# Prove that $x^2 + 1$ divides $f(x) - \frac{(f(i) - f(-i))}{2i} x - \frac{(f(i) + f(-i))}{2}$ in the ring $\mathbb{Q}(i)[x]$.

Let $$\mathbb{Q}(i) = \{a + bi ~|~ a, b \in \mathbb{Q}\}$$. With addition and multiplication as in $$\mathbb{C}$$, $$\mathbb{Q}(i)$$ is a field.
(Note: You are not being asked to prove that $$\mathbb{Q}(i)$$ is a field.)
Let $$f(x) \in \mathbb{Q}(i)[x]$$. Prove that $$x^2 + 1$$ divides \begin{align*} f(x) - \frac{(f(i) - f(-i))}{2i} x - \frac{(f(i) + f(-i))}{2} \end{align*} in the ring $$\mathbb{Q}(i)[x]$$.

$$\textbf{My Attempt:}$$
Since, $$i, i^2, i^3, i^4 \in \mathbb{Q}(i)$$. Also, $$i^5 = i \in \mathbb{Q}(i)$$.
So, for some $$n \in \mathbb{N}$$, $$i^n \in \mathbb{Q}(i)$$.
Since, we let $$f(x) \in \mathbb{Q}(i)[x]$$.
Let $$f(x)$$ has degree of $$n$$ where $$n \in \mathbb{N} \cup \{ 0 \}$$.
Then, $$f(i) = a_n i^n + \cdots + a_0$$ and $$f(-i) = a_n (-i)^n + \cdots + a_0$$. for some $$a_j\in \mathbb{Q}(i)$$.
Since, $$\mathbb{Q}(i)$$ is a field with addition and multiplication as in $$\mathbb{C}$$.
Then, $$f(i) \in \mathbb{Q}(i)$$ and $$f(-i) \in \mathbb{Q}(i)$$.
Also, $$(f(i) - f(-i)) \in \mathbb{Q}(i)$$ and $$(f(i) + f(-i)) \in \mathbb{Q}(i)$$.
Since, notice that $$-\frac{1}{2i} = \frac{1}{2}i \in \mathbb{Q}(i)$$.
Then, $$-\frac{(f(i) - f(-i))}{2i} = (\frac{1}{2}i)(f(i) - f(-i)) \in \mathbb{Q}(i)$$ and $$-\frac{(f(i) + f(-i))}{2} = (-\frac{1}{2})(f(i) + f(-i)) \in \mathbb{Q}(i)$$.

Let $$\alpha = -\frac{(f(i) - f(-i))}{2i}$$ and $$\beta = -\frac{(f(i) + f(-i))}{2}$$. Where $$\alpha, \beta \in \mathbb{Q}(i)$$.
Then, $$f(x) - \frac{(f(i) - f(-i))}{2i}x - \frac{(f(i) + f(-i))}{2} = f(x) + \alpha x + \beta$$.
Since, we let $$f(x) \in \mathbb{Q}(i)[x]$$ and we set $$f(x)$$ has a degree of $$n$$.
So, $$f(x) = b_n x^n + \cdots + b_1 x + b_0$$ where $$b_j \in \mathbb{Q}(i)$$.
Then, $$f(x) + \alpha x + \beta = (b_n x^n + \cdots + b_1 x + b_0) + \alpha x + \beta = b_n x^n + \cdots + (b_1 + \alpha)x + (b_0 + \beta)$$.
Since, $$b_1, b_0, \alpha, \beta \in \mathbb{Q}(i)$$. So, $$(b_1 + \alpha), (b_1 + \beta) \in \mathbb{Q}(i)$$.
Then, we set $$g(x) = b_n x^n + \cdots + (b_1 + \alpha)x + (b_0 + \beta) = c_n x^n + \cdots + c_1 x + c_0$$.
$$~\hspace{10mm}$$ Which $$c_n = b_n$$, ..., $$c_1 = (b_1 + \alpha)$$, $$c_0 = (b_0 + \beta)$$. Where each $$c_j \in \mathbb{Q}(i)$$.
Then, $$g(x)$$ has degree of $$n$$ as well.
Also, $$g(x) = c_n x^n + \cdots + c_1 x + c_0 = f(x) - \frac{(f(i) - f(-i))}{2i}x - \frac{(f(i) + f(-i))}{2}$$.
$$~\hspace{10mm}$$ Where $$g(x)$$ is in the ring $$\mathbb{Q}(i)[x]$$

Next, I want to prove that $$x^2 + 1$$ divides g(x) in the ring $$\mathbb{Q}(i)[x]$$: Which I need to find a $$h(x) \in \mathbb{Q}(i)$$ such that $$g(x) = (x^2 + 1)h(x)$$.
But I don't know how to find it. And I don't think it always work since if $$g(x) = x + 1$$. Then, I cannot find a $$h(x)$$.

Your solution is very hard to read. Note that $$x+i$$ and $$x-i$$ are coprime and so for a polynomial to be divisible by $$(x+i)(x-i) = x^2+1$$ if suffices for it to have $$i$$ and $$-i$$ as roots.

A direct computation shows that

\begin{align*} f(i) - \frac{(f(i) - f(-i))}{2i} i - \frac{(f(i) + f(-i))}{2} = f(i)-f(i) = 0 \end{align*}

and

\begin{align*} f(-i) - \frac{(f(i) - f(-i))}{2i} (-i) - \frac{(f(i) + f(-i))}{2} = f(-i)-f(-i) = 0. \end{align*}

• These computations are abstracted in the general Lagrange interpolation (or CRT) formula - see my answer and its link. Jan 24, 2021 at 13:17

More conceptually, it's a special case of Lagrange interpolation [= a special case of CRT], viz.

\begin{align} &f(x) \,\equiv\ f(a)\, +\, (x-a)\ \dfrac{f(b)-f(a)}{b-a}\ \ \pmod{(x\!-\!a)(x\!-\!b)},\,\ a\neq b\\[.2em] \overset{\textstyle a,b=\pm i}\Longrightarrow\ &f(x) \,\equiv\, f(-i) + (x+i)\ \dfrac{f(i)-f(-i)}{2i}\!\pmod{x^2+1} \end{align}

Or, directly applying the CRT = Chinese Remainder Theorem formula

\begin{align} &\!\begin{array}{l} f(x)\,\equiv\, f(i)\ \ \ \ \pmod{x-i}\\ f(x)\,\equiv\, f(-i)\ \pmod{x+i} \end{array}\\[.4em] \iff\ & f(x) \,\equiv\, \dfrac{x+i}{2i}\, f(i) + \dfrac{x-i}{-2i}\, f(-i)\pmod{(x-i)(x+i)}\\[.6em] &\ \ \ \ \ \ \ \ \equiv\, \dfrac{f(i)+f(-i)}2 + \dfrac{f(i)-f(-i)}{2i}\,x\pmod{\!x^2+1} \end{align}\qquad\quad

See here for a conceptual derivation of the common CRT formula applied above.