I=<f> is maximal iff f is irreducible

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Discussion Overview

The discussion revolves around the characterization of maximal ideals in the polynomial ring $F[x]$ over a field $F$. Participants are exploring the relationship between maximal ideals and irreducible polynomials, specifically whether the ideal generated by a polynomial $f$ is maximal if and only if $f$ is irreducible.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that if the ideal $I=\langle f\rangle$ is maximal, then $f$ must be irreducible, arguing that if $f$ is reducible, it can be factored into non-constant polynomials, leading to a larger ideal.
  • Others question the completeness of the argument, suggesting that more detail is needed to explain why the ideal generated by a factor $g(x)$ is strictly larger than $I$ and remains a proper ideal in $F[x]$.
  • Some participants note that if $f$ is irreducible and $I$ is not maximal, then there exists an ideal $J$ such that $I \subsetneq J \subseteq F[x]$, and they explore how to show that $J$ must be of the form $\langle g(x)\rangle$ for some polynomial $g(x)$.
  • There is a discussion about the conditions under which a polynomial is a unit in $F[x]$, with participants agreeing that only constant polynomials are units.
  • Participants also discuss the implications of assuming $g(x) \in \langle f(x)\rangle$ and the resulting contradictions that arise from degree considerations.

Areas of Agreement / Disagreement

Participants generally agree on the direction of the arguments but express uncertainty about the completeness and clarity of the reasoning. Multiple competing views remain on how to rigorously establish the relationships between the ideals and the properties of the polynomials involved.

Contextual Notes

The discussion highlights limitations in the arguments presented, particularly regarding the need for detailed justifications of why certain ideals are proper or larger than others, as well as the assumptions about the nature of polynomials in $F[x]$.

mathmari
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Hey! :o

Let $F$ be a field, $f\in F[x]$ non-zero and $I=\langle f\rangle$. I want to show that $I$ is a maximal ideal of the ring $F[x]$ iff $f$ is an irreducible polynomial over $F$.

I have done the following:

$\Rightarrow$ :
Suppose that $I$ is maximal. Then if $J$ is an ideal of $F[x]$ and $I\subseteq J\subseteq F[x]$ then either $J=I$ or $J=F[x]$.
We assume that $f(x)$ is not irreducible, so $\exists g(x), h(x) \in F[x]$ such that $f(x)=g(x)h(x)$, where $g(x),h(x)$ are non-constant polynomials and not equal to $f(x)$.
Then $I=\langle f\rangle$ is not maximal, since $f(x)\in \langle g(x)\rangle \Rightarrow I\subseteq \langle g(x)\rangle$.
Therefore, we conclude that $f(x)$ is irreducible.

Is this direction correct? (Wondering) $\Leftarrow$ :
Suppose that $f$ is irreducible. We assume that $I$ is not maximal. Let $J$ be an ideal of $F[x]$ and $I\subsetneq J\subseteq F[x]$.

How could we continue? (Wondering)
 
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mathmari said:
Hey! :o

Let $F$ be a field, $f\in F[x]$ non-zero and $I=\langle f\rangle$. I want to show that $I$ is a maximal ideal of the ring $F[x]$ iff $f$ is an irreducible polynomial over $F$.

I have done the following:

$\Rightarrow$ :
Suppose that $I$ is maximal. Then if $J$ is an ideal of $F[x]$ and $I\subseteq J\subseteq F[x]$ then either $J=I$ or $J=F[x]$.
We assume that $f(x)$ is not irreducible, so $\exists g(x), h(x) \in F[x]$ such that $f(x)=g(x)h(x)$, where $g(x),h(x)$ are non-constant polynomials and not equal to $f(x)$.
Then $I=\langle f\rangle$ is not maximal, since $f(x)\in \langle g(x)\rangle \Rightarrow I\subseteq \langle g(x)\rangle$.
Therefore, we conclude that $f(x)$ is irreducible.

Is this direction correct? (Wondering)
That argument is correct in principle, but could do with a bit more detail, to explain why the ideal $\langle g(x)\rangle$ is (1) strictly larger than $I$, and (2) a proper ideal in $F[x].$

mathmari said:
$\Leftarrow$ :
Suppose that $f$ is irreducible. We assume that $I$ is not maximal. Let $J$ be an ideal of $F[x]$ and $I\subsetneq J\subseteq F[x]$.

