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

• Math Amateur
Now look at k[x,y]/(x,y)^n. This is the same as (k[x,y]/(x,y))/(x,y)^{n-1}. But we saw that (x,y) is maximal, hence (k[x,y]/(x,y)) is a field. But then (x,y)^{n-1} is maximal in k[x,y]/(x,y), so by the same logic as before, (x,y)^n is maximal in k[x,y].Hence (x,y)^n is prime. But then by the correspondence theorem, (x,y)^n is primary in k[x,y]. Hence the ideal (x,y)^n is primary in k
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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 $n \ge 1$, 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

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• Page 682 - D&F - 12th Nov 2013.pdf
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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 $I \subseteq k[x_1, x_2, ... ... , x_n]$ 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

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)##.

1 person

Will now check out the use of the First Isomorphism Theorem

Peter

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.

## 1. What is a maximal ideal?

A maximal ideal is an ideal in a ring that is not contained in any larger ideal, except for the entire ring itself.

## 2. How do you determine if (x,y) is a maximal ideal?

In a polynomial ring, a maximal ideal is generated by a single irreducible polynomial. So, to determine if (x,y) is a maximal ideal, you would need to check if (x,y) is generated by an irreducible polynomial.

## 3. What is the significance of (x,y) being a maximal ideal?

Maximal ideals play an important role in algebraic geometry and commutative algebra. They are used to define the algebraic variety associated with a given ideal and can also be used to define prime ideals.

## 4. What is a primary ideal?

A primary ideal is a generalization of a prime ideal. It is an ideal in a ring where any zero-divisors are contained in the ideal. In other words, if a product of two elements in the ring is in the ideal, then at least one of the elements is in the ideal.

## 5. How does the primary ideal (x,y)^n differ from the maximal ideal (x,y)?

The primary ideal (x,y)^n is the ideal generated by all monomials of the form x^ay^b where a+b ≥ n. This means that the primary ideal contains more elements than the maximal ideal, and is therefore a larger ideal. Additionally, the primary ideal has the property that any zero-divisors in the ring are contained in the ideal, whereas the maximal ideal does not necessarily have this property.

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