Maximal Ideals of K[X] in Commutative Algebra: A Helpful Guide

  • MHB
  • Thread starter Math Amateur
  • Start date
In summary: In Euclidean domains, what is the smallest ring that has a given field as its principal ideal domain?In Euclidean domains, what is the smallest ring that has a given field as its maximal ideal domain?In Euclidean domains, what is the maximal ideal domain of a given field?
  • #1
Math Amateur
Gold Member
MHB
3,998
48
I am reading R.Y.Sharp's book: "Steps in Commutative Algebra.

In Chapter 3: Prime Ideals and Maximal Ideals, Exercise 3.6 reads as follows:

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

Determine all the maximal ideals of the ring K[X],

where K is a field and X is an indeterminate.

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

Can someone please help me get started on this problem.

I suspect we may be able to use the following theorem/lemma:

"Let I be an ideal of the commutative ring R.

Then I is maximal if and only if R/I is a field."

However I am not sure of exactly how to go about utilising this result.

I would very much appreciate some help.

Peter

[This has also been posted on MHF.]
 
Physics news on Phys.org
  • #2
You have asked this question before, although you may not realize it.

First of all, $K[X]$ is a principal ideal domain, because it is Euclidean. Memorize this, if you have to. So ANY ideal is of the form $(f(X))$ for some polynomial $f(X) \in K[X]$.

Now, suppose we have an ideal $(p(X))$, where:

$p(X) = h(X)k(X)$ and neither $h$ nor $k$ is a unit (that is, an element of $K$, viewed as a constant polynomial).

Prove that $K[X]/(p(X))$ has zero divisors. What does this tell you?
 
  • #3
Deveno said:
You have asked this question before, although you may not realize it.

First of all, $K[X]$ is a principal ideal domain, because it is Euclidean. Memorize this, if you have to. So ANY ideal is of the form $(f(X))$ for some polynomial $f(X) \in K[X]$.

Now, suppose we have an ideal $(p(X))$, where:

$p(X) = h(X)k(X)$ and neither $h$ nor $k$ is a unit (that is, an element of $K$, viewed as a constant polynomial).

Prove that $K[X]/(p(X))$ has zero divisors. What does this tell you?

Hi Deveno,

Thanks for the help ... and the reminder :-(

You write:

"Now, suppose we have an ideal [FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]p[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]X[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT], where:

[FONT=MathJax_Math]p[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]X[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]h[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]X[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]X[/FONT][FONT=MathJax_Main])[/FONT] and neither [FONT=MathJax_Math]h[/FONT] nor [FONT=MathJax_Math]k[/FONT] is a unit (that is, an element of [FONT=MathJax_Math]K[/FONT], viewed as a constant polynomial).

Prove that [FONT=MathJax_Math]K[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]X[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]p[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]X[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT] has zero divisors. What does this tell you?"Ok so we hae that there exists an ideal (p(X)) with p(X) = h(X)k(X) where neither h(X) nor k(X) is a unit.

So in K[X]/(p(X)) we have

[h(X) + (p(X))][k(X) + (p(X))] = h(X)k(X) + (p(X)) = p(X) + (p(X)) = 0 + (p(X))

Thus h(X) + (p(X)) and k(X) + (p(X)) are zero divisor of K[X]/(p(X)) so long as we can show that neither of h(X) + (p(X)) nor k(X) + (p(X)) equals zero.

I think that h(X) and k(X) being non-units assures this ... but how to show it?

Certainly if h(X) was a unit of K[X] then it would be a constant polynomial c where \(\displaystyle c \in K \).

Then c. k(X) = p(x) and \(\displaystyle k(X) = c^{-1} \) and in the factor ring K[X]/(p(X)) that is equivalent to zero.

Thus neither of h(X), k(X) can be units ... but does this assure that h(X) + (p(X)) and k(X) + (p(X)) are not zero ... ie is the condition that they are both non-units enough to ensure they are not zero?

Please indicate whether the above analysis is OK ...

Will now reflect on the next step ...

Peter
 
  • #4
Consider the degrees of h and k. Can they even be equal to that of p?
 
  • #5
Deveno said:
Consider the degrees of h and k. Can they even be equal to that of p?

Hmm ... cannot see where you are taking me on this ...

But since p(X) = h(X)K(X) we have

deg (h(X)) + deg (k(X)) = deg(p(X))

and further, since h(X), K(X) are non-units neither of them can be degree zero,so each of them have degree strictly less than p(X) ... but what relevant deductions can we make from this analysis ... ?

I am starting to think that therefore neither of them is divisible by p(X) ...can I make use of this ... well, maybe the answer is that neither h(X), k(X) can be a multiple of p(X) i.e. can not be zero in the factor (or residue class) ring.

Peter
 
Last edited:
  • #6
In the polynomial case, the remainder and quotient polynomials are UNIQUE, so there is a unique polynomial:

$r(X)$ with deg(r) < deg(p) such that:

$(r(X) + (p(X)) = h(X) + (p(X))$.