How could we continue? (Wondering)
The key thing here is that $F[x]$ is a principal ideal domain, so that $J$ must be of the form $\langle g(x)\rangle$ for some $g(x) \in F[x].$
 
Opalg said:
That argument is correct in principle, but could do with a bit more detail, to explain why the ideal $\langle g(x)\rangle$ is (1) strictly larger than $I$, and (2) a proper ideal in $F[x].$

(1):
We have that $f(x)=g(x)h(x)$, where $g(x),h(x)$ are non-constant polynomials and not equal to $f(x)$.

We have that $\langle f(x)\rangle \subsetneq \langle g(x)\rangle$ because of the following:
We have that $g(x)\in \langle g(x)\rangle$ but $g(x)\notin \langle f(x)\rangle$, because if $g(x)\in \langle f(x)\rangle$ that would mean that $g(x)$ is a multiple of $f(x)$ in $F[x]$ and since $f(x)=g(x)h(x)$ we would have that $g(x)=f(x)\frac{1}{h(x)}$ and so $\frac{1}{h(x)}\in F[x]$, a contradiction.

Is this explanation correct? Could I improve something? (Wondering) (2):
To show that $\langle g(x)\rangle$ is a proper ideal in $F[x]$ we have to show that $1\notin \langle g(x)\rangle$, or not? (Wondering)
How could we show that? (Wondering)
Opalg said:
The key thing here is that $F[x]$ is a principal ideal domain, so that $J$ must be of the form $\langle g(x)\rangle$ for some $g(x) \in F[x].$

$\Leftarrow$ :
Suppose that $f$ is irreducible. We assume that $I$ is not maximal. Let $J$ be an ideal of $F[x]$ and $I\subsetneq J\subseteq F[x]$.
Since $F[x]$ is a principal ideal domain we have that $J$ must be of the form $\langle g(x)\rangle$ for some $g(x) \in F[x]$.
We have to show that $1\in \langle g(x)\rangle$ so that $\langle g(x)\rangle=F[x]$, to get a contradiction, or not?
But how could we show that? (Wondering)
 
mathmari said:
(1):
We have that $f(x)=g(x)h(x)$, where $g(x),h(x)$ are non-constant polynomials and not equal to $f(x)$.

We have that $\langle f(x)\rangle \subsetneq \langle g(x)\rangle$ because of the following:
We have that $g(x)\in \langle g(x)\rangle$ but $g(x)\notin \langle f(x)\rangle$, because if $g(x)\in \langle f(x)\rangle$ that would mean that $g(x)$ is a multiple of $f(x)$ in $F[x]$ and since $f(x)=g(x)h(x)$ we would have that $g(x)=f(x)\frac{1}{h(x)}$ and so $\frac{1}{h(x)}\in F[x]$, a contradiction.

Is this explanation correct? Could I improve something? (Wondering)
The expression $\frac{1}{h(x)}$ only makes sense if $h(x)$ is a unit in $F(x).$ Your explanation should aim to say that $f(x)$ cannot be in $\langle g(x)\rangle$ because that would imply that $h(x)$ is a unit (which it isn't).

mathmari said:
(2):
To show that $\langle g(x)\rangle$ is a proper ideal in $F[x]$ we have to show that $1\notin \langle g(x)\rangle$, or not? (Wondering)
How could we show that? (Wondering)
If $1\in \langle g(x)\rangle$ then $1 = g(x)k(x)$ for some $k(x) \in F(x).$ That would mean that $g(x)$ is a unit (which, again, it isn't).

mathmari said:
$\Leftarrow$ :
Suppose that $f$ is irreducible. We assume that $I$ is not maximal. Let $J$ be an ideal of $F[x]$ and $I\subsetneq J\subseteq F[x]$.
Since $F[x]$ is a principal ideal domain we have that $J$ must be of the form $\langle g(x)\rangle$ for some $g(x) \in F[x]$.
We have to show that $1\in \langle g(x)\rangle$ so that $\langle g(x)\rangle=F[x]$, to get a contradiction, or not?
But how could we show that? (Wondering)
If $\langle g(x)\rangle$ contains $I$ then $f(x) \in \langle g(x)\rangle$ and therefore $f(x) = g(x)h(x)$ for some $h(x)\in F(x).$ Now you need to show that neither $g(x)$ nor $h(x)$ is a unit. That contradicts the fact that $f(x)$ is irreducible.
 
