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

[Math Amateur]

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 n \ge 1, the ideal (x,y)^n is primary

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

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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 (x,y)^n 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)

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Proposition 19. Let R be a commutative ring with 1

... ...

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

for some n \ge 1.

Then Q is a primary ideal with rad Q = M

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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 (x,y)^n is primary on the basis of being a power of a maximal ideal.

Would appreciate some help

Peter

 

[Math Amateur]

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 I \subseteq k[x_1, x_2, ... ... , x_n] of the form

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

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

 

[R136a1]

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

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

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 $M=(X,Y)$.

 

[Math Amateur]

Thanks r136a1, most helpful

Will now check out the use of the First Isomorphism Theorem

Peter

 

[mathwonk]

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.