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If finite product of maximal ideals of ring $R$ is zero, then $R$ is Noetherian$iff R$ is Artinian

“$A$ is a ring with zero ideal the product of a finite number of maximal ideals . Then $A$ is Noetherian if and only if $A$ is Artinian.”Atiyah-Macdonald, Exercise 8.3: Artinian iff finite k-algebra.$R$ is Noetherian/Artinian iff $I$ and $R/I$ are“$A$ is a ring with zero ideal the product of a finite number of maximal ideals . Then $A$ is Noetherian if and only if $A$ is Artinian.”Atiyah & MacDonald on local Noetherian and Artinian rings - sanity check.Maximal ideals of commutative Artinian ringsEquivalence between artinian ring and module whose length is finiteIs there a difference between being Artinian/Noetherian as an $R$-module, versus as an $R/I$-module?Showing an Artinian ring, all of whose maximal ideals are principal, is a principal ideal ring.Quotient of Noetherian (Artinian) ring is also Noetherian (Artinian)Consequence of epimorphism from Noetherian $R$-module

0

0

$begingroup$

I'm studying commutative algebra and now I am struggling to understand the following proof:

Proposition. Let $R$ be a commutative ring with $1_R$, $tin Bbb{Z}^+$ and $P_1,dots,P_t in mathrm{mSpec}(R)$ maximal ideals of $R$ such that $P_1dotsb P_t={0_R}$. Then $R$ is Noetherian $iff$ $R$ is Artinian.

My textbook contains the proof, but it's extremely bad-written.

So, could anybody write down clearly and step by step the proof of this proposition?

Notice that I have already studied short exact sequences, I know that if we have a $K-$vector space $V$ space ($K$ is a field), then $dim_K V<infty iff K- $module $V$ is Noetherian $iff K- $module $V$ is Artinian and at last if $IM=0$ then $R/I-$modules $M$ are equivalent to $R-$modules $M$.

PS: This Proposition is also in Atiyah and McDonald's "Introduction to Commutative Algebra", Corollary 6.11, p. 78 and in MSE there.

Thank you.

abstract-algebra commutative-algebra noetherian artinian

share|cite|improve this question

edited 2 days ago

Eric Wofsey

189k14216347

asked Mar 9 at 3:14

ChrisChris

862411

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, in your statement of the proposition, what are the $M_i$?
  $endgroup$
  – Santana Afton
  Mar 9 at 3:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your comment. Where are you reffered to?
  $endgroup$
  – Chris
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No worries! There was a typo I was confused about, and Eric Wofsey corrected it.
  $endgroup$
  – Santana Afton
  4 hours ago
  

  
  
  
  

add a comment | 

0

0

$begingroup$

I'm studying commutative algebra and now I am struggling to understand the following proof:

Proposition. Let $R$ be a commutative ring with $1_R$, $tin Bbb{Z}^+$ and $P_1,dots,P_t in mathrm{mSpec}(R)$ maximal ideals of $R$ such that $P_1dotsb P_t={0_R}$. Then $R$ is Noetherian $iff$ $R$ is Artinian.

My textbook contains the proof, but it's extremely bad-written.

So, could anybody write down clearly and step by step the proof of this proposition?

Notice that I have already studied short exact sequences, I know that if we have a $K-$vector space $V$ space ($K$ is a field), then $dim_K V<infty iff K- $module $V$ is Noetherian $iff K- $module $V$ is Artinian and at last if $IM=0$ then $R/I-$modules $M$ are equivalent to $R-$modules $M$.

PS: This Proposition is also in Atiyah and McDonald's "Introduction to Commutative Algebra", Corollary 6.11, p. 78 and in MSE there.

Thank you.

abstract-algebra commutative-algebra noetherian artinian

share|cite|improve this question

edited 2 days ago

Eric Wofsey

189k14216347

asked Mar 9 at 3:14

ChrisChris

862411

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, in your statement of the proposition, what are the $M_i$?
  $endgroup$
  – Santana Afton
  Mar 9 at 3:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your comment. Where are you reffered to?
  $endgroup$
  – Chris
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No worries! There was a typo I was confused about, and Eric Wofsey corrected it.
  $endgroup$
  – Santana Afton
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I'm studying commutative algebra and now I am struggling to understand the following proof:

Proposition. Let $R$ be a commutative ring with $1_R$, $tin Bbb{Z}^+$ and $P_1,dots,P_t in mathrm{mSpec}(R)$ maximal ideals of $R$ such that $P_1dotsb P_t={0_R}$. Then $R$ is Noetherian $iff$ $R$ is Artinian.

My textbook contains the proof, but it's extremely bad-written.

So, could anybody write down clearly and step by step the proof of this proposition?

