Irreducible Components in a Primary Decomposition

  • #1
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Let R be a Noetherian Ring and I an ideal in R.

Let I = Q_1 n ... n Q_r = J_1 n ... n J_r be two primary decompositions of I.

How can I show the number of irreducible components in each decomposition is the same?
 

Answers and Replies

  • #2
Firstly you shoulnd't have r on both sides of that equality. Secondly, you will need to know the definition of 'primary'. So what is it? What does it imply? Does this seem familiar to any other result you know? Is the result even true? (i.e. think about what a counter example might be and see where it fails to satisfy some hypothesis.)
 
  • #3
Firstly you shoulnd't have r on both sides of that equality. Secondly, you will need to know the definition of 'primary'. So what is it? What does it imply? Does this seem familiar to any other result you know? Is the result even true? (i.e. think about what a counter example might be and see where it fails to satisfy some hypothesis.)

Matt, thank you very much for your reply. Yes the result is true. I'm assuming all uniqueness theorems found in say, Dummit and Foote. I figured it would be easier to prove with some theory in place, hence why I have r on both sides. Whether I need this or not I don't know, but I figured it might make things easier.

Any help would be greatly appreciated.
 
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  • #4
The point was that by writing r on both sides you were assuming the result you wanted to show. And you haven't written out the definition of primary/prime and what it implies in a Noetherian ring. The result will be deducible from the definitions, so what are the definitions? I gave you plenty of help and you ignored it all.
 
  • #5
i think you need also to assume the decompositions are irredundant, or it is not true.
 

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