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Show that any non-zero prime ideal of $R$ is invertible.

Ideals in a Dedekind domain localized at a prime idealA proof in Janusz Algebraic Number FieldAnother Error in Neukirch's Algebraic Number Theory?Proving that a ring has only finitely many prime idealsNoetherian ring and prime ideal contained in an invertible maximal ideal.Define, $p^-1 = x in K: xp subset D$. Then show that there exists a non zero $c in D$ such that $cp^-1 subset D$.A non-zero and non-invertible element in a noetherian integral domain has a decomposition into irreducible elementsIn a Noetherian integral domain, a principal prime ideal can't have proper non-zero prime idealsUnique minimal prime over an idealIdeals of ring finitely generated if and only if ring Noetherian

1

1

$begingroup$

Theorem $:$

Let $R$ be an integral domain such that $R$ is Noetherian, integrally closed and every non-zero prime ideal of $R$ is maximal with quotient field $K.$ Then every non-zero prime ideal of $R$ is invertible.

Proof $:$

Consider a non-zero prime ideal $P$ of $R.$ Let us consider the $P' = x in K : xP subseteq R .$ Then I know that $P' supsetneq R.$ So $exists y in P' setminus R.$ Let $x$ be any non-zero element of $P.$

Then $P= RP subset P'P subset R.$ So either $P'P=P'$ or $P'P=R.$ If $P'P=P$ then $(P')^n P = P, text for all n geq 1.$ Then $xy^n in P$ for all $n geq 1.$ But then $x R[y] subseteq R.$ Then $x R[y]$ is an ideal of $R.$ Since $R$ is Noetherian so $x R[y]$ is finitely generated. Let $x R[y]$ is generated by $a_1,a_2, cdots ,a_n.$ Then $R[y]$ is generated by $x^-1a_1,x^-1a_2, cdots , x^-1 a_n.$ But then $y$ is integral over $R,$ a contradiction to the fact that $R$ is integrally closed. So $P'P neq P.$ Therefore $P'P = R.$ So $P$ is invertible, as required.

I can't understand the sentence in bold letters in the above proof. Would anybody please help me understanding this? Any help will be highly appreciated.

Thank you very much.

algebraic-number-theory noetherian

share|cite|improve this question

edited Mar 15 at 16:26

Dbchatto67

asked Mar 15 at 16:18

Dbchatto67Dbchatto67

2,166320

$endgroup$

add a comment | 

1

1

$begingroup$

Theorem $:$

Let $R$ be an integral domain such that $R$ is Noetherian, integrally closed and every non-zero prime ideal of $R$ is maximal with quotient field $K.$ Then every non-zero prime ideal of $R$ is invertible.

Proof $:$

Consider a non-zero prime ideal $P$ of $R.$ Let us consider the $P' = x in K : xP subseteq R .$ Then I know that $P' supsetneq R.$ So $exists y in P' setminus R.$ Let $x$ be any non-zero element of $P.$

Then $P= RP subset P'P subset R.$ So either $P'P=P'$ or $P'P=R.$ If $P'P=P$ then $(P')^n P = P, text for all n geq 1.$ Then $xy^n in P$ for all $n geq 1.$ But then $x R[y] subseteq R.$ Then $x R[y]$ is an ideal of $R.$ Since $R$ is Noetherian so $x R[y]$ is finitely generated. Let $x R[y]$ is generated by $a_1,a_2, cdots ,a_n.$ Then $R[y]$ is generated by $x^-1a_1,x^-1a_2, cdots , x^-1 a_n.$ But then $y$ is integral over $R,$ a contradiction to the fact that $R$ is integrally closed. So $P'P neq P.$ Therefore $P'P = R.$ So $P$ is invertible, as required.

I can't understand the sentence in bold letters in the above proof. Would anybody please help me understanding this? Any help will be highly appreciated.

Thank you very much.

algebraic-number-theory noetherian

share|cite|improve this question

edited Mar 15 at 16:26

Dbchatto67

asked Mar 15 at 16:18

Dbchatto67Dbchatto67

2,166320

$endgroup$

add a comment | 

$begingroup$

Theorem $:$

Let $R$ be an integral domain such that $R$ is Noetherian, integrally closed and every non-zero prime ideal of $R$ is maximal with quotient field $K.$ Then every non-zero prime ideal of $R$ is invertible.

