I=<f> is maximal iff f is irreducible

[mathmari]

Hey!

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)

 

[Opalg]

Hey!

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 $F[x].$

$\Leftarrow$ :
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 $g(x) \in F[x].$

 

[mathmari]

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)

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)

 

[Opalg]

(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).

(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).

$\Leftarrow$ :
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.

 

[mathmari]

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)

 

[Opalg]

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.

 

[mathmari]

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)

 

[mathmari]

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)

 

[Opalg]

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 $1$).

 

[mathmari]

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)

 

[Opalg]

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.

 

[mathmari]

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)

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)

 

[mathmari]

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)