In a polynomial ring $F[x]$ where $F$ is a field let $f\in F[x] ,a\in F$. How can we show that $f\mod{\langle x-a\rangle }=f(a)$. Doing some simple examples I can grasp why this is true but I can't formally "prove" it.
Could some give me a hint how to show this ? or give me an explanation why this is true ?
Notice that $[x]=[x-a]+[a]=[a]$. Therefore, if we denote $f(x)=\sum\limits_{k=0}^n f_kx^k$, we have the desired identity as $$[f(x)] = \left[\sum_{k=0}^n f_k x^k\right]= \sum_{k=0}^n f_k [x]^k= \sum_{k=0}^n f_k [a]^k= \left[\sum_{k=0}^n f_k a^k\right]=[f(a)]$$ This is equivalent to the fact that $\exists g(x)\in F[x]$ such that $f(x)=(x-a)g(x)+f(a)$ since $a$ is a root of $f(x)-f(a)$.