Since deg(h) < deg(p), it must be that r = h, and similarly for k.

It is clear we cannot have:

$h(X) + (p(X)) = 0 + (p(X))$ because if we did:

$p(X)|h(X)$ so:

$h(X) = p(X)q(X)$ leading to:

$p(X) = h(X)k(X) = p(X)q(X)k(X)$

That is: $q(X)k(X) = 1$ (since Euclidean domains are integral domains, and thus cancellative), contradicting the suppostion that $k(X)$ is a non-unit.

The upshot of this can be re-stated like so:

If $p(X)$ is reducible, $R[X]/(p(X))$ is not even an integral domain, so definitely NOT a field.

The only other case is that $p(X)$ is irreducible. In a Euclidean domain (which are UFDs) irreducible elements are prime.

What kind of ideals do prime elements generate in a Euclidean domain (which is both a PID and a UFD)?

Let me ask you some similar questions:

We have, for the Euclidean domain $\Bbb Z$, that $\Bbb Z/(n)$ is a ring, called the integers modulo $n$.

1) When is $(n)$ maximal? Is (30) maximal? Why not?

2) When is $\Bbb Z/(n)$ a field?

3) What are the units of $\Bbb Z$, and when is $n \in \Bbb Z$ irreducible?
 
  • #7
Deveno said:
In the polynomial case, the remainder and quotient polynomials are UNIQUE, so there is a unique polynomial:

$r(X)$ with deg(r) < deg(p) such that:

$(r(X) + (p(X)) = h(X) + (p(X))$.

Since deg(h) < deg(p), it must be that r = h, and similarly for k.

It is clear we cannot have:

$h(X) + (p(X)) = 0 + (p(X))$ because if we did:

$p(X)|h(X)$ so:

$h(X) = p(X)q(X)$ leading to:

$p(X) = h(X)k(X) = p(X)q(X)k(X)$

That is: $q(X)k(X) = 1$ (since Euclidean domains are integral domains, and thus cancellative), contradicting the suppostion that $k(X)$ is a non-unit.

The upshot of this can be re-stated like so:

If $p(X)$ is reducible, $R[X]/(p(X))$ is not even an integral domain, so definitely NOT a field.

The only other case is that $p(X)$ is irreducible. In a Euclidean domain (which are UFDs) irreducible elements are prime.

What kind of ideals do prime elements generate in a Euclidean domain (which is both a PID and a UFD)?

Let me ask you some similar questions:

We have, for the Euclidean domain $\Bbb Z$, that $\Bbb Z/(n)$ is a ring, called the integers modulo $n$.

1) When is $(n)$ maximal? Is (30) maximal? Why not?

2) When is $\Bbb Z/(n)$ a field?

3) What are the units of $\Bbb Z$, and when is $n \in \Bbb Z$ irreducible?
Thanks for the help ... and the questions ...

Your first question is as follows:

"What kind of ideals do prime elements generate in a Euclidean domain (which is both a PID and a UFD)?"

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

First we need to define prime elements:

On page 284, Dummit and Foote define irreducible elements, prime elements and associate elements for an integral domain. [Just why D&F define these terms for an integral domain and not the more general structure of a ring, I am not sure ... ??]

D&F's definition f a prime element is as follows:

----------------------------------------------------------------------------------
Definition. Let R be an integral domain.
The nonzero element \(\displaystyle p \in R \) is called prime in R if the ideal (p) generated by p is a prime ideal. In other words, a non-zero element p is a prime if it is not a unit and whenever p | ab for any \(\displaystyle a, b \in R \) then either p | a or
p | b. [why does this definition not deal with 0 and hence cover the case of (0) - the zero prime ideal?]
-----------------------------------------------------------------------------------

So, as I understand it, by definition, prime elements generate prime ideals.

But a Euclidean domain is also a PID and by Proposition 7 (D&F, page 280) every nonzero prime ideal in a PID is a maximal ideal.

So in a Euclidean Domain, prime elements generate prime ideals which in turn are maximal ideals.

Can you indicate whether the above is correct or not?

I will now attend to your 3 other questions. (Thanks again for these questions!)

Peter
 
  • #8
Deveno said:
In the polynomial case, the remainder and quotient polynomials are UNIQUE, so there is a unique polynomial:

$r(X)$ with deg(r) < deg(p) such that:

$(r(X) + (p(X)) = h(X) + (p(X))$.

Since deg(h) < deg(p), it must be that r = h, and similarly for k.

It is clear we cannot have:

$h(X) + (p(X)) = 0 + (p(X))$ because if we did:

$p(X)|h(X)$ so:

$h(X) = p(X)q(X)$ leading to:

$p(X) = h(X)k(X) = p(X)q(X)k(X)$

That is: $q(X)k(X) = 1$ (since Euclidean domains are integral domains, and thus cancellative), contradicting the suppostion that $k(X)$ is a non-unit.

