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

[mathmari]

Hey!

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)

 

[Deveno]

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 $F[x]/(x)$)?

 

[mathmari]

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)

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)

 

[Deveno]

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?

 

[mathmari]

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)

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)

 

[Deveno]

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?

 

[mathmari]

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)

 

[Deveno]

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).

 

[mathmari]

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

I don't really have an idea... Could you give me a hint? (Wondering)

 

[Deveno]

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 $x - 1$?

 

[mathmari]

What are the possible remainders of $f(x)$ upon division by $x - 1$?

How could we find that? I got stuck right now... (Wondering)

 

[Deveno]

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?

 

[mathmari]

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)

 

[Deveno]

$r(x)$ must be a constant, right? (Wondering)

Yes, that is-$r \in F$. Can you continue?

 

[mathmari]

Yes, that is-$r \in F$. Can you continue?

Does this mean that $F[x]/(x-1)$ is isomorphic to $F$ ? (Wondering)

 

[mathmari]

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)

 

[Deveno]

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).

 

[mathmari]

$(x)$ 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)

 

[Deveno]

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 $a = 0$.

 

[mathmari]

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)

 

[Deveno]

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.

 

[mathmari]

Yes, that's it.

Thank you very much! (Yes)