Maximal Ideals K[X]: Solving Exercise 3.6 w/ R.Y. Sharpe

[Math Amateur]

I am reading R.Y. Sharpe: Steps in Commutative Algebra Chapter 3: Prime Ideals and Maximal Ideals

Exercise 3.6 on page 39 reads as follows:

Determine all the maximal ideals of K[X], where K is a field and X is an indeterminate.

Can someone please help me get started on this problem.

Peter

[This has also been posted on MHF]

 

[Deveno]

Hint number 1: suppose f(x),g(x) are in a given ideal.

Is it true that gcd(f,g) is also in this ideal?

Hint number 2: suppose h(x)|k(x). Can you describe the relationship between (h(x)) and (k(x))?

 

[Math Amateur]

Hint number 1: suppose f(x),g(x) are in a given ideal.

Is it true that gcd(f,g) is also in this ideal?

Hint number 2: suppose h(x)|k(x). Can you describe the relationship between (h(x)) and (k(x))?

Deveno,

Thanks for the hints ... I think I can answer your questions, but as for the more creative element of putting your hints together to construct a solution ... ...

The answer to your first question - namely "Is it true that gcd(f,g) is also in this ideal?" is yes ... indeed, if $$ f(x) $$ and $$ g(x) \in K[x] $$ then the ideal generated by the two ring elements is the ideal generated by their gcd is equal to the ideal generated by both ring elements ...

i.e. $$ (d(x)) = (f(x), g(x)) $$

(to be honest this is a guess ... it is true for integers, and I am guessing it is true for polynomials .. )

The answer to your second question - namely: "suppose h(x)|k(x). Can you describe the relationship between (h(x)) and (k(x))?"

In this case we have that if $$ h(x) | k(x) $$ then $$ k(x) \in (h(x) $$

BUT ... how to use these relationships to determine the maximal ideals of K[x]?

Peter

 

[Deveno]

Yes, it is true for the ring $K[X]$, where $K$ is a field. Can you PROVE it?

It should be obvious that if d = gcd(f,g) we have:

$(f(x),g(x)) \subset (d(x))$ (why? see hint #2...)

so it only remains to show that $d(x) \in (f(x),g(x))$. If only we knew that:

$d(x) = a(x)f(x) + b(x)g(x)$ for some polynomials a and b...

Does this imply every ideal in $K[X]$ is principal...?

Suppose that $m(x)$ is irreducible...what ideals could possibly contain it...?

 

[Math Amateur]

Yes, it is true for the ring $K[X]$, where $K$ is a field. Can you PROVE it?

It should be obvious that if d = gcd(f,g) we have:

$(f(x),g(x)) \subset (d(x))$ (why? see hint #2...)

so it only remains to show that $d(x) \in (f(x),g(x))$. If only we knew that:

$d(x) = a(x)f(x) + b(x)g(x)$ for some polynomials a and b...

Does this imply every ideal in $K[X]$ is principal...?

Suppose that $m(x)$ is irreducible...what ideals could possibly contain it...?

To show $$ (d(x)) = (f(x), g(x)) $$ ... ... ... (1)

where $$ d(x) = $$ gcd $$ (f(x), g(x)) $$

Firstly, by definition of a gcd, we have

$$ d(x)|f(x) $$ and $$ d(x)|g(x) $$

so $$ f(x) = a(x)d(x) $$ and $$ g(x) = b(x)d(x) $$

Now to show that

$$ (f(x), g(x)) \subseteq (d(x)) $$ ... ... ... (2)

Let $$ h(x) \in (f(x), g(x)) $$

Then $$ h(x) = s(x)f(x) + t(x)g(x) $$ (definition of $$ (f(x), g(x)) $$

$$ \Longrightarrow h(x) = s(x)a(x)d(x) + t(x)b(x)d(x) $$

$$ \Longrightarrow h(x) = (s(x)a(x) + t(x)b(x))d(x) $$

$$ \Longrightarrow h(x) = c(x)d(x) $$

$$ \Longrightarrow h(x) \in d(x) $$

$$ \Longrightarrow (f(x), g(x)) \subseteq (d(x)) $$

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

Now to show $$ (d(x)) \subseteq (f(x), g(x)) $$ ... ... ... (3)

Note first that since K[x] is a Euclidean Domain and since d(x) is the gcd(f(x),g(x))

we have by Dummit and Foote, Theorem 4 page 275 that

$$ d(x) = a(x)f(x) + b(x)g(x) $$

Now to show (3) let $$ h(x) \in (d(x)) $$

Then $$ h(x) = s(x)d(x) $$ for some $$ s(x) \in K[X] $$

$$ \Longrightarrow h(x) = s(x)[a(x)f(x) + b(x)g(x)] $$

$$ \Longrightarrow h(x) = s(x)a(x)f(x) + s(x)b(x)g(x) $$

$$ \Longrightarrow h(x) = l(x)f(x) + m(x)g(x) $$

$$ \Longrightarrow h(x) \in (f(x),g(x)) $$

$$ \Longrightarrow (d(x) \subseteq (f(x),g(x)) $$

Thus we have shown that (2), (3) are the case and therefore $$ (d(x)) = (f(x), g(x)) $$

Can you please confirm that the above proof is correct ...

