I have a polynomial $m(x)= x^2 + x + 2$ that's irreducible over $F=\mathbb{Z}/3\mathbb{Z}$. I need to calculate the multiplicative inverse of the polynomial $2x+1$ in $F/(m(x))$.
I'd normally use division with two polynomials like this, but it doesn't work here. So how can I get the inverse?
On
Working in $(\mathbb{Z}/3\mathbb{Z})[[x]]/\langle x^2 + x + 1 \rangle$, one (awful) way to do it is to notice that $2x + 1 = 1 - x$, so $$(2x+1)^{-1} = \frac{1}{1-x} = 1 + x + x^2 + x^3 + \dotsb. $$ But $1 + x + x^2 = -1 = 2$ and, since $x^2 = -x - 2 = 1-x$, $$x^3 + \dotsb = \frac{x^3}{1-x} = \frac{x^3}{x^2} = x,$$ giving $(2x+1)^{-1} = 2 + x$.
You can just let the unknown inverse be $ax+b$ and then multiply $(ax+b)(2x+1)=2ax^2+(a+2b)x+b.$ Equate this to $1$ in $Z_3[x]/(m)$ which means using $x^2=-x-2$ that the term $2ax^2$ becomes $2a(-x-2)=-2ax-4,$ which is then added to $(a+2b)x+b.$ Try from there.
I'm pretty sure this method should work, though in a comment, the user Unit already gave the result.