Multiplication in $mathbbZ$ corresponds to multiplication in rings?

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So I was trying to prove that the characteristic of an integral domain is either $0$ or prime. I got stuck, so I searched for a proof and I came across the following proof online

Now I almost want to accept this proof, except for the following (possibly silly and pedantic) issue. In the above proof we know that $n_0 in mathbbN subseteq mathbbZ$, so since $n$ is not prime, we factorize $n = m cdot k$ (where $cdot$ represents multiplication on the integers in the ring $(mathbbZ, +, cdot)$). Now in the above proof the following is asserted $$n(1_D) = underbrace1_D + dots + 1_D_n text times =0_D implies mcdot k(1_D) = underbraceleft(1_D + dots + 1_Dright)_m text times bullet underbraceleft(1_D + dots + 1_Dright)_k text times$$

Now seemingly it seems that multiplication of $m$ and $k$ in $mathbbZ$ is inducing (ring) multiplication of elements in $D$ (when I thought we'd only end up with addition of the $1_D$'s $mk$ times).

Is there a reason why this happens?

abstract-algebra ring-theory

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asked Jul 22 at 9:44

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3,51411039

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  For any commutative ring $R$, the map $;beginaligned[t]mathbf Z&to R\n&mapsto ncdot 1_Rendaligned$ is a ring homomorphism.
  – Bernard
  Jul 22 at 9:49
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

So I was trying to prove that the characteristic of an integral domain is either $0$ or prime. I got stuck, so I searched for a proof and I came across the following proof online

Now I almost want to accept this proof, except for the following (possibly silly and pedantic) issue. In the above proof we know that $n_0 in mathbbN subseteq mathbbZ$, so since $n$ is not prime, we factorize $n = m cdot k$ (where $cdot$ represents multiplication on the integers in the ring $(mathbbZ, +, cdot)$). Now in the above proof the following is asserted $$n(1_D) = underbrace1_D + dots + 1_D_n text times =0_D implies mcdot k(1_D) = underbraceleft(1_D + dots + 1_Dright)_m text times bullet underbraceleft(1_D + dots + 1_Dright)_k text times$$

Now seemingly it seems that multiplication of $m$ and $k$ in $mathbbZ$ is inducing (ring) multiplication of elements in $D$ (when I thought we'd only end up with addition of the $1_D$'s $mk$ times).

Is there a reason why this happens?

abstract-algebra ring-theory

share|cite|improve this question

asked Jul 22 at 9:44

Perturbative

3,51411039

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  For any commutative ring $R$, the map $;beginaligned[t]mathbf Z&to R\n&mapsto ncdot 1_Rendaligned$ is a ring homomorphism.
  – Bernard
  Jul 22 at 9:49
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

So I was trying to prove that the characteristic of an integral domain is either $0$ or prime. I got stuck, so I searched for a proof and I came across the following proof online

Now I almost want to accept this proof, except for the following (possibly silly and pedantic) issue. In the above proof we know that $n_0 in mathbbN subseteq mathbbZ$, so since $n$ is not prime, we factorize $n = m cdot k$ (where $cdot$ represents multiplication on the integers in the ring $(mathbbZ, +, cdot)$). Now in the above proof the following is asserted $$n(1_D) = underbrace1_D + dots + 1_D_n text times =0_D implies mcdot k(1_D) = underbraceleft(1_D + dots + 1_Dright)_m text times bullet underbraceleft(1_D + dots + 1_Dright)_k text times$$

Now seemingly it seems that multiplication of $m$ and $k$ in $mathbbZ$ is inducing (ring) multiplication of elements in $D$ (when I thought we'd only end up with addition of the $1_D$'s $mk$ times).

Is there a reason why this happens?

abstract-algebra ring-theory

share|cite|improve this question

asked Jul 22 at 9:44

Perturbative

3,51411039

So I was trying to prove that the characteristic of an integral domain is either $0$ or prime. I got stuck, so I searched for a proof and I came across the following proof online

Now I almost want to accept this proof, except for the following (possibly silly and pedantic) issue. In the above proof we know that $n_0 in mathbbN subseteq mathbbZ$, so since $n$ is not prime, we factorize $n = m cdot k$ (where $cdot$ represents multiplication on the integers in the ring $(mathbbZ, +, cdot)$). Now in the above proof the following is asserted $$n(1_D) = underbrace1_D + dots + 1_D_n text times =0_D implies mcdot k(1_D) = underbraceleft(1_D + dots + 1_Dright)_m text times bullet underbraceleft(1_D + dots + 1_Dright)_k text times$$

Now seemingly it seems that multiplication of $m$ and $k$ in $mathbbZ$ is inducing (ring) multiplication of elements in $D$ (when I thought we'd only end up with addition of the $1_D$'s $mk$ times).

