Htydjtk

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 $y_1, y_2$?

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
$$y_1=1,$$
$$y_2=(1-x^2).$$

Bu applying $ abbmod ac, =, a(bbmod 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 $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$

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).