Proof F[x] is integral domain

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Given $F$ is integral domain, prove $F[X]$ is integral domain

Need to prove:(I did not use the condition: $F$ is integral domain)

Proof: $f(x)g(x) = 0 Leftrightarrow f(x) = 0 text or g(x) = 0 $

Here is my proof, can anyone check whether it make scene or not
$$
beginalign*
deg(f(x)g(x)) &= deg(f(x)) + deg(g(x)) \
f(x)g(x) &= 0 \
implies deg(f(x)g(x)) &= deg(f(x)) + deg(g(x)) = deg(0) \
because deg(f(x)) + deg(g(x)) &= - infty \
implies deg(f(x)) &= -infty text or deg(g(x)) = -infty \
implies f(x) &= 0 text or g(x) = 0
endalign*
$$

polynomials

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edited Jul 15 at 1:47

asked Jul 15 at 1:40

Aron Lee

133

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up vote
1
down vote

favorite

1

Given $F$ is integral domain, prove $F[X]$ is integral domain

Need to prove:(I did not use the condition: $F$ is integral domain)

Proof: $f(x)g(x) = 0 Leftrightarrow f(x) = 0 text or g(x) = 0 $

Here is my proof, can anyone check whether it make scene or not
$$
beginalign*
deg(f(x)g(x)) &= deg(f(x)) + deg(g(x)) \
f(x)g(x) &= 0 \
implies deg(f(x)g(x)) &= deg(f(x)) + deg(g(x)) = deg(0) \
because deg(f(x)) + deg(g(x)) &= - infty \
implies deg(f(x)) &= -infty text or deg(g(x)) = -infty \
implies f(x) &= 0 text or g(x) = 0
endalign*
$$

polynomials

share|cite|improve this question

edited Jul 15 at 1:47

asked Jul 15 at 1:40

Aron Lee

133

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

Given $F$ is integral domain, prove $F[X]$ is integral domain

Need to prove:(I did not use the condition: $F$ is integral domain)

Proof: $f(x)g(x) = 0 Leftrightarrow f(x) = 0 text or g(x) = 0 $

Here is my proof, can anyone check whether it make scene or not
$$
beginalign*
deg(f(x)g(x)) &= deg(f(x)) + deg(g(x)) \
f(x)g(x) &= 0 \
implies deg(f(x)g(x)) &= deg(f(x)) + deg(g(x)) = deg(0) \
because deg(f(x)) + deg(g(x)) &= - infty \
implies deg(f(x)) &= -infty text or deg(g(x)) = -infty \
implies f(x) &= 0 text or g(x) = 0
endalign*
$$

polynomials

share|cite|improve this question

edited Jul 15 at 1:47

asked Jul 15 at 1:40

Aron Lee

133

Given $F$ is integral domain, prove $F[X]$ is integral domain

Need to prove:(I did not use the condition: $F$ is integral domain)

Proof: $f(x)g(x) = 0 Leftrightarrow f(x) = 0 text or g(x) = 0 $

Here is my proof, can anyone check whether it make scene or not
$$
beginalign*
deg(f(x)g(x)) &= deg(f(x)) + deg(g(x)) \
f(x)g(x) &= 0 \
implies deg(f(x)g(x)) &= deg(f(x)) + deg(g(x)) = deg(0) \
because deg(f(x)) + deg(g(x)) &= - infty \
implies deg(f(x)) &= -infty text or deg(g(x)) = -infty \
implies f(x) &= 0 text or g(x) = 0
endalign*
$$

polynomials

share|cite|improve this question

edited Jul 15 at 1:47

asked Jul 15 at 1:40

Aron Lee

133

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 1:47

edited Jul 15 at 1:47

edited Jul 15 at 1:47

asked Jul 15 at 1:40

Aron Lee

133

asked Jul 15 at 1:40

Aron Lee

133

asked Jul 15 at 1:40

Aron Lee

133

133

add a comment | 

add a comment | 

1 Answer
1

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up vote
2
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accepted

You need that $F$ is integral domain when you use $deg (fg)=deg(f) + deg(g)$. Take for example $g(x)=f(x)=2x$ in $F[X]$, where $F= mathbb Z/ 4mathbb Z$, which is not integral domain. We have $fg =4x^2 = 0$ and therefore $-infty =deg(fg)neq deg(f)+deg(g)=2$.

Let's prove this formula holds for integral domains. Let's suppose $fneq 0$ and $gneq 0$. If $f = a_nx^n + cdots+a_0$ and $g(x)= b_mx^m + cdots + b_0$, where $a_nneq 0$ and $b_mneq 0$, then we have
$$fg = a_nb_m x^n+m+cdots + a_0b_0.$$

Since $F$ is integral domain $a_n b_m neq 0$ and, therefore,
$$deg(fg) = n+m = deg(f)+deg(g).$$

If $f=0$ or $g=0$ then the formula is also true because both sides of the identity equal $-infty$.

share|cite|improve this answer

edited Jul 15 at 1:56

answered Jul 15 at 1:51

Hugocito

1,6451019

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The rest of your proof is correct.
  – Hugocito
  Jul 15 at 1:58
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

You need that $F$ is integral domain when you use $deg (fg)=deg(f) + deg(g)$. Take for example $g(x)=f(x)=2x$ in $F[X]$, where $F= mathbb Z/ 4mathbb Z$, which is not integral domain. We have $fg =4x^2 = 0$ and therefore $-infty =deg(fg)neq deg(f)+deg(g)=2$.