Notice that I have already studied short exact sequences, I know that if we have a $K-$vector space $V$ space ($K$ is a field), then $dim_K V<infty iff K- $module $V$ is Noetherian $iff K- $module $V$ is Artinian and at last if $IM=0$ then $R/I-$modules $M$ are equivalent to $R-$modules $M$.

PS: This Proposition is also in Atiyah and McDonald's "Introduction to Commutative Algebra", Corollary 6.11, p. 78 and in MSE there.

Thank you.

abstract-algebra commutative-algebra noetherian artinian

share|cite|improve this question

edited 2 days ago

Eric Wofsey

189k14216347

asked Mar 9 at 3:14

ChrisChris

862411

$endgroup$

share|cite|improve this question

edited 2 days ago

Eric Wofsey

189k14216347

asked Mar 9 at 3:14

ChrisChris

862411

share|cite|improve this question

edited 2 days ago

Eric Wofsey

189k14216347

asked Mar 9 at 3:14

ChrisChris

862411

edited 2 days ago

Eric Wofsey

189k14216347

edited 2 days ago

Eric Wofsey

189k14216347

asked Mar 9 at 3:14

ChrisChris

862411

asked Mar 9 at 3:14

ChrisChris

862411

1 Answer
1

active

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votes

1

$begingroup$

Suppose $R$ is Noetherian. For $r=0,dots,t$, let $I_r=P_1dots P_r$ (for $r=0$, this means $I_r=R$). We will prove by reverse induction on $r$ that $I_r$ is an Artinian $R$-module, so that in the last case $r=0$ we conclude $I_0=R$ is Artinian. In the base case $r=t$, $I_t=0$ by hypothesis is trivially Artinian.

Now suppose $r<t$ and $I_{r+1}$ is Artinian; we will prove $I_r$ is Artinian. Note that $I_rsupseteq I_rP_{r+1}=I_{r+1}$ so we have a short exact sequence $$0to I_{r+1}to I_rto I_r/I_{r+1}to 0.$$ Since $I_{r+1}$ is Artinian, it suffices to show that $I_r/I_{r+1}$ is also Artinian. Now since $I_{r+1}=I_rP_{r+1}$, the module $I_r/I_{r+1}$ is annihilated by $P_{r+1}$, and so $I_r/I_{r+1}$ can be considered as a a module over the field $K=R/P_{r+1}$, with $R$-submodules of $I_r/I_{r+1}$ being the same thing as $K$-submodules of $I_r/I_{r+1}$. Since $R$ is Noetherian, $I_r$ and $I_{r+1}$ are Noetherian and hence $I_r/I_{r+1}$ is Noetherian as an $R$-module and thus also as a $K$-module. But since $K$ is a field, this implies $I_r/I_{r+1}$ is Artinian as a $K$-module, and hence also as an $R$-module. By the short exact sequence above we conclude $I_r$ is Artinian, completing the induction step.

That proves that if $R$ is Noetherian, then $R$ is Artinian. The proof of the converse is identical except you just swap the words "Noetherian" and "Artinian" everywhere.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Eric WofseyEric Wofsey

189k14216347

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your well written answer. I understood it. Also check professor's Morandi old page sierra.nmsu.edu/morandi/oldwebpages/Math582Spring2013/… :)
  $endgroup$
  – Chris
  5 hours ago
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

Suppose $R$ is Noetherian. For $r=0,dots,t$, let $I_r=P_1dots P_r$ (for $r=0$, this means $I_r=R$). We will prove by reverse induction on $r$ that $I_r$ is an Artinian $R$-module, so that in the last case $r=0$ we conclude $I_0=R$ is Artinian. In the base case $r=t$, $I_t=0$ by hypothesis is trivially Artinian.

Now suppose $r<t$ and $I_{r+1}$ is Artinian; we will prove $I_r$ is Artinian. Note that $I_rsupseteq I_rP_{r+1}=I_{r+1}$ so we have a short exact sequence $$0to I_{r+1}to I_rto I_r/I_{r+1}to 0.$$ Since $I_{r+1}$ is Artinian, it suffices to show that $I_r/I_{r+1}$ is also Artinian. Now since $I_{r+1}=I_rP_{r+1}$, the module $I_r/I_{r+1}$ is annihilated by $P_{r+1}$, and so $I_r/I_{r+1}$ can be considered as a a module over the field $K=R/P_{r+1}$, with $R$-submodules of $I_r/I_{r+1}$ being the same thing as $K$-submodules of $I_r/I_{r+1}$. Since $R$ is Noetherian, $I_r$ and $I_{r+1}$ are Noetherian and hence $I_r/I_{r+1}$ is Noetherian as an $R$-module and thus also as a $K$-module. But since $K$ is a field, this implies $I_r/I_{r+1}$ is Artinian as a $K$-module, and hence also as an $R$-module. By the short exact sequence above we conclude $I_r$ is Artinian, completing the induction step.

