MHB Are \( x-a \) and \( x-b \) Co-primes?

[evinda]

Hello! :)

Knowing that $K$ is a field, $a,b \in K$ different from each other,show that $x-a,x-b$ co-primes.

We suppose that $\exists f(x) \in K(x)$ such that:

$f(x)|x-a$ and $f(x)|x-b$Then $deg f(x) \leq 1$

So, $deg f(x)=0 \text{ or } 1$

If $deg f(x)=1$:

$f(x)=x-c,c\in K$

$x-c|x-a , x-c|x-b \Rightarrow x-c| x-a-x+b \Rightarrow x-c|b-a$ contradiction,because that would mean $deg(x-c) \leq deg(b-a) \Rightarrow 1 \leq 0$

If $deg f(x)=0, f(x)=c,c \in K$

$c|x-a, c|x-b \Rightarrow c|b-a \Rightarrow c|(b-a)^{-1}(b-a) \Rightarrow c|1 \Rightarrow c=\pm 1$

So, $gcd(x-a,x-b)=1$.Do we suppose that $f(x)$ is the greatest common divisor of $x-a$ and $x-b$ or just a common divisor ? Also,at the case when $deg f(x)=1$,why do we take $f(x)=x-c$ ??

 

[Deveno]

Any polynomial in $K[x]$ of degree 1 is of the form:

$c_1x + c_0$.

Since it is of degree 1 (and not 0), we must have $c_1 \neq 0$, that is: $c_1 \in K^{\ast}$, so that $c_1$ is a unit (since $K$ is a field). Hence $c_1^{-1}$ exists in $K$, so we have:

$c_1x + c_0 = c_1(x + c_0c_1^{-1})$.

Since multiplying by a unit makes no difference to divisibility, we have:

$(c_1x + c_0)|g(x) \iff (x + c_0c_1^{-1})|g(x)$.

Letting $c = -c_0c_1^{-1}$, we see it suffices to consider $x - c$.

******

Yes, we only assume we have a common divisor, there is not need to suppose we have the greatest common divisor.

Recall as well, that the greatest common divisor is only unique up to a unit, so from:

$c|1$ all we can really conclude is that $c$ is a unit, which is all we need

(For example, in the field $\Bbb Q$ we have $4|1$ since: $1 = 4(\frac{1}{4})$).

 

[I like Serena]

Hey! (Sun)

$c|x-a, c|x-b \Rightarrow c|b-a \Rightarrow c|(b-a)^{-1}(b-a) \Rightarrow c|1 \Rightarrow c=\pm 1$

Huh?? (Wondering)
How did you get that $c=\pm 1$?
I think the only thing you can say is that $c \in K^*$, which was already implicit.

Do we suppose that $f(x)$ is the greatest common divisor of $x-a$ and $x-b$ or just a common divisor ?

Just any divisor.
The point is that we prove that f(x) will have to be a unit in K.

Also,at the case when $deg f(x)=1$,why do we take $f(x)=x-c$ ??

What alternative would you be able to pick?

 

[ThePerfectHacker]

If $a\not = b$ then $b-a\not = 0$ and so,

$$ \tfrac1{b-a} \cdot (x-a) +(-\tfrac1{b-a}) \cdot (x-b) = 1$$

 

[evinda]

How did you get that $c=\pm 1$?
I think the only thing you can say is that $c \in K^*$, which was already implicit.

How do we conclude that $c \in K^*$ ?? (Thinking)

Just any divisor.
The point is that we prove that f(x) will have to be a unit in K.
What alternative would you be able to pick?

I would pick $f(x)=ex+d$,but from this we could take $e^{-1}f(x)=x+e^{-1}d$,right? And then we have to set $-c=e^{-1}d$ ? (Thinking)

 

[evinda]

Any polynomial in $K[x]$ of degree 1 is of the form:

$c_1x + c_0$.

Since it is of degree 1 (and not 0), we must have $c_1 \neq 0$, that is: $c_1 \in K^{\ast}$, so that $c_1$ is a unit (since $K$ is a field). Hence $c_1^{-1}$ exists in $K$, so we have:

$c_1x + c_0 = c_1(x + c_0c_1^{-1})$.

Since multiplying by a unit makes no difference to divisibility, we have:

$(c_1x + c_0)|g(x) \iff (x + c_0c_1^{-1})|g(x)$.

Letting $c = -c_0c_1^{-1}$, we see it suffices to consider $x - c$.

******

Yes, we only assume we have a common divisor, there is not need to suppose we have the greatest common divisor.

I understand.. :)

Recall as well, that the greatest common divisor is only unique up to a unit, so from:

$c|1$ all we can really conclude is that $c$ is a unit, which is all we need

(For example, in the field $\Bbb Q$ we have $4|1$ since: $1 = 4(\frac{1}{4})$).

I haven't understood why from the relation $c|1$,we conclude that $c$ is a unit..Could you explain it further to me?

- - - Updated - - -

If $a\not = b$ then $b-a\not = 0$ and so,

$$ \tfrac1{b-a} \cdot (x-a) +(-\tfrac1{b-a}) \cdot (x-b) = 1$$

And how do we conclude from this relation that the greatest common divisor is $1$? Because we can find $h_1(x),h_2(x) \in K(x)$ such that $h_1(x)(x-a)+h_2(x)(x-b)=1$ ??(Thinking)

 

[ThePerfectHacker]

I
And how do we conclude from this relation that the greatest common divisor is $1$? Because we can find $h_1(x),h_2(x) \in K(x)$ such that $h_1(x)(x-a)+h_2(x)(x-b)=1$ ??(Thinking)

Yes. Two polynomials $f(x),g(x)$ are relatively prime if and only if $a(x)f(x) + b(x)g(x) = 1$.

 

[evinda]

Yes. Two polynomials $f(x),g(x)$ are relatively prime if and only if $a(x)f(x) + b(x)g(x) = 1$.

I understand..thanks a lot! :)

 

[evinda]

$c|1$ all we can really conclude is that $c$ is a unit, which is all we need

Is it an identity that if $c|1$,it is a unit??

 

[I like Serena]

Is it an identity that if $c|1$,it is a unit??

Let's see what they both mean.

$c$ is a unit if it has an inverse.
In a field every element except 0 has an inverse.

$c|1$ if there is a number $k$ in the field such that $kc=1$.
We can pick $k=c^{-1}$ satisfying the equation, which works for every element except 0.

So yes, if $c$ is an element of a field, then $c|1$ iff $c$ is a unit.

 

[evinda]

Let's see what they both mean.

$c$ is a unit if it has an inverse.
In a field every element except 0 has an inverse.

$c|1$ if there is a number $k$ in the field such that $kc=1$.
We can pick $k=c^{-1}$ satisfying the equation, which works for every element except 0.

So yes, if $c$ is an element of a field, then $c|1$ iff $c$ is a unit.

I understand..Thank you very much! (Nod)