Question:
After proving that it exists, find $f(x)\in\Bbb Q[x]$ of degree at most 2 such that $$f(x)\equiv2 \text{ mod }x+1\ \text{and }\ f(x)\equiv x+1\text{ mod }x^2+1$$
My attempt:
Since $(x+1)$ and $(x^2+1)$ are coprime ideals of $\Bbb Q[x]$ by the Chinese reminder theorem $$\Bbb Q[x]/(x+1)\times\Bbb Q[x]/(x^2+1)\cong\Bbb Q[x]/(x^3+x^2+x+1)$$ and there is a unique $f(x)\in \Bbb Q[x]/(x^3+x^2+x+1)$ that satisfies the property.
To find it let's first ovserve that $${1\over2}(x^2+1)+[-{1\over2}(x-1)](x+1)=1$$
By a theorem the polynomial $f(x)$ should be
$-{1\over2}(x-1)(x+1)^2+{1\over2}(x^2+1)\cdot2\\=-{1\over2}(x^3+x^2-x-1)+x^2+1\text{ mod } (x^3+x^2+x+1)\\=-{1\over2}(-2x-2)+x^2+1\\=x^2-x$
And I find that $x^2-x\equiv 2$ mod $x+1$ but $x^2-x\equiv -x-1$ mod $x^2+1$ instead of $x^2-x\equiv x+1$ mod $x^2+1$ Or is that ok?
$(x+1)(x-2)+2=x^2-x\ $ $\ (x^2+1)-x-1=x^2-x$
There exists a ring isomorphism
$$f:\Bbb Q[x]/(x+1)\times\Bbb Q[x]/(x^2+1)\to\Bbb Q[x]/(x^3+x^2+x+1)$$
f(a,b)=au+bv where $u = f(1,0)$ and $v = f(0,1)$.
If $u=f(1,0)$, then $u \equiv 1 \pmod{x+1}$ and $u \equiv 0 \pmod{x^2+1}$
Note that \begin{align} x &\equiv -1 \pmod{x+1} \\ x^2 &\equiv -x \pmod{x+1} \\ &\equiv 1 \pmod{x+1} \\ x^2+1 &\equiv 2 \pmod{x+1} \\ \dfrac 12 (x^2+1) &\equiv 1 \pmod{x+1} \end{align}
So $u = \dfrac 12 (x^2+1)$
If $v=f(0,1)$, then $v \equiv 0 \pmod{x+1}$ and $v \equiv 1 \pmod{x^2+1}$
Note that \begin{align} x+1 &\equiv x+1 \pmod{x^2+1} \\ (x+1)^2 &\equiv x^2+2x+1 \pmod{x^2+1} \\ &\equiv 2x \pmod{x^2+1} \\ (x+1)^3 &\equiv 2x^2+2x \pmod{x^2+1} \\ &\equiv 2x^2+2x-2(x^2+1) \pmod{x^2+1} \\ &\equiv 2x-2 \pmod{x^2+1} \\ (x+1)^2 - (x+1)^3 &\equiv 2 \pmod{x^2+1} \\ \dfrac 12((x+1)^2 - (x+1)^3) &\equiv 1 \pmod{x^2+1} \\ -\dfrac 12(x + 2 x^2 + x^3) &\equiv 1 \pmod{x^2+1} \\ \end{align}
So $v = -\dfrac 12(x^3 + 2x^2 + x)$
\begin{align} f(2,x+1) &\equiv 2(\dfrac 12 (x^2+1)) + (x+1)(-\dfrac 12(x^3 + 2x^2 + x)) \pmod{x^3+x^2+x+1} \\ f(2,x+1) &\equiv -\dfrac 12 x^4 - \dfrac 32 x^3 - \dfrac 12 x^2 - \dfrac 12 x + 1 \pmod{x^3+x^2+x+1} \end{align}
Note that \begin{align} x^4 &\equiv x^4 \pmod{x^3+x^2+x+1} \\ x^4 &\equiv x^4 - x(x^3+x^2+x+1) \pmod{x^3+x^2+x+1} \\ x^4 &\equiv -x^3-x^2-x \pmod{x^3+x^2+x+1} \end{align}
\begin{align} f(2,x+1) &\equiv -\dfrac 12 x^4 - \dfrac 32 x^3 - \dfrac 12 x^2 - \dfrac 12 x + 1 \pmod{x^3+x^2+x+1} \\ &\equiv -\dfrac 12 (-x^3-x^2-x) - \dfrac 32 x^3 - \dfrac 12 x^2 - \dfrac 12 x + 1 \pmod{x^3+x^2+x+1} \\ &\equiv 1-x^3 \pmod{x^3+x^2+x+1} \end{align}
CHECK:
$1 - x^3 = (-x^2 + x - 1)(x + 1) + 2$
$1 - x^3 = -x(x^2 + 1) + x + 1$