The principal ideal is maximal in but it is not the unique maximal ideal

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In summary: Well, clearly $x \in \text{ker }\phi$, so $(x) \subseteq \text{ker }\phi$.Suppose $f(0) = 0$. Does $x|f(x)$?Yes, if $f(0) = 0$ then $f(x) = xg(x)$ for some $g(x) \in F[x]$.Yes, if $f(0) = 0$ then $f(x) = xg(x)$ for some $g(x) \in F[x]$. Right. So if $f(0) = 0$ then $f(x) \in (x)$.What is $\text{ker }\phi$
  • #1
mathmari
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Hey! :eek:

I want to show that if $F$ is a field then the principal ideal $(x)$ is maximal in $F[x]$ but it is not the unique maximal ideal.

We have that if $F$ is a field then $F[x]$ is a P.I.D..
So, every ideal of $F[x]$ is principal, say $(x)$.
To show that this is not the unique maximal do we have to assume that there is an other ideal, say $I$, so that $(x)\subseteq I$ and find a contradiction? (Wondering)
 
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  • #2
Another maximal ideal cannot contain $(x)$ or $(x)$ would not be maximal.

What can you say about the ideal $(x - 1)$?

Also, how do you know $(x)$ is maximal (hint: what is $F[x]/(x)$)?
 
  • #3
Deveno said:
Another maximal ideal cannot contain $(x)$ or $(x)$ would not be maximal.

What can you say about the ideal $(x - 1)$?

Do you mean which the relation between the two ideals is? (Wondering)
Deveno said:
Also, how do you know $(x)$ is maximal (hint: what is $F[x]/(x)$)?

Do we have to show that the mapping $F[x]\rightarrow F$ is an homomorphism with the kernel $(x)$ ? (Wondering)
 
  • #4
mathmari said:
Do you mean which the relation between the two ideals is? (Wondering)

Indeed.


Do we have to show that the mapping $F[x]\rightarrow F$ is an homomorphism with the kernel $(x)$ ? (Wondering)

Yes, but how does that show $(x)$ is maximal?
 
  • #5
Deveno said:
Indeed.

We have that $y\in (x) \Rightarrow y=ax$, for some $a\in F$ and that $z\in (x-1) \Rightarrow z=b(x-1)$, for some $b\in F$, right? (Wondering)
Deveno said:
Yes, but how does that show $(x)$ is maximal?

$(x)$ is maximal ideal in $F[x]$ when $F[x]/(x)$ is a field, right? So, when we have that $F[x]\rightarrow F$ is an homomorphism with kernel $(x)$ and since $F$ is a field, we conclude that $(x)$ is maximal, right? (Wondering)
 
  • #6
Two points for part 2 of your answer-no score for the first; you've just restated the definitions. How do $(x)$ and $(x - 1)$ compare?
 
  • #7
Deveno said:
How do $(x)$ and $(x - 1)$ compare?

Do we not have the following?

$y\in (x-1)$

$y=a(x-1) \Rightarrow y=ax-a\Rightarrow y\notin (x)$ $z\in (x)$

$z=bx\Rightarrow z=bx-b+b \Rightarrow z=b(x-1)+1\Rightarrow z\notin (x-1)$

(Wondering)
 
  • #8
Good, so neither one is contained in the other (they are said to be "incomparable" via inclusion).

Now...what is $F[x]/(x-1)$ isomorphic to?

(Note: I could have used $a \neq 0 \in F$ instead of 1, but some fields are finite, and "quite small", for example we might have $F = GF_2$ which only has one non-zero element).
 
  • #9
Deveno said:
Now...what is $F[x]/(x-1)$ isomorphic to?
I don't really have an idea... Could you give me a hint? (Wondering)
 
  • #10
mathmari said:
I don't really have an idea... Could you give me a hint? (Wondering)

What are the possible remainders of $f(x)$ upon division by $x - 1$?
 
  • #11
Deveno said:
What are the possible remainders of $f(x)$ upon division by $x - 1$?

How could we find that? I got stuck right now... (Wondering)
 
  • #12
For a field $F$, the ring $F[x]$ is a Euclidean domain, so we can write for any polynomial $f(x)$:

$f(x) = q(x)(x - 1) + r(x)$, where either $r(x) = 0$, or the degree of $r$ is less than the degree of $x - 1$.

So what can you say about what $r(x)$ is?
 
  • #13
Deveno said:
For a field $F$, the ring $F[x]$ is a Euclidean domain, so we can write for any polynomial $f(x)$:

$f(x) = q(x)(x - 1) + r(x)$, where either $r(x) = 0$, or the degree of $r$ is less than the degree of $x - 1$.

So what can you say about what $r(x)$ is?

$r(x)$ must be a constant, right? (Wondering)
 
  • #14
mathmari said:
$r(x)$ must be a constant, right? (Wondering)

Yes, that is-$r \in F$. Can you continue?
 
  • #15
Deveno said:
Yes, that is-$r \in F$. Can you continue?

Does this mean that $F[x]/(x-1)$ is isomorphic to $F$ ? (Wondering)
 
  • #16
Since $F[x]/(x-1)$ is isomorphic to $F$ we have that since $F$ is a field $(x-1)$ is also a maximal ideal of $F[x]$.
The two maximal ideals are incomparable.
Therefore, $(x)$ is not the unique maximal ideal.

