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Question1: Proof:

If $d(x)|f(x),d(x)|g(x)$, $d(x)$ is an combination of $f(x)$ and $g(x)$, then $d(x)$ is $\text{GCD}(f(x),g(x))$.

My proof 1

$d(x)=u(x)f(x)+v(x)g(x)$. If $d(x)$ is not $\text{GCD}(f(x),g(x))$ ,

then $d(x)h(x)=\text{GCD}(f(x),g(x))=u'(x)f(x)+v'(x)g(x)$.

$d(x)=[\text{step1} ]u'(x)\frac{f(x)}{h(x)}+v'(x)\frac{g(x)}{h(x)}$, where $h(x)|f(x)$ and $h(x)|g(x)$.

Consider the case $u'(x)=v'(x)=1$. if $h(x)\neq 1$, then $d(x)$ is not a combination of $f(x)$ and $g(x)$.

so $h(x)=1$.

Is my proof right? How to prove along my thread?

If the above is wrong, one of my thougts is to prove $d(x)$ does not divides both $f(x)$ and $g(x)$.

update:

As @Marc van Leeuwen mentioned about invertible. Seems step1 is wrong, I cannot divide $h(x)$ to make the equality true and also, $h(x)|u'(x)$ could holds so that I cannot conclude $h(x)|f(x)$.

My proof 2

$d(x)=\text{CD}(f(x),g(x))=u(x)f(x)+v(x)g(x)$. If $h(x)$ is one $\text{CD}(f(x),g(x))$ ,

then $h(x)$ divides any combinations of $f(x)$ and $g(x)$, so $h(x)|d(x)$, then $d(x)$ is $\text{GCD}(f(x),g(x))$

Question2: Proof:

$\text{GCD}(f(x)h(x),g(x)h(x))=\text{GCD}(f(x),g(x))h(x)$ (coefficient of $h(x)$ 's first term is $1$).

My proof

$$d(x) = \text{GCD}(f(x)h(x),g(x)h(x))=u(x)f(x)h(x)+v(x)g(x)h(x)=h(x)(u(x)f(x)+g(x)v(x))=h(x)d_1(x)~~~(1)$$

to prove (1), we should show $\left.d_1(x)\right|f(x), \left.d_1(x)\right|g(x)$

because $d(x)|f(x)h(x)$ and $d(x)|g(x)h(x)$, so(1) is true by the result of question1.

Is the proof right? Seems wrong, because I haven't used the condition [...coefficient is 1] , what does that mean. Or what's the right first response for seeing this condition.

HyperGroups
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  • There is always the nasty question of invertible elements. Since you are (in view of all those tiring $(x)$'s) working with polynomials, the invertibles are the constant polynomials. So instead of $h(x)=1$ you can only conclude $h(x)$ is a constant. This might be implicit in your reasoning, but you have to be careful since not all expressions manipulated can be taken "up to a scalar multiple". – Marc van Leeuwen Jun 22 '13 at 08:19
  • @MarcvanLeeuwen ah, have I got your point? seems [step] 1 is wrong. – HyperGroups Jun 23 '13 at 05:56

1 Answers1

1

1) There might not be an $x$ such that $u'(x)=1$.

Abbrevieate $d=d(x)$ etc.

If $d=uf+vg$ then $\gcd(f,g)|d$ and $h=1$.

2) $\gcd(fh,gh)=h\gcd(f,g)$ which is not necessarily $fgh$ because $f$ and $g$ might share a factor. Was there a typo in the question?

Angela Pretorius
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