Specifically, how would I find [a+bx]-1 in R[x]/(x 2 + 1)?
I'm assuming because they are they are irreducible and therefore relatively prime that it is possible, I just don't understand how to use Euclid's algorithm to get there.
My attempt was:
x2+1=($\frac{x}{a}$-$\frac{b}{a}$)*(ax+b) + 1 - $\frac{b^2}{a^2}$
but I don't know if this is the right approach or what to do with what the information here, because the solution should be ($\frac{a}{c}$-$\frac{b}{c}$) for c = $a^2 + b^2$ if a + bx != 0.
In this extension we have $x^2+1=0$, so $x^2=-1$. That means we can write polynomials as just $a+bx$ with no need for higher terms. Let's find the inverse to $(a+bx)$ by assuming there's an inverse we can write as $(c+dx)$. Multiply them together and use $x^2=-1$. We get $(a+bx)(c+dx)=ac+(bc+ad)x+bdx^2=(ac-bd)+(bc+ad)x$. Setting this equal to 1 gives us the system of equations $ac-bd=1$ and $bc+ad=0$ which we can solve and then get the solution you mentioned.
This field extension is (perhaps one should say 'is isomorphic to') the complex numbers where we relabel $x$ as $i$.