Is this correct? (Wondering)
 
  • #17
Yep. It's not a "hard" problem, just one to test your *understanding* of the basic concepts involved.

Another useful theorem:

$F[x]/(p(x))$ is a field if and only if $p(x)$ is irreducible over $F$. Any linear polynomial over a field is irreducible (up to a unit factor).
 
  • #18
mathmari said:
$(x)$ is maximal ideal in $F[x]$ when $F[x]/(x)$ is a field, right? So, when we have that $F[x]\rightarrow F$ is an homomorphism with kernel $(x)$ and since $F$ is a field, we conclude that $(x)$ is maximal, right? (Wondering)

We consider the mapping $\phi : F[x]\rightarrow F$ with $f(x)\mapsto f(0)$.

We have that $\phi$ is an homomorphism:
Let $f(x)\in F[x]$ then $f(x)=\sum_{i=0}^nc_ix^i$.
We have that $f_1(x)+f_2(x)=\sum_{i=0}^{n_1}a_ix^i+\sum_{i=0}^{n_2}b_ix^i=:\sum_{i=0}^nc_ix^i=:f(x)$
then $\phi (f_1(x)+f_2(x))=\phi (f(x))=f(0)=f_1(0)+f_2(0)=\phi (f_1(x))+\phi (f_2(x))$
and
$f_1(x)\cdot f_2(x)=\left (\sum_{i=0}^{n_1}a_ix^i\right )\left (\sum_{i=0}^{n_2}b_ix^i\right )=:\sum_{i=0}^nc_ix^i=:f(x)$
then $\phi (f_1(x)f_2(x))=\phi (f(x))=f(0)=f_1(0)f_2(0)=\phi (f_1(x))\phi (f_2(x))$

$\phi$ is onto:
Let $a\in F$.
Then $f(x)=ax^0\in F[x]$ and $\phi (f(x))=f(0)=a$. So, we have that $\phi$ is an epimorphism.

We have from the first isomorphism theorem that $F[x]/\ker\phi \cong F$, or not? (Wondering)

If this is correct, then it is left to show that $\ker\phi =(x)$, right? (Wondering)
 
  • #19
Well, clearly $x \in \text{ker }\phi$, so $(x) \subseteq \text{ker }\phi$.

Suppose $f(0) = 0$. Does $x|f(x)$?

It may be helpful to remember that:

$f(x) = q(x)(x - a) + f(a)$, and take $a = 0$.
 
  • #20
Deveno said:
Well, clearly $x \in \text{ker }\phi$, so $(x) \subseteq \text{ker }\phi$.

Suppose $f(0) = 0$. Does $x|f(x)$?

It may be helpful to remember that:

$f(x) = q(x)(x - a) + f(a)$, and take $a = 0$.

We have that if $f(x)=xg(x)\in (x)$ then $\phi (f(x))=0$.
So, $(x)\subseteq \ker\phi$.

If $f(x)\in \ker\phi$, then $\phi (f(x))=f(0)=0$.
We have that $f(x)=xq(x)+r\Rightarrow f(0)=r \Rightarrow r=0$.
So, $f(x)=xq(x)\Rightarrow f(x)\in (x)$.
Therefore, $\ker\phi =(x)$. Is everything correct? (Wondering)

So, we have that $\ker\phi =(x)$. Is this correct? (Wondering)
 
  • #21
mathmari said:
We have that if $f(x)=xg(x)\in (x)$ then $\phi (f(x))=0$.
So, $(x)\subseteq \ker\phi$.

If $f(x)\in \ker\phi$, then $\phi (f(x))=f(0)=0$.
We have that $f(x)=xq(x)+r\Rightarrow f(0)=r \Rightarrow r=0$.
So, $f(x)=xq(x)\Rightarrow f(x)\in (x)$.
Therefore, $\ker\phi =(x)$. Is everything correct? (Wondering)

So, we have that $\ker\phi =(x)$. Is this correct? (Wondering)

Yes, that's it.
 
  • #22
Deveno said:
Yes, that's it.

Thank you very much! (Yes)
 

FAQ: The principal ideal is maximal in but it is not the unique maximal ideal

What does it mean for a principal ideal to be maximal?

A principal ideal is considered maximal if it is not a proper subset of any other ideal in a given ring. In other words, there are no larger ideals that contain the principal ideal.

Why is the principal ideal maximal in some cases?

This is because the principal ideal contains all elements that can be generated by a single element, making it impossible for a larger ideal to contain it.

Why is the principal ideal not always the unique maximal ideal?

The principal ideal may not be the unique maximal ideal in a ring because there can be other ideals that are not principal but are also maximal. This means they cannot be contained in any larger ideal.

What is an example of a ring where the principal ideal is maximal but not unique?

An example of this is the ring of integers, where the principal ideal generated by the number 2 is maximal, but there are other maximal ideals such as the ideal generated by the number 3.

What implications does this have for the properties of the ring?

If a ring has multiple maximal ideals, it means that it does not have unique factorization of elements. This can also affect other algebraic properties of the ring, such as being a principal ideal domain.

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