The upshot of this can be re-stated like so:

If $p(X)$ is reducible, $R[X]/(p(X))$ is not even an integral domain, so definitely NOT a field.

The only other case is that $p(X)$ is irreducible. In a Euclidean domain (which are UFDs) irreducible elements are prime.

What kind of ideals do prime elements generate in a Euclidean domain (which is both a PID and a UFD)?

Let me ask you some similar questions:

We have, for the Euclidean domain $\Bbb Z$, that $\Bbb Z/(n)$ is a ring, called the integers modulo $n$.

1) When is $(n)$ maximal? Is (30) maximal? Why not?

2) When is $\Bbb Z/(n)$ a field?

3) What are the units of $\Bbb Z$, and when is $n \in \Bbb Z$ irreducible?
Thanks again for the guidance and the questions ,,,

You write:

"We have, for the Euclidean domain $\Bbb Z$, that $\Bbb Z/(n)$ is a ring, called the integers modulo $n$.

1) When is $(n)$ maximal? Is (30) maximal? Why not?

2) When is $\Bbb Z/(n)$ a field?

3) What are the units of $\Bbb Z$, and when is $n \in \Bbb Z$ irreducible?"

Thus we are considering the Euclidean Domain \(\displaystyle \mathbb{Z} \)

and the ring \(\displaystyle \mathbb{Z}/(n) = \mathbb{Z}/n\mathbb{Z} \)

Now consider Question 1

"When is \(\displaystyle n \mathbb{Z} = (n) \) maximal. Is (30) maximal? Why
not?

----------------------------------------------------------------------------------
The ideal \(\displaystyle n \mathbb{Z} = (n) \) of \(\displaystyle n \mathbb{Z} \) is a maximal ideal if an only if \(\displaystyle \mathbb{Z}/n \mathbb{Z} = \mathbb{Z}/(n)\) is a field.

(30) is not maximal because it is contained in a number of other ideals of \(\displaystyle \mathbb{Z} \), such as (15), (6), (5) and (2). Of course this type of inclusion cannot happen when n is a prime integer.
-----------------------------------------------------------------------------------
Now consider Question 2

When is
\(\displaystyle \mathbb{Z}/n \mathbb{Z} = \mathbb{Z}/(n)\) a field.
------------------------------------------
-----------------------------------------
\(\displaystyle \mathbb{Z}/n \mathbb{Z} = \mathbb{Z}/(n)\) is a field if and only if n is a prime integer.
Now consider Question 3

What are the units of \(\displaystyle \mathbb{Z} \), and when is
\(\displaystyle n \in \mathbb{Z} \) irreducible.
---------------------------------------------------------------------------------
The units of \(\displaystyle \mathbb{Z} \) are +1 and -1.

\(\displaystyle \mathbb{Z} \) is a PID, and in a PID a nonzero element is irreducible if and only if the element is prime.

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

Can you please indicate if the above is correct?

Peter
 
Last edited:
  • #9
Good...now, how many of these questions generalize to ANY Euclidean domain?

And...which polynomials are prime in $K[X]$?
 

Related to Maximal Ideals of K[X] in Commutative Algebra: A Helpful Guide

What are maximal ideals of K[X]?

Maximal ideals of K[X] are the largest possible ideals of the polynomial ring K[X], where K is a field. They are defined as proper ideals that are not contained in any other proper ideal of K[X].

How do you determine if an ideal is maximal?

An ideal in K[X] is maximal if and only if the quotient ring K[X]/I is a field, where I is the ideal in question. This means that every non-zero element in the quotient ring has a multiplicative inverse.

Can a polynomial ring have more than one maximal ideal?

Yes, a polynomial ring can have multiple maximal ideals. For example, the polynomial ring R[x] has infinitely many maximal ideals where R is a ring. However, a polynomial ring over a field K, denoted by K[x], has only two maximal ideals: the zero ideal and the ideal generated by a single irreducible polynomial.

What is the relationship between maximal ideals and irreducible polynomials?

In a polynomial ring K[x], a polynomial is irreducible if and only if it generates a maximal ideal. This means that every maximal ideal in K[x] is generated by a single irreducible polynomial.

How are maximal ideals related to the algebraic structure of a polynomial ring?

Maximal ideals play a crucial role in the study of the algebraic structure of a polynomial ring. They are used to define quotient rings, which are important tools in understanding the structure of polynomial rings. Additionally, maximal ideals are closely related to the concept of prime ideals, which are used to define the notion of a prime element in a polynomial ring.

Similar threads

  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
8
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
2K
  • Linear and Abstract Algebra
Replies
16
Views
3K
  • Linear and Abstract Algebra
Replies
3
Views
4K
  • Linear and Abstract Algebra
Replies
3
Views
1K
Replies
3
Views
2K
  • Linear and Abstract Algebra
Replies
5
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
2K
  • Linear and Abstract Algebra
Replies
6
Views
2K
Back
Top