Is there a reason why this happens?

abstract-algebra ring-theory

share|cite|improve this question

asked Jul 22 at 9:44

Perturbative

3,51411039

share|cite|improve this question

share|cite|improve this question

asked Jul 22 at 9:44

Perturbative

3,51411039

asked Jul 22 at 9:44

Perturbative

3,51411039

asked Jul 22 at 9:44

Perturbative

3,51411039

3,51411039

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  For any commutative ring $R$, the map $;beginaligned[t]mathbf Z&to R\n&mapsto ncdot 1_Rendaligned$ is a ring homomorphism.
  – Bernard
  Jul 22 at 9:49
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  For any commutative ring $R$, the map $;beginaligned[t]mathbf Z&to R\n&mapsto ncdot 1_Rendaligned$ is a ring homomorphism.
  – Bernard
  Jul 22 at 9:49
  
  

  
  
  
  

4

4

For any commutative ring $R$, the map $;beginaligned[t]mathbf Z&to R\n&mapsto ncdot 1_Rendaligned$ is a ring homomorphism.
– Bernard
Jul 22 at 9:49

For any commutative ring $R$, the map $;beginaligned[t]mathbf Z&to R\n&mapsto ncdot 1_Rendaligned$ is a ring homomorphism.
– Bernard
Jul 22 at 9:49

add a comment | 

2 Answers
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1
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This happens simply because by induction and definition of $kcdot 1_D$, one may prove $(ncdot 1_D)times (mcdot 1_D) = nmcdot 1_D$.

The proof by induction on, say, $m$ is easy : $(ncdot 1_D)((m+1)cdot 1_D) = (ncdot 1_D)(mcdot 1_D + 1_D) =...$

share|cite|improve this answer

answered Jul 22 at 14:34

Max

9,8651635

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The key property being the distributive law.
  – Dustan Levenstein
  Jul 22 at 14:41
  

  
  
  
  

add a comment | 

up vote
0
down vote

As @Bernard pointed out in the comment above, for any commutative ring $R$ the map $varphi : mathbbZ to R$ defined by $$varphi(n) = ncdot 1_R = underbrace1_R + dots + 1_R_n text times$$
is a ring homomorphism. Thus in my problem above where we have as our ring the integral domain $(D, +, bullet)$ since we can factorize $n = mcdot k$ in the ring $(mathbbZ, +, cdot)$ then we have
beginalign*
mn cdot 1_R &= varphi(m cdot n) \
&= varphi(m) bullet varphi(n) \
&= (mcdot 1_R) bullet (ncdot 1_R) \
&= underbraceleft(1_D + dots + 1_Dright)_m text times bullet underbraceleft(1_D + dots + 1_Dright)_k text times
endalign*

share|cite|improve this answer

answered Jul 22 at 16:35

Perturbative

3,51411039

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

This happens simply because by induction and definition of $kcdot 1_D$, one may prove $(ncdot 1_D)times (mcdot 1_D) = nmcdot 1_D$.

The proof by induction on, say, $m$ is easy : $(ncdot 1_D)((m+1)cdot 1_D) = (ncdot 1_D)(mcdot 1_D + 1_D) =...$

share|cite|improve this answer

answered Jul 22 at 14:34

Max

9,8651635

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The key property being the distributive law.
  – Dustan Levenstein
  Jul 22 at 14:41
  

  
  
  
  

add a comment | 

up vote
1
down vote

This happens simply because by induction and definition of $kcdot 1_D$, one may prove $(ncdot 1_D)times (mcdot 1_D) = nmcdot 1_D$.

The proof by induction on, say, $m$ is easy : $(ncdot 1_D)((m+1)cdot 1_D) = (ncdot 1_D)(mcdot 1_D + 1_D) =...$

share|cite|improve this answer

answered Jul 22 at 14:34

Max

9,8651635

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The key property being the distributive law.
  – Dustan Levenstein
  Jul 22 at 14:41
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

This happens simply because by induction and definition of $kcdot 1_D$, one may prove $(ncdot 1_D)times (mcdot 1_D) = nmcdot 1_D$.

The proof by induction on, say, $m$ is easy : $(ncdot 1_D)((m+1)cdot 1_D) = (ncdot 1_D)(mcdot 1_D + 1_D) =...$

share|cite|improve this answer

answered Jul 22 at 14:34

Max

9,8651635

This happens simply because by induction and definition of $kcdot 1_D$, one may prove $(ncdot 1_D)times (mcdot 1_D) = nmcdot 1_D$.

