Show that $p(x)|f(x)$: Proving the Divisibility

[mathmari]

Hey!

Let $K$ be a field, $p(x)\in K[x]$ irreducible and $L$ an extension of $K$, that contains the root $\rho$ of $p(x)$.
Let $f(x)\in K[x]$ has also the root $\rho$.
I want to show that $p(x)\mid f(x)$.

I have done the following:

Let $f(x)=g(x)p(x)+r(x)$, where $g(x), r(x)\in K[x]$ and $\deg r(x)<\deg p(x)$.
For $x=\rho$ we have that $$f(\rho)=g(\rho)p(\rho)+r(\rho) \Rightarrow r(\rho)=0$$

Does this help? (Wondering)

How could we continue? (Wondering)

 

[Deveno]

Suppose $m(x)$ is the minimal polynomial of $\rho$ over $K$.

Show $m|p$, then apply what you have shown.

 

[mathmari]

Suppose $m(x)$ is the minimal polynomial of $\rho$ over $K$.

Show $m|p$, then apply what you have shown.

The minimal polynomial $m(x)$ of $\rho$ is the monic polynomial of least degree among all polynomials in $K[x]$ having $\rho$ as a root. So, we have that $m(\rho)=p(\rho)=0$ and $\deg m\leq \deg p$.
How exactly can we show that $m\mid p$ ? Could you give me some hints? (Wondering)

 

[Deveno]

The minimal polynomial $m(x)$ of $\rho$ is the monic polynomial of least degree among all polynomials in $K[x]$ having $\rho$ as a root. So, we have that $m(\rho)=p(\rho)=0$ and $\deg m\leq \deg p$.
How exactly can we show that $m\mid p$ ? Could you give me some hints? (Wondering)

Write $p(x) = q(x)m(x) + r(x)$, as before. Then:

$0 = p(\rho) = q(\rho)m(\rho) + r(\rho) = q(\rho)\cdot 0 + r(\rho) = r(\rho)$. Why can we not have $r(x) \neq 0$?

Now, $p$ is irreducible, so...

 

[mathmari]

Write $p(x) = q(x)m(x) + r(x)$, as before. Then:

$0 = p(\rho) = q(\rho)m(\rho) + r(\rho) = q(\rho)\cdot 0 + r(\rho) = r(\rho)$. Why can we not have $r(x) \neq 0$?

Suppose that $r(x)\neq 0$.

From the eucildean division, we have that $\deg r<\deg m$.

We have that $m(x)$ is the monic polynomial of least degree among all polynomials in $K[x]$ having $\rho$ as a root. So $\deg m\leq \deg r$. A contradiction.

Therefore, it must be $r(x)=0$. So, $p(x)=q(x)m(x)\Rightarrow m(x)\mid p(x)$.

Is this correct? (Wondering)

 

[Deveno]

Suppose that $r(x)\neq 0$.

From the eucildean division, we have that $\deg r<\deg m$.

We have that $m(x)$ is the monic polynomial of least degree among all polynomials in $K[x]$ having $\rho$ as a root. So $\deg m\leq \deg r$. A contradiction.

Therefore, it must be $r(x)=0$. So, $p(x)=q(x)m(x)\Rightarrow m(x)\mid p(x)$.

Is this correct? (Wondering)

Indeed, in fact $m(x)$ divides any polynomial having $\rho$ as a root. Now, finish it up...

 

[mathmari]

Now, $p$ is irreducible, so...

How can we use this fact? (Wondering)

 

[Deveno]

What kind of factors do irreducible polynomials have?

 

[mathmari]

What kind of factors do irreducible polynomials have?

Just the polynomials themselves, or not? (Wondering)

 

[Deveno]

In a ring $R$, we say that $a \in R$ is irreducible if:

$a = bc \implies b$ or $c$ is a unit.

What are the units of $K[x]$?

 

[mathmari]

In a ring $R$, we say that $a \in R$ is irreducible if:

$a = bc \implies b$ or $c$ is a unit.

What are the units of $K[x]$?

The units of $K[x]$ are the constant elements of $K[x]$ that are units of $K$, or not? (Wondering)

 

[Deveno]

The units of $K[x]$ are the constant elements of $K[x]$ that are units of $K$, or not? (Wondering)

Yes, that is correct.

So we know the following:

1. $m(x)|p(x)$

2. If $p(x) = a(x)m(x)$, then $a(x)$ is a unit, since $p$ is irreducible.

3. The units of $K[x]$ are the elements of $K^{\ast}$.

4. $m(x)|f(x)$

 

[mathmari]

2. If $p(x) = a(x)m(x)$, then $a(x)$ is a unit, since $p$ is irreducible.

Why is then $a(x)$ a unit and not $m(x)$ ? (Wondering)

4. $m(x)|f(x)$

Why does this hold? (Wondering) I got stuck right now...