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

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 $$p(x) | f(x)$$ it follows that $$d(x) = p(x)$$ ...My rough reasoning is as follows:We have that $$d(x) = (p,f)$$ ...

so that $$d|p$$ and $$d|f$$ ...

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

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

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

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

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

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

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

So it must be that $$p = d$$ ...
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 $$d(x)$$ 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 $$d(x)$$ 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 $q(x)=1$/