933 bronze badges

New contributor

Sanu Adhikary 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 8 hours ago

egreg

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asked 8 hours ago

Sanu Adhikary Sanu Adhikary

3933 bronze badges

New contributor

Sanu Adhikary 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 8 hours ago

egreg

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asked 8 hours ago

Sanu Adhikary Sanu Adhikary

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Check out our Code of Conduct.

edited 8 hours ago

egreg

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edited 8 hours ago

egreg

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asked 8 hours ago

Sanu Adhikary Sanu Adhikary

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asked 8 hours ago

Sanu Adhikary Sanu Adhikary

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2 Answers
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5

$begingroup$

The ring of integers $mathcal O_K$ of a number field $K$ is defined as the set of roots of monic polynomials with integer coefficients to mimic the relationship of $Bbb Z$ to $Bbb Q$: $a in Bbb Q$ is in $Bbb Z$ iff it is the root of a (monic) polynomial of the form $x - b$ with $b in Bbb Z$.

If we relaxed the condition to allow arbitrary polynomials in $Bbb Z[x]$, then any rational number $r/s$, being a root of $sx - r$, would be an "integer" and the ring of integers would coincide with the field it lies in (since $K = operatornameFrac(mathcal O_K$)), making the definition useless.

This motivates the definition of an integral element for arbitrary ring extensions too.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Lukas KoflerLukas Kofler

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$endgroup$

add a comment | 

1

$begingroup$

Well, if you have monic polynomial $f(x)in R[x]$ of minimal degree with root $s$ and there is another polynomial $h(x)in R[x]$ which has root $s$, then you can divide $h(x)$ by $f(x)$ as the leading coefficient of $f(x)$ is $1$ (or a unit in $R$). Otherwise, this would not be possible.

share|cite|improve this answer

answered 8 hours ago

WuestenfuxWuestenfux

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$endgroup$

add a comment | 

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5

$begingroup$

The ring of integers $mathcal O_K$ of a number field $K$ is defined as the set of roots of monic polynomials with integer coefficients to mimic the relationship of $Bbb Z$ to $Bbb Q$: $a in Bbb Q$ is in $Bbb Z$ iff it is the root of a (monic) polynomial of the form $x - b$ with $b in Bbb Z$.

If we relaxed the condition to allow arbitrary polynomials in $Bbb Z[x]$, then any rational number $r/s$, being a root of $sx - r$, would be an "integer" and the ring of integers would coincide with the field it lies in (since $K = operatornameFrac(mathcal O_K$)), making the definition useless.

This motivates the definition of an integral element for arbitrary ring extensions too.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Lukas KoflerLukas Kofler

2,44422 gold badges99 silver badges2626 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

The ring of integers $mathcal O_K$ of a number field $K$ is defined as the set of roots of monic polynomials with integer coefficients to mimic the relationship of $Bbb Z$ to $Bbb Q$: $a in Bbb Q$ is in $Bbb Z$ iff it is the root of a (monic) polynomial of the form $x - b$ with $b in Bbb Z$.

If we relaxed the condition to allow arbitrary polynomials in $Bbb Z[x]$, then any rational number $r/s$, being a root of $sx - r$, would be an "integer" and the ring of integers would coincide with the field it lies in (since $K = operatornameFrac(mathcal O_K$)), making the definition useless.

This motivates the definition of an integral element for arbitrary ring extensions too.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Lukas KoflerLukas Kofler

2,44422 gold badges99 silver badges2626 bronze badges

$endgroup$

add a comment | 

$begingroup$

The ring of integers $mathcal O_K$ of a number field $K$ is defined as the set of roots of monic polynomials with integer coefficients to mimic the relationship of $Bbb Z$ to $Bbb Q$: $a in Bbb Q$ is in $Bbb Z$ iff it is the root of a (monic) polynomial of the form $x - b$ with $b in Bbb Z$.

If we relaxed the condition to allow arbitrary polynomials in $Bbb Z[x]$, then any rational number $r/s$, being a root of $sx - r$, would be an "integer" and the ring of integers would coincide with the field it lies in (since $K = operatornameFrac(mathcal O_K$)), making the definition useless.

This motivates the definition of an integral element for arbitrary ring extensions too.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Lukas KoflerLukas Kofler

2,44422 gold badges99 silver badges2626 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Lukas KoflerLukas Kofler

2,44422 gold badges99 silver badges2626 bronze badges

answered 8 hours ago

Lukas KoflerLukas Kofler

2,44422 gold badges99 silver badges2626 bronze badges

answered 8 hours ago

Lukas KoflerLukas Kofler

2,44422 gold badges99 silver badges2626 bronze badges

1

$begingroup$

Well, if you have monic polynomial $f(x)in R[x]$ of minimal degree with root $s$ and there is another polynomial $h(x)in R[x]$ which has root $s$, then you can divide $h(x)$ by $f(x)$ as the leading coefficient of $f(x)$ is $1$ (or a unit in $R$). Otherwise, this would not be possible.

share|cite|improve this answer

answered 8 hours ago

WuestenfuxWuestenfux

10.5k22 gold badges66 silver badges1717 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Well, if you have monic polynomial $f(x)in R[x]$ of minimal degree with root $s$ and there is another polynomial $h(x)in R[x]$ which has root $s$, then you can divide $h(x)$ by $f(x)$ as the leading coefficient of $f(x)$ is $1$ (or a unit in $R$). Otherwise, this would not be possible.

share|cite|improve this answer

answered 8 hours ago

WuestenfuxWuestenfux

10.5k22 gold badges66 silver badges1717 bronze badges

$endgroup$

add a comment | 

$begingroup$

Well, if you have monic polynomial $f(x)in R[x]$ of minimal degree with root $s$ and there is another polynomial $h(x)in R[x]$ which has root $s$, then you can divide $h(x)$ by $f(x)$ as the leading coefficient of $f(x)$ is $1$ (or a unit in $R$). Otherwise, this would not be possible.

share|cite|improve this answer

answered 8 hours ago

WuestenfuxWuestenfux

10.5k22 gold badges66 silver badges1717 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

WuestenfuxWuestenfux

10.5k22 gold badges66 silver badges1717 bronze badges

answered 8 hours ago

WuestenfuxWuestenfux

10.5k22 gold badges66 silver badges1717 bronze badges

answered 8 hours ago

WuestenfuxWuestenfux

10.5k22 gold badges66 silver badges1717 bronze badges