Maximal ideal (x,y) - and then primary ideal (x,y)^n

  • #1
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Example (2) on page 682 of Dummit and Foote reads as follows: (see attached)

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(2) For any field k, the ideal (x) in k[x,y] is primary since it is a prime ideal.

For any [itex] n \ge 1 [/itex], the ideal [itex] (x,y)^n [/itex] is primary

since it is a power of the maximal ideal (x,y)

-------------------------------------------------------------------------------------------

My first problem with this example is as follows:

How can we demonstrate the the ideal (x,y) in k[x,y] is maximal?



Then my second problem with the example is as follows:

How do we rigorously demonstrate that the ideal [itex] (x,y)^n [/itex] is primary.

D&F say that this is because it is the power of a maximal ideal - but where have they developed that theorem/result?

The closest result they have to that is the following part of Proposition 19 (top of page 682 - see attachment)

------------------------------------------------------------------------------------------------------------------

Proposition 19. Let R be a commutative ring with 1

... ...

(5) Suppose M is a maximal ideal and Q is an ideal with [itex] M^n \subseteq Q \subseteq M [/itex]

for some [itex] n \ge 1[/itex].

Then Q is a primary ideal with rad Q = M

--------------------------------------------------------------------------------------------------------------------

Now if my suspicions are correct and Proposition 19 is being used, then can someone explain (preferably demonstrate formally and rigorously)

how part (5) of 19 demonstrates that the ideal [itex] (x,y)^n [/itex] is primary on the basis of being a power of a maximal ideal.

Would appreciate some help

Peter
 

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Answers and Replies

  • #2
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I have been doing some reflecting and reading around the two issues/problems mentioned in my post above.

First problem/issue was as follows:

"My first problem with this example is as follows:

How can we demonstrate the the ideal (x,y) in k[x,y] is maximal"

In the excellent book "Ideals, Varieties and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative Algebra" by David Cox, John Little and Donal O'Shea we find the following theorem (and its proof) on pages 201-202.

Proposition 9. If k is any field, an ideal [itex] I \subseteq k[x_1, x_2, ... ... , x_n] [/itex] of the form

[itex] I = (x_1 - a_1, x_2 - a_2, ... ... x_n - a_n) [/itex] where [itex] a_1, a_2, ... ... , a_n \in k [/itex]

is maximal.


Now (x, y) is of the form mentioned in Cox et al Proposition 9 since [itex (x,y) = (x-0, y-0) [/itex] and so by Cox et al Proposition 9, (x,y) is maximal

Can someone confirm that this is correct.

Now reflecting on my second problem/issue.

Peter
 
  • #3
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For the first issue, the thing you read in Cox is correct. But you can easily see it directly as follows. The idea is to calculate ##k[X,Y]/(X,Y)##. You can do this by using the first isomorphism theorem and the function

[tex]\Phi:k[X,Y]\rightarrow k: P(X,Y)\rightarrow P(0,0)[/tex]

Thus you see that ##k[X,Y]/(X,Y)## is a field and thus ##(X,Y)## is maximal.

For the second issue, just use the theorem you found with ##Q=M^n## and ##M=(X,Y)##.
 
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  • #4
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Thanks r136a1, most helpful

Will now check out the use of the First Isomorphism Theorem

Peter
 
  • #5
mathwonk
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if you take any polynomial in x,y, and set equal to zero every term with an x or a y in it, you have left only the constant term. Thus modding out k[x,y] by the ideal (x,y) leaves you with just the constant field. thus (x,y) is maximal.
 

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