Now, I need to reflect further on the original problem ...

... but I am still not sure how to proceed ... ...

Peter

 

[Deveno]

Your proof is a good one...do you understand it?

It really is quite similar to the same proof for the ring of integers, for a very good reason: both rings are Euclidean domains, so we can leverage the division algorithm to our advantage.

What I am driving at in all of this is this:

A maximal ideal in $K[X]$ is generated by an irreducible (prime) polynomial (prime = irreducible in all GCD domains, and Euclidean domains are certainly all GCD domains).

In general, maximal ideals are prime ideals, in any ring. In a PID, which ideals are prime ideals? The ones generated by prime elements. For $K[X]$, this means the ones generated by irreducible polynomials.

Do you understand this? All of the definitions of the items above:

Ideals
Maximal ideals
Prime ideals
Prime elements
Principal ideals

exist because they are USEFUL ones. Ring theory, unlike its baby brother group theory, grew organically, out of an attempt to understand "number systems".

You are probably quite comfortable with the idea that prime integers are important because of the utility of factoring. It turns out we can recover the same infrastructure by considering the ideals of the ring of integers, and that considering ideals in other rings is often more elegant than "drilling down to the element level".

For example, polynomial factors are only unique "up to units". By replacing the factorization of actual polynomials with a factorization of ideals, we can say things a bit more "cleanly". We trade the divisibility partial order for a partial order by inclusion of lattices (which is an "upside-down" version of the divisibility lattice, the ideal $(1) = K[X]$ is now at the top, instead of the bottom).

In other words, if we really want to understand a ring, especially a "nice" ring like a Euclidean domain, we look at its lattice of ideals. The ideals are the subsets capturing "how our nice ring behaves", and perhaps more importantly, determine which other rings we can possibly have homomorphisms onto.

A short look ahead, since you are reading a book on commutative algebra:

Rings have a defect: unlike the case in groups, where kernels of a homomorphism are themselves groups, ring homomorphism kernels are not always rings (they have a bad habit of not preserving the ring identity, either).

The good news is: there is a structure that fixes this-and this structure is called a module. Many things that seem confusing become more clear when stated in the context of modules. For example, if a ring $R$ is considered as an $R$-module over itself, than an ideal $J$ of $R$ is just an $R$-submodule.

 

[Math Amateur]

Your proof is a good one...do you understand it?

It really is quite similar to the same proof for the ring of integers, for a very good reason: both rings are Euclidean domains, so we can leverage the division algorithm to our advantage.

What I am driving at in all of this is this:

A maximal ideal in $K[X]$ is generated by an irreducible (prime) polynomial (prime = irreducible in all GCD domains, and Euclidean domains are certainly all GCD domains).

In general, maximal ideals are prime ideals, in any ring. In a PID, which ideals are prime ideals? The ones generated by prime elements. For $K[X]$, this means the ones generated by irreducible polynomials.

Do you understand this? All of the definitions of the items above:

Ideals
Maximal ideals
Prime ideals
Prime elements
Principal ideals

exist because they are USEFUL ones. Ring theory, unlike its baby brother group theory, grew organically, out of an attempt to understand "number systems".

You are probably quite comfortable with the idea that prime integers are important because of the utility of factoring. It turns out we can recover the same infrastructure by considering the ideals of the ring of integers, and that considering ideals in other rings is often more elegant than "drilling down to the element level".

For example, polynomial factors are only unique "up to units". By replacing the factorization of actual polynomials with a factorization of ideals, we can say things a bit more "cleanly". We trade the divisibility partial order for a partial order by inclusion of lattices (which is an "upside-down" version of the divisibility lattice, the ideal $(1) = K[X]$ is now at the top, instead of the bottom).

In other words, if we really want to understand a ring, especially a "nice" ring like a Euclidean domain, we look at its lattice of ideals. The ideals are the subsets capturing "how our nice ring behaves", and perhaps more importantly, determine which other rings we can possibly have homomorphisms onto.

A short look ahead, since you are reading a book on commutative algebra:

Rings have a defect: unlike the case in groups, where kernels of a homomorphism are themselves groups, ring homomorphism kernels are not always rings (they have a bad habit of not preserving the ring identity, either).

The good news is: there is a structure that fixes this-and this structure is called a module. Many things that seem confusing become more clear when stated in the context of modules. For example, if a ring $R$ is considered as an $R$-module over itself, than an ideal $J$ of $R$ is just an $R$-submodule.

Hi Deveno,

Tanks so much for the helpful post.

