Using the Euclidean algorithm in a field extension to find an inverse

Question

In a course on Galois theory which I'm taking, the following example problem has come up.

Let $\frac{\mathbb{Q}[x]}{\mathbb{Q}[x]f} = \mathbb{Q}(t)$ be a field where $f = t^3 -2 = 0$. Find the inverse of $t^2 + 1$ in $\mathbb{Q}(t)$.

The solution given to this example problem starts by saying:

"The Euclidean algorithm gives $5 = (x^2+1)(-x^2+2x+1)+(x^3-2)(x-2)$"

I understand the structure of the field, and I understand the problem. I thought I had an understanding of how the Euclidean algorithm works, but I don't understand how we have arrived at this answer.

I can see that if $g = t^2 + 1 \in \mathbb{Q}(t)$, then $g$ and $f$ are coprime, so we have $1 = rg + sf$ for some $r,s$. That's about as far as I got.

EDIT:

To be clear, I'm not asking why this is the answer. I'm asking how we got there in the first place, via the Euclidean algorithm. How does the Euclidean algorithm give the above?

Answer 1

If $$rg+sf=1$$ in $\mathbb{Q}\left(x\right)$, then $$\overline{r}\cdot\overline{g}+\overline{s}\cdot\overline{f}=\overline{1}$$ in $\mathbb{Q}\left(x\right)/f\mathbb{Q}\left(x\right)$, where $\overline{h}=h+f\mathbb{Q}\left(x\right)$. Since $\overline{f}=\overline{0}$, we have $$\overline{r}\cdot\overline{g}=\overline{1}$$ which means that $\overline{r}$ is the inverse of $\overline{g}$, as required.

Answer 2

Here is the calculation, maybe I will typeset the good part later, the fractions $p_j/q_j$

$$ \begin{array}{cccccccccccc} & & x & & -x+2 & & (-x-2)/5 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{x}{1} & & \frac{-x^2 + 2x + 1}{-x+2} & & \frac{\frac{x^3 - 2}{5}}{\frac{x^2 +1}{5}} \end{array} $$ This is the continued fractions style of writing that I like, easy to remember. The litle two by two cross product of consecutive "fractions" is $\pm 1$ These are $$ 0 \cdot 0 - 1 \cdot 1 = -1 $$ $$ 1 \cdot 1 - x \cdot 0 = 1 $$ $$ x \cdot (-x+2) - (-x^2+2x+1) \cdot -1 = 1 $$ $$ (-x^2+2x+1) \cdot \left(\frac{x^2 +1}{5}\right) - \left(\frac{x^3 - 2}{5}\right) \cdot (-x+2) = -1 $$ Mukltiply the last one through by $5$ and you have the identity you asked about.

parisize = 4000000, primelimit = 500509
? f0 = x^3 - 2
%1 = x^3 - 2
? f1 = x^2 + 1
%2 = x^2 + 1
? 
? d2 = x
%3 = x
? f2 = f0 - d2 * f1
%4 = -x - 2
? 
? d3 = -x + 2
%5 = -x + 2
? f3 = f1 - d3 * f2
%6 = 5
? 
? d4 = f2 / f3 
%7 = -1/5*x - 2/5
? f4 = f2 - d4 * f3
%8 = 0
? 
? 
? p0 = 0
%9 = 0
? q0 = 1
%10 = 1
? 
? p1 = 1
%11 = 1
? q1 = 0
%12 = 0
? 
? p2 = d2 * p1 + p0
%13 = x
? q2 = d2 * q1 + q0
%14 = 1
? 
? p3 = d3 * p2 + p1
%15 = -x^2 + 2*x + 1
? q3 = d3 * q2 + q1
%16 = -x + 2
? 
? p4 = d4 * p3 + p2
%17 = 1/5*x^3 - 2/5
? q4 = d4 * q3 + q2
%18 = 1/5*x^2 + 1/5
? 
?