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

[Math Amateur]

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:

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

where K is a field and X is an indeterminate.

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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.]

 

[Deveno]

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?

 

[Math Amateur]

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

 

[Deveno]

Consider the degrees of h and k. Can they even be equal to that of p?

 

[Math Amateur]

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

 

[Deveno]

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?

 

[Math Amateur]

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)?"

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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:

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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?]
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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

 

[Math Amateur]

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?
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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.
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Now consider Question 2

When is \(\displaystyle \mathbb{Z}/n \mathbb{Z} = \mathbb{Z}/(n)\) a field.
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\(\displaystyle \mathbb{Z}/n \mathbb{Z} = \mathbb{Z}/(n)\) 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.
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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.

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Can you please indicate if the above is correct?

Peter

 

[Deveno]

Good...now, how many of these questions generalize to ANY Euclidean domain?

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