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Let K be a field, $a,b \in K[X]$ and $i,j \in K$ with $i \not= 0$ and $j \not= 0$.

Prove that $gcd(a,b) = gcd(a*i,b*j)$

Could you please help me proving this. I have the following idea:

For any common divisor $d$ of $a$ and $b$ is: $d | a$ and $d | b$ and $d | a*i$ and $d | b*j$

For any common divisor $d$ of $a*i$ and $b*i$ is: $d | a$ and $d | b$ and $d | a*i$ and $d | b*j$

Thus every common divisor of a and b is an common divisor of $a*i$ and $b*j$. But how to prove it is the biggest divisor? Thanks!

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Generally, in any gcd domain, if $\,i,j\,$ are units (invertibles) then

$$ d\mid a,b \iff d\mid ai,bj$$

Thus $\,a,b\,$ and $\,ai,bi\,$ have the same set $\,S\,$ of common divisors $\,d,\,$ therefore they also have the same greatest common divisor - any greatest element of $S\,$ (= greatest degree polynomial in your case, usually normalized to be monic, i.e. have leading coefficient $=1)$,

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    Yes. Thats also the idea I had, but my math teacher wants me to explicitly show that the greatest common divisor is equal. – 2017-02-23
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    @user3162981 Hint: by the above, both gcds have equal degree, and they divide each other, so they are equal up to a unit constant factor $\,c;\,$ necessarily $c=1$ since both are monic. – 2017-02-23
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    Ok. So this is the full proof? Thanks! – 2017-02-23
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    @user3162981 Yes, but that's a question you should answer, i.e. *you* should understand why the proof is complete. – 2017-02-23