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When working in the field $\mathbb{Q}[X] / \langle X^3 + 3X + 3 \rangle$, let $a$ represent the image of $X$ under the natural quotient mapping.

I am trying to understand the range of strategies that I might use to represent the element $\frac{1}{a^2 +1}$ in the form $k_2 a^2 + k_1 a + k_0$ with $k_i \in \mathbb{Q}$. I have already determined the following:

  1. $\frac{1}{a}$ is $-\frac{1}{3} a^2 - 1$, an answer which I found by rearranging the equation $a^3 + 3a + 3 = 0$.
  2. $\frac{1}{a+1}$ is $a^2 - a + 4$, which was found by dividing $a^3 + 3a + 4$ by $a+1$.

In order to find $\frac{1}{a^2 + 1}$, I am attempting to use the same method. I started by rearranging $a^3 + 3a + 3 = 0$ to form the equation $\frac{1}{a^2 + 1} = \frac {a}{-2a-3}$ and proceeding with more division.

I have checked my work with Sage. The correct representation is $\frac{2}{13}a^2 - \frac{3}{13}a + \frac{4}{13}$, but the correct path forward is eluding me at the moment. Any possible strategy hints would be welcome. Thank you so much for your time!

Bill Dubuque
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    I would apply the Euclidean algorithm to the polynomials $x^2+1$ and $x^3+3x+3$. – Ted Shifrin Nov 30 '23 at 00:12
  • See here in the dupe for how to very simply compute the inverse via the (forward) extended Euclidean algorithm. – Bill Dubuque Nov 30 '23 at 00:17
  • Or as here we can use inverse reciprocity, to reduce to computing an inverse in $\Bbb Q[x]/(x^2+1)\cong\Bbb Q[i]$ by $\rm\color{#0a0}{rationalizing}$ denominators. Write the Bezout equation as $\tag{}$ $\qquad\qquad 1 = ag + bf = a(x^3+3x+3) + b(x^2+1)$ $\tag{}$ $!\bmod, f!=!\color{#c00}{x^2}+1!:,\ a\equiv \dfrac{1}{g}\equiv \dfrac{1}{3+3x+x\color{#c00}{x^2}}\equiv \dfrac{1}{3+2x}\equiv \dfrac{3-2x}{\color{#0a0}{13}},\ $ so $\tag*{}$ $!\bmod g!:\ \dfrac{1}f \equiv b\equiv \dfrac{1-ag}f\equiv \dfrac{2x^2-3x+4}{13}\ \ $ – Bill Dubuque Nov 30 '23 at 01:08
  • See also quotient reciprocity $\ \ $ – Bill Dubuque Nov 30 '23 at 01:09
  • We can continue your inverse calculation from $,\color{#c00}{b\equiv x/(-2x-3)},$ as below

    $$\begin{align} [![1]!]\quad\ \ \ (x^2+1), b&,\equiv, 1\ [![2]!]\quad \color{#c00}{{-}(2x+3),b} &,\equiv\ \color{#c00}x\ (2x!-!3):![![2]!]+4!:[![1]!]\qquad\qquad\ 13,b &,\equiv, (2x!-!3)x+4 \end{align}\quad$$

    It's the forward extended Euclidean algorithm presented in the first dupe, but we omitted one column.

     – Bill Dubuque Nov 30 '23 at 04:45
  • The prior calculation can be expressed naturally in fractional form, e.g. see the 2nd section here, which is a polynomial analog of the integer case. $\ \ $ – Bill Dubuque Nov 30 '23 at 08:02

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