That proves that if $R$ is Noetherian, then $R$ is Artinian. The proof of the converse is identical except you just swap the words "Noetherian" and "Artinian" everywhere.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Eric WofseyEric Wofsey

189k14216347

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your well written answer. I understood it. Also check professor's Morandi old page sierra.nmsu.edu/morandi/oldwebpages/Math582Spring2013/… :)
  $endgroup$
  – Chris
  5 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose $R$ is Noetherian. For $r=0,dots,t$, let $I_r=P_1dots P_r$ (for $r=0$, this means $I_r=R$). We will prove by reverse induction on $r$ that $I_r$ is an Artinian $R$-module, so that in the last case $r=0$ we conclude $I_0=R$ is Artinian. In the base case $r=t$, $I_t=0$ by hypothesis is trivially Artinian.

Now suppose $r<t$ and $I_{r+1}$ is Artinian; we will prove $I_r$ is Artinian. Note that $I_rsupseteq I_rP_{r+1}=I_{r+1}$ so we have a short exact sequence $$0to I_{r+1}to I_rto I_r/I_{r+1}to 0.$$ Since $I_{r+1}$ is Artinian, it suffices to show that $I_r/I_{r+1}$ is also Artinian. Now since $I_{r+1}=I_rP_{r+1}$, the module $I_r/I_{r+1}$ is annihilated by $P_{r+1}$, and so $I_r/I_{r+1}$ can be considered as a a module over the field $K=R/P_{r+1}$, with $R$-submodules of $I_r/I_{r+1}$ being the same thing as $K$-submodules of $I_r/I_{r+1}$. Since $R$ is Noetherian, $I_r$ and $I_{r+1}$ are Noetherian and hence $I_r/I_{r+1}$ is Noetherian as an $R$-module and thus also as a $K$-module. But since $K$ is a field, this implies $I_r/I_{r+1}$ is Artinian as a $K$-module, and hence also as an $R$-module. By the short exact sequence above we conclude $I_r$ is Artinian, completing the induction step.

That proves that if $R$ is Noetherian, then $R$ is Artinian. The proof of the converse is identical except you just swap the words "Noetherian" and "Artinian" everywhere.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Eric WofseyEric Wofsey

189k14216347

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your well written answer. I understood it. Also check professor's Morandi old page sierra.nmsu.edu/morandi/oldwebpages/Math582Spring2013/… :)
  $endgroup$
  – Chris
  5 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $R$ is Noetherian. For $r=0,dots,t$, let $I_r=P_1dots P_r$ (for $r=0$, this means $I_r=R$). We will prove by reverse induction on $r$ that $I_r$ is an Artinian $R$-module, so that in the last case $r=0$ we conclude $I_0=R$ is Artinian. In the base case $r=t$, $I_t=0$ by hypothesis is trivially Artinian.

Now suppose $r<t$ and $I_{r+1}$ is Artinian; we will prove $I_r$ is Artinian. Note that $I_rsupseteq I_rP_{r+1}=I_{r+1}$ so we have a short exact sequence $$0to I_{r+1}to I_rto I_r/I_{r+1}to 0.$$ Since $I_{r+1}$ is Artinian, it suffices to show that $I_r/I_{r+1}$ is also Artinian. Now since $I_{r+1}=I_rP_{r+1}$, the module $I_r/I_{r+1}$ is annihilated by $P_{r+1}$, and so $I_r/I_{r+1}$ can be considered as a a module over the field $K=R/P_{r+1}$, with $R$-submodules of $I_r/I_{r+1}$ being the same thing as $K$-submodules of $I_r/I_{r+1}$. Since $R$ is Noetherian, $I_r$ and $I_{r+1}$ are Noetherian and hence $I_r/I_{r+1}$ is Noetherian as an $R$-module and thus also as a $K$-module. But since $K$ is a field, this implies $I_r/I_{r+1}$ is Artinian as a $K$-module, and hence also as an $R$-module. By the short exact sequence above we conclude $I_r$ is Artinian, completing the induction step.

That proves that if $R$ is Noetherian, then $R$ is Artinian. The proof of the converse is identical except you just swap the words "Noetherian" and "Artinian" everywhere.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Eric WofseyEric Wofsey

189k14216347

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Eric WofseyEric Wofsey

189k14216347

answered 2 days ago

Eric WofseyEric Wofsey

189k14216347

answered 2 days ago

Eric WofseyEric Wofsey

189k14216347