I=<f> is maximal iff f is irreducible

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In summary, the conversation discusses proving that a given ideal $I$ is maximal in the ring $F[x]$ if and only if the generating element $f$ is an irreducible polynomial over the field $F$. The first part of the conversation provides a proof for the direction $\Rightarrow$, while the second part discusses how to continue with the proof for the direction $\Leftarrow$. The conversation also includes clarifications and explanations for the proof steps.
  • #1
mathmari
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Hey! :eek:

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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  • #2
mathmari said:
Hey! :eek:

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 $g(x) \in F[x].$
 
  • #3
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)
 
  • #4
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.
 
  • #5
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)
 
  • #6
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.
 
  • #7
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)
 
  • #8
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)
 
  • #9
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)
 

1. What does "maximal" mean in this context?

In the context of algebra, "maximal" refers to a property of an ideal, which is a subset of a ring. A maximal ideal is an ideal that is not contained in any other strictly larger ideal. In simpler terms, it is an ideal that cannot be extended any further.

2. What is the significance of an ideal being maximal?

Maximal ideals are important in the study of rings and fields because they are a key tool in understanding the structure and properties of these mathematical objects. They also play a crucial role in the classification of rings and fields.

3. How is the concept of "maximal" related to irreducibility?

In the context of algebra, an irreducible element is one that cannot be factored into smaller pieces. In this case, "smaller pieces" refers to elements that are not units (i.e. elements with multiplicative inverses) and are not irreducible themselves. The statement "I= is maximal iff f is irreducible" means that the ideal generated by the irreducible element f is maximal, and vice versa.

4. Can you provide an example to illustrate this statement?

Consider the ring Z[x], which consists of polynomials with integer coefficients. Let f(x) = x and I = = {a * x | a is an integer}. It can be shown that this ideal is maximal, and f(x) is an irreducible element. This means that the ideal generated by the irreducible element f(x) is maximal, and vice versa.

5. How does this statement apply to real-life situations?

The concept of maximal ideals and irreducible elements is mainly used in abstract algebra, which has applications in various fields such as computer science, physics, and engineering. For example, in coding theory, the factorization of polynomials over finite fields is used to construct error-correcting codes. The study of maximal ideals and irreducible elements helps in understanding the structure of these codes and their properties.

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