Polynomials Over a Field - Rotman, Lemma 3.67

[Math Amateur]

I am reading Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) ...

I am currently focused on Section 3.5 From Numbers to Polynomials ...

I need help with an aspect of the proof of Lemma 3.67 ...

The relevant text from Rotman's book is as follows:View attachment 4640In the proof of the above Lemma, we read the following:

" ... ... If \(\displaystyle p(x) | f(x)\), then \(\displaystyle d(x) = p(x)\), for \(\displaystyle p(x)\) is monic. ... ... "Can someone please explain how \(\displaystyle p(x) | f(x)\) and \(\displaystyle p(x)\) being monic implies that \(\displaystyle d(x) = p(x)\) ... ...

Hope someone can help ... ...

Peter*** EDIT ***

I have just realized that I do not understand how the second line of the proof works ... so if someone can also help me with that, i would appreciate it very much ...

 

[Deveno]

$p$ is irreducible (this is like being a prime integer).

So if $d(x)|p(x)$, the ONLY things $d$ can be (if $d$ is monic) are $1$ and $p(x)$.

If $p(x)|f(x)$, then $d(x) = p(x)$, since $\text{deg}(p) > \text{deg}(1) = 0$ (remember, irreducible elements are BY DEFINITION non-units-this is why we say $1$ and $-1$ are not prime, in the integers). Remember $d$ is the monic polynomial OF GREATEST DEGREE that divides both $p$ and $f$.

On the other hand, if $p(x)\not\mid f(x)$, then $d(x) \neq p(x)$, so $d(x)$ must be $1$.

 

[Math Amateur]

$p$ is irreducible (this is like being a prime integer).

So if $d(x)|p(x)$, the ONLY things $d$ can be (if $d$ is monic) are $1$ and $p(x)$.

If $p(x)|f(x)$, then $d(x) = p(x)$, since $\text{deg}(p) > \text{deg}(1) = 0$ (remember, irreducible elements are BY DEFINITION non-units-this is why we say $1$ and $-1$ are not prime, in the integers). Remember $d$ is the monic polynomial OF GREATEST DEGREE that divides both $p$ and $f$.

On the other hand, if $p(x)\not\mid f(x)$, then $d(x) \neq p(x)$, so $d(x)$ must be $1$.

Thanks Deveno ... but (apologies) I need a bit more help ...

You write:

" ... If $p(x)|f(x)$, then $d(x) = p(x)$, since $\text{deg}(p) > \text{deg}(1) = 0$..."

I am still struggling to see how this follows ... can you be more explicit as to why this follows ...

(Note: I can see that $\text{deg}(p) > \text{deg}(1) = 0$ since an irreducible in a ring is by definition not a unit ... )

Sorry to be slow on this matter ...

Peter

 

[caffeinemachine]

Thanks Deveno ... but (apologies) I need a bit more help ...

You write:

" ... If $p(x)|f(x)$, then $d(x) = p(x)$, since $\text{deg}(p) > \text{deg}(1) = 0$..."

I am still struggling to see how this follows ... can you be more explicit as to why this follows ...

(Note: I can see that $\text{deg}(p) > \text{deg}(1) = 0$ since an irreducible in a ring is by definition not a unit ... )

Sorry to be slow on this matter ...

Peter

I do not mean to hijack this thread. but I think I can help you with this Peter.

Suppose $p(x)|f(x)$. Write $D(x)=p(x)$. Note that $D(x)$ divides both $p(x)$ and $f(x)$.

We claim that $D(x)|d(x)$. This is because we can write $d(x)=p(x)a(x)+f(x)b(x)$ for some polynomials $a(x)$ and $b(x)$.

From this we conclude that $d(x)=p(x)q(x)$ for some polynomial $q(x)$. But since $d(x)|p(x)$, we must have $q(x)=1$ (since $d(x)$ is a monic.)

 

[Math Amateur]

$p$ is irreducible (this is like being a prime integer).

