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Is there a mistake in the proof of Theorem 4 in Kaplansky's Commutative Rings?

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0

$begingroup$

Here is a theorem (Theorem 4) in Kaplansky's Commutative Rings.

Let $R$ be an integral domain. Let $S$ be the set of all elements in $R$ expressible as a product of principal primes. Then $S$ is a saturated multiplicatively closed set.

The definition of a saturated multiplicatively closed set is as follows:

A multiplicatively closed set $S$ is saturated if $xin S$ implies that $S$ contains all the divisors of $x$.

When the author was proving the theorem, he said that

To prove that $S$ is saturated we assume $abin S$ and have to prove that $a$ and $b$ lie in $S$.

My Question:
I am wondering if $a=p$ is a prime and $b=1$, then $ab=pcdot 1in S$ implies that $a=pin S$ and $b=1in S$. But $1$ should not be in $S$. Maybe I misunderstand something or how to fix the problem.

Thanks a lot.

abstract-algebra commutative-algebra

share|cite|improve this question

edited Mar 22 at 2:17

Saad

20.3k92352

asked Mar 21 at 4:40

bfhahabfhaha

1,5751125

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Why should 1 not be in $S$? In fact, some authors require that 1 be in every multiplicative set, but even if this book does not, certainly no one disallows it.
  $endgroup$
  – Ted
  Mar 21 at 4:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ted I think $1$ is not a product of primes.
  $endgroup$
  – bfhaha
  Mar 21 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  It's an empty product, of zero primes.
  $endgroup$
  – Qiaochu Yuan
  Mar 21 at 5:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here is a question related to empty product
  $endgroup$
  – J. W. Tanner
  Mar 21 at 11:40
  

  
  
  
  

add a comment | 

0

$begingroup$

Here is a theorem (Theorem 4) in Kaplansky's Commutative Rings.

Let $R$ be an integral domain. Let $S$ be the set of all elements in $R$ expressible as a product of principal primes. Then $S$ is a saturated multiplicatively closed set.

The definition of a saturated multiplicatively closed set is as follows:

A multiplicatively closed set $S$ is saturated if $xin S$ implies that $S$ contains all the divisors of $x$.

When the author was proving the theorem, he said that

To prove that $S$ is saturated we assume $abin S$ and have to prove that $a$ and $b$ lie in $S$.

My Question:
I am wondering if $a=p$ is a prime and $b=1$, then $ab=pcdot 1in S$ implies that $a=pin S$ and $b=1in S$. But $1$ should not be in $S$. Maybe I misunderstand something or how to fix the problem.

Thanks a lot.

abstract-algebra commutative-algebra

share|cite|improve this question

edited Mar 22 at 2:17

Saad

20.3k92352

asked Mar 21 at 4:40

bfhahabfhaha

1,5751125

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Why should 1 not be in $S$? In fact, some authors require that 1 be in every multiplicative set, but even if this book does not, certainly no one disallows it.
  $endgroup$
  – Ted
  Mar 21 at 4:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ted I think $1$ is not a product of primes.
  $endgroup$
  – bfhaha
  Mar 21 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  It's an empty product, of zero primes.
  $endgroup$
  – Qiaochu Yuan
  Mar 21 at 5:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here is a question related to empty product
  $endgroup$
  – J. W. Tanner
  Mar 21 at 11:40
  

  
  
  
  

add a comment | 

$begingroup$

Here is a theorem (Theorem 4) in Kaplansky's Commutative Rings.

Let $R$ be an integral domain. Let $S$ be the set of all elements in $R$ expressible as a product of principal primes. Then $S$ is a saturated multiplicatively closed set.

The definition of a saturated multiplicatively closed set is as follows:

A multiplicatively closed set $S$ is saturated if $xin S$ implies that $S$ contains all the divisors of $x$.

When the author was proving the theorem, he said that

To prove that $S$ is saturated we assume $abin S$ and have to prove that $a$ and $b$ lie in $S$.

My Question:
I am wondering if $a=p$ is a prime and $b=1$, then $ab=pcdot 1in S$ implies that $a=pin S$ and $b=1in S$. But $1$ should not be in $S$. Maybe I misunderstand something or how to fix the problem.

Thanks a lot.

abstract-algebra commutative-algebra

share|cite|improve this question

edited Mar 22 at 2:17

Saad

20.3k92352

asked Mar 21 at 4:40

bfhahabfhaha

1,5751125

$endgroup$

share|cite|improve this question

edited Mar 22 at 2:17

Saad

20.3k92352

asked Mar 21 at 4:40

bfhahabfhaha

1,5751125

share|cite|improve this question

edited Mar 22 at 2:17

Saad

20.3k92352

asked Mar 21 at 4:40

bfhahabfhaha

1,5751125

edited Mar 22 at 2:17

Saad

20.3k92352

edited Mar 22 at 2:17

Saad

20.3k92352

asked Mar 21 at 4:40

bfhahabfhaha

1,5751125

asked Mar 21 at 4:40

bfhahabfhaha

1,5751125