Remainder Theorem in complex numbers. 

Is the Remainder Theorem works in complex numbers? I have read many articles but they just say linear function , not include whether complex numbers are allowed. Can someone check my understanding please. Thank you!


 A: CRT can be formulated as a ring isomorphism.
Let $I_1, ..., I_n$ be two-sided ideals of a commutative ring $R$ and let $I$ be their intersection. If the ideals are pairwise coprime, we have the following isomorphism:
$R/I\rightarrow R/I_{1}\times \cdots \times R/I_{n}: 
x \bmod I\mapsto (x\bmod I_{1},\,\ldots ,\,\bmod I_{n})$
CRT comes in when you want to find a preimage of an $n$-tuple $(b_1\bmod I_1,\ldots,b_n\bmod I_n)$, i.e., find $x\in R$ such that
$x\equiv b_j\bmod I_j$ for $i=1,\ldots,n$.
The solution $x$ is uniquely determined modulo $I$.
A: I assume you refer to the following theorem.
Let $P$ be a polynomial in $x$ with coefficients in $\mathbb{C}$, and suppose $a \in \mathbb{C}$. Then the remainder when dividing $P$ by $x - a$ is equal to $P(a)$.
Let's walk through the proof.
After we finish dividing $P$ by $x - a$, we will have a quotient polynomial (call it $Q$) and a numerical remainder (call it $r$). These satisfy the property that $P = (x - a) Q + r$ by the definition of being quotient and remainder. Both sides in this equation are a polynomials in $x$, so substitute in $a$ for $x$ to get $P(a) = (a - a) Q(a) + r = 0 \cdot Q(a) + r = 0 + r = r$. This is the very thing we sought to prove.
Notice that many of the "interesting" properties of $\mathbb{C}$ are totally irrelevant here. The only ones that mattered were the basic properties of addition, subtraction, and multiplication. Thus, the remainder theorem is extremely general and holds for other "rings" - sets where addition, subtraction, and multiplication are nicely defined.