Opalg said:
The expression $\frac{1}{h(x)}$ only makes sense if $h(x)$ is a unit in $F(x).$ Your explanation should aim to say that $f(x)$ cannot be in $\langle g(x)\rangle$ because that would imply that $h(x)$ is a unit (which it isn't).

Does it hold that $h(x)\in F[x]$ is a unit only when $h(x)$ is a constant polynomial? (Wondering)
 
mathmari said:
Does it hold that $h(x)\in F[x]$ is a unit only when $h(x)$ is a constant polynomial? (Wondering)
Yes. The units in $F[x]$ are the nonzero constants.
 
Opalg said:
Yes. The units in $F[x]$ are the nonzero constants.

Suppose that $h(x)$ is a unit in $F[x]$, then there is a polynomial in $F[x]$, say $g(x)$ such that $g(x)h(x)=1$.
Since $F$ is a field, so also an integral domain we have that $\deg (gh)=\deg (1) \Rightarrow \deg (g)+\deg (h)=0 \Rightarrow \deg (g)=\deg (h)=0$, since $\deg \geq 0$.

Is this justification correct? (Wondering)
 
Opalg said:
Your explanation should aim to say that $f(x)$ cannot be in $\langle g(x)\rangle$ because that would imply that $h(x)$ is a unit (which it isn't).

Do we not have to show that $g(x)\notin \langle f(x)\rangle$, so that $\langle f(x)\rangle \subsetneq \langle g(x)\rangle$ ? (Wondering)
 
mathmari said:
Do we not have to show that $g(x)\notin \langle f(x)\rangle$, so that $\langle f(x)\rangle \subsetneq \langle g(x)\rangle$ ? (Wondering)
My mistake – yes, you want to show that $g(x)\notin \langle f(x)\rangle$. To do that, suppose that $g(x) = f(x)k(x)$. Then $f(x) = g(x)h(x) = f(x)\bigl(k(x)h(x)\bigr).$ But that is not possible because the degree of the left side is less than the degree of the right side (since $h(x)$ is not a constant, so its degree is at least $1$).
 
  • #10
Opalg said:
My mistake – yes, you want to show that $g(x)\notin \langle f(x)\rangle$. To do that, suppose that $g(x) = f(x)k(x)$. Then $f(x) = g(x)h(x) = f(x)\bigl(k(x)h(x)\bigr).$ But that is not possible because the degree of the left side is less than the degree of the right side (since $h(x)$ is not a constant, so its degree is at least $1$).

Do we have that $\deg (f(x)) =\deg ( f(x)\bigl(k(x)h(x)\bigr))=\deg (f(x))+\deg (k(x))+\deg (h(x))$, since $F$ is a field? (Wondering)
 
  • #11
mathmari said:
Do we have that $\deg (f(x)) =\deg ( f(x)\bigl(k(x)h(x)\bigr))=\deg (f(x))+\deg (k(x))+\deg (h(x))$, since $F$ is a field? (Wondering)
Yes.
 
  • #12
Opalg said:
The key thing here is that $F[x]$ is a principal ideal domain, so that $J$ must be of the form $\langle g(x)\rangle$ for some $g(x) \in F[x].$

How do we know that $F[x]$ is a principal ideal domain? I got stuck right now... (Wondering)
Opalg said:
If $\langle g(x)\rangle$ contains $I$ then $f(x) \in \langle g(x)\rangle$ and therefore $f(x) = g(x)h(x)$ for some $h(x)\in F(x).$ Now you need to show that neither $g(x)$ nor $h(x)$ is a unit. That contradicts the fact that $f(x)$ is irreducible.

Do we have that neither $g(x)$ nor $h(x)$ is a unit because they cannot be constant polynomials? (Wondering)
 
  • #13
Opalg said:
If $\langle g(x)\rangle$ contains $I$ then $f(x) \in \langle g(x)\rangle$ and therefore $f(x) = g(x)h(x)$ for some $h(x)\in F(x).$ Now you need to show that neither $g(x)$ nor $h(x)$ is a unit. That contradicts the fact that $f(x)$ is irreducible.

Do you maybe mean $h(x)\in F[x]$ ? (Wondering)
 

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