The proof by induction on, say, $m$ is easy : $(ncdot 1_D)((m+1)cdot 1_D) = (ncdot 1_D)(mcdot 1_D + 1_D) =...$

share|cite|improve this answer

answered Jul 22 at 14:34

Max

9,8651635

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 14:34

Max

9,8651635

answered Jul 22 at 14:34

Max

9,8651635

answered Jul 22 at 14:34

Max

9,8651635

9,8651635

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The key property being the distributive law.
  – Dustan Levenstein
  Jul 22 at 14:41
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The key property being the distributive law.
  – Dustan Levenstein
  Jul 22 at 14:41
  

  
  
  
  

2

2

The key property being the distributive law.
– Dustan Levenstein
Jul 22 at 14:41

The key property being the distributive law.
– Dustan Levenstein
Jul 22 at 14:41

add a comment | 

up vote
0
down vote

As @Bernard pointed out in the comment above, for any commutative ring $R$ the map $varphi : mathbbZ to R$ defined by $$varphi(n) = ncdot 1_R = underbrace1_R + dots + 1_R_n text times$$
is a ring homomorphism. Thus in my problem above where we have as our ring the integral domain $(D, +, bullet)$ since we can factorize $n = mcdot k$ in the ring $(mathbbZ, +, cdot)$ then we have
beginalign*
mn cdot 1_R &= varphi(m cdot n) \
&= varphi(m) bullet varphi(n) \
&= (mcdot 1_R) bullet (ncdot 1_R) \
&= underbraceleft(1_D + dots + 1_Dright)_m text times bullet underbraceleft(1_D + dots + 1_Dright)_k text times
endalign*

share|cite|improve this answer

answered Jul 22 at 16:35

Perturbative

3,51411039

add a comment | 

up vote
0
down vote

As @Bernard pointed out in the comment above, for any commutative ring $R$ the map $varphi : mathbbZ to R$ defined by $$varphi(n) = ncdot 1_R = underbrace1_R + dots + 1_R_n text times$$
is a ring homomorphism. Thus in my problem above where we have as our ring the integral domain $(D, +, bullet)$ since we can factorize $n = mcdot k$ in the ring $(mathbbZ, +, cdot)$ then we have
beginalign*
mn cdot 1_R &= varphi(m cdot n) \
&= varphi(m) bullet varphi(n) \
&= (mcdot 1_R) bullet (ncdot 1_R) \
&= underbraceleft(1_D + dots + 1_Dright)_m text times bullet underbraceleft(1_D + dots + 1_Dright)_k text times
endalign*

share|cite|improve this answer

answered Jul 22 at 16:35

Perturbative

3,51411039

add a comment | 

up vote
0
down vote

up vote
0
down vote

As @Bernard pointed out in the comment above, for any commutative ring $R$ the map $varphi : mathbbZ to R$ defined by $$varphi(n) = ncdot 1_R = underbrace1_R + dots + 1_R_n text times$$
is a ring homomorphism. Thus in my problem above where we have as our ring the integral domain $(D, +, bullet)$ since we can factorize $n = mcdot k$ in the ring $(mathbbZ, +, cdot)$ then we have
beginalign*
mn cdot 1_R &= varphi(m cdot n) \
&= varphi(m) bullet varphi(n) \
&= (mcdot 1_R) bullet (ncdot 1_R) \
&= underbraceleft(1_D + dots + 1_Dright)_m text times bullet underbraceleft(1_D + dots + 1_Dright)_k text times
endalign*

share|cite|improve this answer

answered Jul 22 at 16:35

Perturbative

3,51411039

As @Bernard pointed out in the comment above, for any commutative ring $R$ the map $varphi : mathbbZ to R$ defined by $$varphi(n) = ncdot 1_R = underbrace1_R + dots + 1_R_n text times$$
is a ring homomorphism. Thus in my problem above where we have as our ring the integral domain $(D, +, bullet)$ since we can factorize $n = mcdot k$ in the ring $(mathbbZ, +, cdot)$ then we have
beginalign*
mn cdot 1_R &= varphi(m cdot n) \
&= varphi(m) bullet varphi(n) \
&= (mcdot 1_R) bullet (ncdot 1_R) \
&= underbraceleft(1_D + dots + 1_Dright)_m text times bullet underbraceleft(1_D + dots + 1_Dright)_k text times
endalign*

share|cite|improve this answer

answered Jul 22 at 16:35

Perturbative

3,51411039

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 16:35

Perturbative

3,51411039

answered Jul 22 at 16:35

Perturbative

3,51411039

answered Jul 22 at 16:35

Perturbative

3,51411039

3,51411039

add a comment | 

add a comment | 

 

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