3
$\begingroup$

Division Algorithm for $F[x]$ Theorem: Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ and $g(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_0$ be two elements of $F[x]$ with $a_n$ and $b_m$ both nonzero of $F$ and $m>0$. Then there are unique polynomials $q(x)$ and $r(x)$ in $F[x]$ such that $f(x)=g(x)q(x)+r(x)$ where $r(x)=0$ or the degree of $r(x)$ is less than the degree of $g(x)$.

In this theorem, there's a part on the proof that I don't really understand. The proof goes like this, Consider $S=\{f(x)g(x)s(x)|s(x)\in\ F[x]\}$

In Case 1, it consider $0\in\S$ While in Case 2, $0\notin\S$ So, we let $r(x)$ be the element with the smallest exponent and $r(x)=c_tx^t+c_{t-1}+\cdots+c_0$ where $c_t$ nonzero of $F$ We also have $f(x)-g(x)q(x)=r(x)$

Now, we need to show that $t<m$ By contradiction, assume that $ t>=m$ then, $$f(x)-g(x)q(x)-\frac{c_t}{b_m}x^{t-m}g(x)=r(x)-\frac{c_t}{b_m}x^{t-m}g(x)$$ This is the part where I got stuck, because I don't know how thus $\frac{c_t}{b_m}x^{t-m}g(x)$ was derived. Any help on understanding or deriving this is much appreciated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.