Proof $:$

Consider a non-zero prime ideal $P$ of $R.$ Let us consider the $P' = x in K : xP subseteq R .$ Then I know that $P' supsetneq R.$ So $exists y in P' setminus R.$ Let $x$ be any non-zero element of $P.$

Then $P= RP subset P'P subset R.$ So either $P'P=P'$ or $P'P=R.$ If $P'P=P$ then $(P')^n P = P, text for all n geq 1.$ Then $xy^n in P$ for all $n geq 1.$ But then $x R[y] subseteq R.$ Then $x R[y]$ is an ideal of $R.$ Since $R$ is Noetherian so $x R[y]$ is finitely generated. Let $x R[y]$ is generated by $a_1,a_2, cdots ,a_n.$ Then $R[y]$ is generated by $x^-1a_1,x^-1a_2, cdots , x^-1 a_n.$ But then $y$ is integral over $R,$ a contradiction to the fact that $R$ is integrally closed. So $P'P neq P.$ Therefore $P'P = R.$ So $P$ is invertible, as required.

I can't understand the sentence in bold letters in the above proof. Would anybody please help me understanding this? Any help will be highly appreciated.

Thank you very much.

algebraic-number-theory noetherian

share|cite|improve this question

edited Mar 15 at 16:26

Dbchatto67

asked Mar 15 at 16:18

Dbchatto67Dbchatto67

2,166320

$endgroup$

share|cite|improve this question

edited Mar 15 at 16:26

Dbchatto67

asked Mar 15 at 16:18

Dbchatto67Dbchatto67

2,166320

share|cite|improve this question

edited Mar 15 at 16:26

Dbchatto67

asked Mar 15 at 16:18

Dbchatto67Dbchatto67

2,166320

asked Mar 15 at 16:18

Dbchatto67Dbchatto67

2,166320

asked Mar 15 at 16:18

Dbchatto67Dbchatto67

2,166320

1 Answer
1

active

oldest

votes

1

$begingroup$

Since $R[y]$ is finitely generated over $R$ as a module, it is integral over $R$, so $y$ is integral over $R$.

share|cite|improve this answer

answered Mar 15 at 16:26

asdqasdq

1,8681518

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do I prove that?
  $endgroup$
  – Dbchatto67
  Mar 15 at 16:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that if $y$ is integral over $R$ then $R[y]$ is finitely generated. But how do I prove the reverse part?
  $endgroup$
  – Dbchatto67
  Mar 15 at 16:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dbchatto67 It's a classic result in commutative algebra. You should have seen it in your book, or during class. The proof is quite tricky. Click here for a proof. (Theorem 2.1.4)
  $endgroup$
  – Crostul
  Mar 15 at 16:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you provide me some link at least where I can able to find this proof? Thank you very much.
  $endgroup$
  – Dbchatto67
  Mar 15 at 16:38
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

Since $R[y]$ is finitely generated over $R$ as a module, it is integral over $R$, so $y$ is integral over $R$.

share|cite|improve this answer

answered Mar 15 at 16:26

asdqasdq

1,8681518

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do I prove that?
  $endgroup$
  – Dbchatto67
  Mar 15 at 16:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that if $y$ is integral over $R$ then $R[y]$ is finitely generated. But how do I prove the reverse part?
  $endgroup$
  – Dbchatto67
  Mar 15 at 16:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dbchatto67 It's a classic result in commutative algebra. You should have seen it in your book, or during class. The proof is quite tricky. Click here for a proof. (Theorem 2.1.4)
  $endgroup$
  – Crostul
  Mar 15 at 16:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you provide me some link at least where I can able to find this proof? Thank you very much.
  $endgroup$
  – Dbchatto67
  Mar 15 at 16:38
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Since $R[y]$ is finitely generated over $R$ as a module, it is integral over $R$, so $y$ is integral over $R$.

share|cite|improve this answer

answered Mar 15 at 16:26

asdqasdq

1,8681518

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do I prove that?
  $endgroup$
  – Dbchatto67
  Mar 15 at 16:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that if $y$ is integral over $R$ then $R[y]$ is finitely generated. But how do I prove the reverse part?
  $endgroup$
  – Dbchatto67
  Mar 15 at 16:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dbchatto67 It's a classic result in commutative algebra. You should have seen it in your book, or during class. The proof is quite tricky. Click here for a proof. (Theorem 2.1.4)
  $endgroup$
  – Crostul
  Mar 15 at 16:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you provide me some link at least where I can able to find this proof? Thank you very much.
  $endgroup$
  – Dbchatto67
  Mar 15 at 16:38
  
  

  
  
  
  

add a comment | 

$begingroup$

Since $R[y]$ is finitely generated over $R$ as a module, it is integral over $R$, so $y$ is integral over $R$.

share|cite|improve this answer

answered Mar 15 at 16:26

asdqasdq

1,8681518

$endgroup$

share|cite|improve this answer

answered Mar 15 at 16:26

asdqasdq

1,8681518

answered Mar 15 at 16:26

asdqasdq

1,8681518

answered Mar 15 at 16:26

asdqasdq

1,8681518