Adapting the Chinese Remainder Theorem (CRT) for integers to polynomials

Question

I did a few examples using the CRT to solve congruences where everything was in terms of integers. I'm trying to use the same technique for polynomials over $\mathbb{Q}$, but I'm getting stuck.

---

Here's an example with integers:

$\begin{cases}x \equiv 1 \, (\mathrm{mod} \, 5) \\ x \equiv 2 \, (\mathrm{mod} \, 7) \\ x \equiv 3 \, (\mathrm{mod} \, 9) \\ x \equiv 4 \, (\mathrm{mod} \, 11). \end{cases}$

Since all the moduli are pairwise relatively prime, we can use the CRT. Here's some notation I'm using:

$\bullet \, M$ denotes the product of the moduli (in this case, $M = 5 \cdot7 \cdot 9 \cdot 11$)

$\bullet \, m_i $ denotes the modulus in the $i^{\mathrm{th}}$ congruence

$\bullet \, M_i$ denotes $\dfrac{M}{m_i}$

$\bullet \, y_i$ denotes the inverse of $M_i$ (mod $m_i$), i.e. $y_i$ satisfies $y_i M_i \equiv 1$ (mod $m_i$).

Then $x = \displaystyle \sum_{i = 1}^n a_iM_iy_i$, and this solution is unique (mod $M$).

---

Now I want to apply the same technique to the following:

$\begin{cases} f(x) \equiv 1 \, (\mathrm{ mod } \, x^2 + 1) \\ f(x) \equiv x \, (\mathrm{mod} \, x^4), \end{cases}$

where $f(x) \in \mathbb{Q}(x)$. Having checked that the moduli are relatively prime, we should be able to use the CRT. Using the notation above, I have the following:

$M = (x^4)(x^2 + 1)$

$M_1 = x^4$

$M_2 = x^2 + 1$

Here's where I run into a problem. I need to find $y_1, y_2$ such that

$\begin{cases} y_1 (x^4) \equiv 1 \, (\mathrm{mod} \, x^2 + 1) \\ y_2 (x^2+1) \equiv 1 \, (\mathrm{mod} \, x^4). \end{cases}$

But how does one find $y_1, y_2$?

Answer 1

To find $y_1$ and $y_2$ consider solving the problem $$y_1x^4+y_2(x^2+1)=1.$$ This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares $$(x^2-1)(x^2+1)=x^4-1,$$ hence $$x^4+[(-1)(x^2-1)](x^2+1)=1.$$ This tells us that we can choose $$y_1=1,$$ $$y_2=(1-x^2).$$

Answer 2

Bu applying $\ ab\bmod ac\, =\, a(b\bmod c)\ $ [Mod Distributive Law] $ $ it is a bit simpler:

$ f-x\,\bmod\, {x^{\large 4}(x^{\large 2}\!+\!1)}\, =\, x^{\large 4}\underbrace{{\left[\dfrac{\color{#c00}f-x}{\color{#0a0}{x^{\large 4}}}\bmod {x^{\large 2}\!+\!1}\right]}}_{\large \color{#0a0}{x^{\Large 4}} \ \equiv\ 1\ \ \ {\rm by}\ \ \ x^{\Large 2}\ \equiv\ \ -1 } =\, x^{\large 4}[1-x],\ $ by $\,\color{#c00}f\equiv 1\pmod{\!x^{\large 2}\!+\!1}$

Remark $ $ Here are further examples done using MDL (an operational form of CRT).

You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).