Question

Prove or disprove: If $\mathit{p}\left(x\right)$is relatively prime to $\mathit{k}\left(x\right)$and $\mathit{f}\left(x\right)\mathit{k}\left(x\right)\mathbf{\equiv }\mathit{g}\left(x\right)\mathit{k}\left(x\right)\left(modp\left(x\right)\right)$ , then $\mathit{f}\left(x\right)\mathbf{\equiv }\mathit{g}\left(x\right)\left(modp\left(x\right)\right)$

Short answer

It is proved$f\left(x\right)\equiv g\left(x\right)\left(modp\left(x\right)\right)$

Step-by-step solution

Write the theorem

Assume that F is a field and $a\left(x\right)b\left(x\right)c\left(x\right)$ belong to the field; $a\left(x\right)$divides $c\left(x\right)$if$a\left(x\right),b\left(x\right),c\left(x\right)$ and$a\left(x\right),b\left(x\right)$ are relatively primes

Proof

It is given that $p\left(x\right)$is relatively prime to $k\left(x\right)$, and $f\left(x\right)k\left(x\right)\equiv g\left(x\right)k\left(x\right)\left(modp\left(x\right)\right)$

Then, by definition $f\left(x\right)k\left(x\right)-g\left(x\right)k\left(x\right)=\left(f\left(x\right)-g\left(x\right)\right)k\left(x\right)$, we have $p\left(x\right)/f\left(x\right)-g\left(x\right)k\left(x\right)$

Using the theorem stated in step 1, we have

$p\left(x\right)/\left(f\left(x\right)-g\left(x\right)\right)$

That implies,

$f\left(x\right)\equiv g\left(x\right)\left(modp\left(x\right)\right)$

Hence, it is proved

$f\left(x\right)\equiv g\left(x\right)\left(modp\left(x\right)\right)$