Do I understand my proof ... well at one level, yes ... but I suspect not completely ... since although I see links, I do not fully understand how my proof leads directly to your statement that "A maximal ideal in $K[X]$ is generated by an irreducible (prime) polynomial (prime = irreducible in all GCD domains, and Euclidean domains are certainly all GCD domains)." - can you make the link more explicit ... sorry if I am being slow and cautious ...

I am also somewhat troubled by the fact that while prime ideals are maximal ideals in a PID (Proposition 7, D&F Section 8.2, page 280) the converse is not true in general (also D&F, page 280) ... so there may be maximal ideals that are not prime ideals in a PID ... but maybe if we consider ideals that are generated by prime ideals we get all maximal ideals in a PID and in k[X] in general ... can you clarify?

I am now reflecting on the rest of your post which contains material/ideas that are most interesting and helpful to the understanding of commutative algebra!

Thanks again for your guidance and help.

Peter

 

[Deveno]

I think you have it backwards: maximal ideals are always prime (in a commutative ring with unity).

Let $R$ be a commutative ring with unity, and suppose $M$ is a maximal ideal.

Then as a consequence of the fundamental homomorphism theorem, we have that any ideal of $R/M$ is an ideal of $R$ containing $M$.

As the only such ideals are $R$ and $M$, $R/M$ has only the ideals $R/M$ and $M$ (the 0-element of $R/M$).

Thus $R/M$ is a field, and thus a fortiori an integral domain.

It follows that if for any $a,b \in R$ we have:

$ab \in M$ so that $ab + M = M = 0 + M$, then:

$(a + M)(b + M) = ab + M = M = 0 + M$

and since $R/M$ is an integral domain, either $a + M = 0 + M$, or $b + M = 0 + M$, that is:

either $a \in M$, or $b \in M$, so $M$ is a prime ideal.

It is, however, true, that prime ideals are not necessarily maximal, even in a PID. But in PID's the only non-maximal prime ideal is the zero ideal. I think you are mis-reading the content of Dummit and Foote on page 280.

Finally, although in general prime elements and irreducible elements are not the same thing (the standard example is 3 in $\Bbb Z[\sqrt{-5}]$) in GCD domains (which includes the more restrictive class UFD domains) they are. Euclidean domains are, of course, both UFDs and GCDs, being an even further restricted class of integral domain.

So the situation we have in $K[X]$ is this:

$K[X]$ is a Euclidean domain, so:

Ideal = principal ideal
Maximal ideal = prime ideal
Prime ideal = ideal generated by prime element
Prime element = irreducible element

and thus:

Maximal ideal = ideal generated by irreducible element

As an example, in $\Bbb Q[X]$, the ideal generated by $X^2 - 2$ is maximal, and thus:

$\Bbb Q[X]/\langle X^2 - 2\rangle$ is a field, isomorphic to $\Bbb Q(\sqrt{2})$.

 

[Math Amateur]

I think you have it backwards: maximal ideals are always prime (in a commutative ring with unity).

Let $R$ be a commutative ring with unity, and suppose $M$ is a maximal ideal.

Then as a consequence of the fundamental homomorphism theorem, we have that any ideal of $R/M$ is an ideal of $R$ containing $M$.

As the only such ideals are $R$ and $M$, $R/M$ has only the ideals $R/M$ and $M$ (the 0-element of $R/M$).

Thus $R/M$ is a field, and thus a fortiori an integral domain.

It follows that if for any $a,b \in R$ we have:

$ab \in M$ so that $ab + M = M = 0 + M$, then:

$(a + M)(b + M) = ab + M = M = 0 + M$

and since $R/M$ is an integral domain, either $a + M = 0 + M$, or $b + M = 0 + M$, that is:

either $a \in M$, or $b \in M$, so $M$ is a prime ideal.

It is, however, true, that prime ideals are not necessarily maximal, even in a PID. But in PID's the only non-maximal prime ideal is the zero ideal. I think you are mis-reading the content of Dummit and Foote on page 280.

Finally, although in general prime elements and irreducible elements are not the same thing (the standard example is 3 in $\Bbb Z[\sqrt{-5}]$) in GCD domains (which includes the more restrictive class UFD domains) they are. Euclidean domains are, of course, both UFDs and GCDs, being an even further restricted class of integral domain.

So the situation we have in $K[X]$ is this:

$K[X]$ is a Euclidean domain, so:

Ideal = principal ideal
Maximal ideal = prime ideal
Prime ideal = ideal generated by prime element
Prime element = irreducible element

and thus:

Maximal ideal = ideal generated by irreducible element

As an example, in $\Bbb Q[X]$, the ideal generated by $X^2 - 2$ is maximal, and thus:

$\Bbb Q[X]/\langle X^2 - 2\rangle$ is a field, isomorphic to $\Bbb Q(\sqrt{2})$.

Thanks Deveno! that clears up a lot of issues for me.

Also, I need to re-read D&F Section 8.2 again

Thank you again for all the help on this exercise

Peter