Let's prove this formula holds for integral domains. Let's suppose $fneq 0$ and $gneq 0$. If $f = a_nx^n + cdots+a_0$ and $g(x)= b_mx^m + cdots + b_0$, where $a_nneq 0$ and $b_mneq 0$, then we have
$$fg = a_nb_m x^n+m+cdots + a_0b_0.$$

Since $F$ is integral domain $a_n b_m neq 0$ and, therefore,
$$deg(fg) = n+m = deg(f)+deg(g).$$

If $f=0$ or $g=0$ then the formula is also true because both sides of the identity equal $-infty$.

share|cite|improve this answer

edited Jul 15 at 1:56

answered Jul 15 at 1:51

Hugocito

1,6451019

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The rest of your proof is correct.
  – Hugocito
  Jul 15 at 1:58
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

You need that $F$ is integral domain when you use $deg (fg)=deg(f) + deg(g)$. Take for example $g(x)=f(x)=2x$ in $F[X]$, where $F= mathbb Z/ 4mathbb Z$, which is not integral domain. We have $fg =4x^2 = 0$ and therefore $-infty =deg(fg)neq deg(f)+deg(g)=2$.

Let's prove this formula holds for integral domains. Let's suppose $fneq 0$ and $gneq 0$. If $f = a_nx^n + cdots+a_0$ and $g(x)= b_mx^m + cdots + b_0$, where $a_nneq 0$ and $b_mneq 0$, then we have
$$fg = a_nb_m x^n+m+cdots + a_0b_0.$$

Since $F$ is integral domain $a_n b_m neq 0$ and, therefore,
$$deg(fg) = n+m = deg(f)+deg(g).$$

If $f=0$ or $g=0$ then the formula is also true because both sides of the identity equal $-infty$.

share|cite|improve this answer

edited Jul 15 at 1:56

answered Jul 15 at 1:51

Hugocito

1,6451019

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The rest of your proof is correct.
  – Hugocito
  Jul 15 at 1:58
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

You need that $F$ is integral domain when you use $deg (fg)=deg(f) + deg(g)$. Take for example $g(x)=f(x)=2x$ in $F[X]$, where $F= mathbb Z/ 4mathbb Z$, which is not integral domain. We have $fg =4x^2 = 0$ and therefore $-infty =deg(fg)neq deg(f)+deg(g)=2$.

Let's prove this formula holds for integral domains. Let's suppose $fneq 0$ and $gneq 0$. If $f = a_nx^n + cdots+a_0$ and $g(x)= b_mx^m + cdots + b_0$, where $a_nneq 0$ and $b_mneq 0$, then we have
$$fg = a_nb_m x^n+m+cdots + a_0b_0.$$

Since $F$ is integral domain $a_n b_m neq 0$ and, therefore,
$$deg(fg) = n+m = deg(f)+deg(g).$$

If $f=0$ or $g=0$ then the formula is also true because both sides of the identity equal $-infty$.

share|cite|improve this answer

edited Jul 15 at 1:56

answered Jul 15 at 1:51

Hugocito

1,6451019

You need that $F$ is integral domain when you use $deg (fg)=deg(f) + deg(g)$. Take for example $g(x)=f(x)=2x$ in $F[X]$, where $F= mathbb Z/ 4mathbb Z$, which is not integral domain. We have $fg =4x^2 = 0$ and therefore $-infty =deg(fg)neq deg(f)+deg(g)=2$.

Let's prove this formula holds for integral domains. Let's suppose $fneq 0$ and $gneq 0$. If $f = a_nx^n + cdots+a_0$ and $g(x)= b_mx^m + cdots + b_0$, where $a_nneq 0$ and $b_mneq 0$, then we have
$$fg = a_nb_m x^n+m+cdots + a_0b_0.$$

Since $F$ is integral domain $a_n b_m neq 0$ and, therefore,
$$deg(fg) = n+m = deg(f)+deg(g).$$

If $f=0$ or $g=0$ then the formula is also true because both sides of the identity equal $-infty$.

share|cite|improve this answer

edited Jul 15 at 1:56

answered Jul 15 at 1:51

Hugocito

1,6451019

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 1:56

edited Jul 15 at 1:56

edited Jul 15 at 1:56

answered Jul 15 at 1:51

Hugocito

1,6451019

answered Jul 15 at 1:51

Hugocito

1,6451019

answered Jul 15 at 1:51

Hugocito

1,6451019

1,6451019

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The rest of your proof is correct.
  – Hugocito
  Jul 15 at 1:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The rest of your proof is correct.
  – Hugocito
  Jul 15 at 1:58
  

  
  
  
  

The rest of your proof is correct.
– Hugocito
Jul 15 at 1:58

The rest of your proof is correct.
– Hugocito
Jul 15 at 1:58

add a comment | 

 

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