So if $d(x)|p(x)$, the ONLY things $d$ can be (if $d$ is monic) are $1$ and $p(x)$.

If $p(x)|f(x)$, then $d(x) = p(x)$, since $\text{deg}(p) > \text{deg}(1) = 0$ (remember, irreducible elements are BY DEFINITION non-units-this is why we say $1$ and $-1$ are not prime, in the integers). Remember $d$ is the monic polynomial OF GREATEST DEGREE that divides both $p$ and $f$.

On the other hand, if $p(x)\not\mid f(x)$, then $d(x) \neq p(x)$, so $d(x)$ must be $1$.

Hi Deveno,

I have been reflecting on what you have said and I think I now follow why if \(\displaystyle p(x) | f(x)\) it follows that \(\displaystyle d(x) = p(x)\) ...My rough reasoning is as follows:We have that \(\displaystyle d(x) = (p,f)\) ...

so that \(\displaystyle d|p\) and \(\displaystyle d|f\) ...

We also have that \(\displaystyle p(x)\) is a monic irreducible polynomial ...

Now, p irreducible implies that if \(\displaystyle p = uv\) then one of \(\displaystyle u, v\) is a unit ... take it to be \(\displaystyle u\) ...

But \(\displaystyle p\) is monic, so if the divisors of \(\displaystyle p\) are monic (as \(\displaystyle d\) is!) then \(\displaystyle u = 1\) and \(\displaystyle v = p\)

Thus the only monic divisors of p(x) are 1 and d(x)Now, suppose \(\displaystyle p(x) |f(x)\) ...

We know that \(\displaystyle p(x)\), being irreducible, is not a unit so \(\displaystyle \text{deg } p \gt 0 \) ...

But we also know that \(\displaystyle d(x) | f(x)\) and, further that \(\displaystyle d\) is the monic polynomial of greatest degree that divides both \(\displaystyle p\) and \(\displaystyle f\) ...

So, if \(\displaystyle p \neq d\) then since \(\displaystyle p|f\) and \(\displaystyle d|f\) ... we have that \(\displaystyle d\) is greater degree than \(\displaystyle p\) ... ... which cannot be true as we also have that \(\displaystyle d|p\) ...

So it must be that \(\displaystyle p = d\) ...
Can you please confirm that my reasoning is OK ... ... or alternatively, point out deficiencies or errors ...

Peter

 

[Math Amateur]

I do not mean to hijack this thread. but I think I can help you with this Peter.

Suppose $p(x)|f(x)$. Write $D(x)=p(x)$. Note that $D(x)$ divides both $p(x)$ and $f(x)$.

We claim that $D(x)|d(x)$. This is because we can write $d(x)=p(x)a(x)+f(x)b(x)$ for some polynomials $a(x)$ and $b(x)$.

From this we conclude that $d(x)=p(x)q(x)$ for some polynomial $q(x)$. But since $d(x)|p(x)$, we must have $q(x)=1$ (since $d(x)$ is a monic.)

Hi caffeinemachine,

Thanks for your help ... most welcome ...... Yes follow that $d(x)=p(x)q(x)$ for some polynomial $q(x)$ ...

But do not quite see how \(\displaystyle d(x)\) being monic and $d(x)|p(x)$ mean that $q(x)=1$ ...

Can you help?

Peter

 

[caffeinemachine]

Hi caffeinemachine,

Thanks for your help ... most welcome ...... Yes follow that $d(x)=p(x)q(x)$ for some polynomial $q(x)$ ...

But do not quite see how \(\displaystyle d(x)\) being monic and $d(x)|p(x)$ mean that $q(x)=1$ ...

Can you help?

Peter

Since $d(x)|p(x)$, we know there is a polynomial $a(x)$ such that $d(x)a(x)=p(x)$, which gives $p(x)q(x)a(x)=p(x)$. This leads to $q(x)a(x)=1$. Thus $q(x)$ must be a constant polynomial. Now since $d(x)$ is moinc we must have $q(x)=1$/