# Thread: Inverses over a field F

1. ## Inverses over a field F

I'm given the field $\displaystyle Z_2[t]/(t^3 + t + 1)$ for which I need to calculate the inverse of each element.

I assume the way to do this is similar to the Extended Euclidean Algorithm and this is what I have for computing the inverse of $\displaystyle t^2 + 1$. I'm a little bit skeptical with my reasoning however and this is partially due to my lack of familiarity of working with polynomials over $\displaystyle Z_p[x]$.

1) First use the Division algorithm to obtain $\displaystyle t^3 + t + 1 = t*(t^2 + 1) + 1$

From the above, this gives us $\displaystyle -1*(t^3 + t + 1) + 1 = -t*(t^2 + 1)$.

On the other hand, as we are working over $\displaystyle Z_2[t]$, it follows that we must have $\displaystyle (t^3 + t + 1) + 1 = t*(t^2 + 1)$ since $\displaystyle -1 \equiv 1$. Hence the inverse of $\displaystyle t^2 + 1$ is just $\displaystyle t$.

Is this correct?

2. Originally Posted by jamix
I'm given the field $\displaystyle Z_2[t]/(t^3 + t + 1)$ for which I need to calculate the inverse of each element.

I assume the way to do this is similar to the Extended Euclidean Algorithm and this is what I have for computing the inverse of $\displaystyle t^2 + 1$. I'm a little bit skeptical with my reasoning however and this is partially due to my lack of familiarity of working with polynomials over $\displaystyle Z_p[x]$.

1) First use the Division algorithm to obtain $\displaystyle t^3 + t + 1 = t*(t^2 + 1) + 1$

From the above, this gives us $\displaystyle -1*(t^3 + t + 1) + 1 = -t*(t^2 + 1)$.

On the other hand, as we are working over $\displaystyle Z_2[t]$, it follows that we must have $\displaystyle (t^3 + t + 1) + 1 = t*(t^2 + 1)$ since $\displaystyle -1 \equiv 1$. Hence the inverse of $\displaystyle t^2 + 1$ is just $\displaystyle t$.

Is this correct?
there's a much neater way to solve this problem: let $\displaystyle F=\mathbb{Z}_2[t]/(t^3 + t + 1)$ and for simplicity let's show the image of $\displaystyle t$ in $\displaystyle F$ by $\displaystyle t.$ then $\displaystyle t^3=t+1$ and so $\displaystyle t^6=t^2+1 \neq 1.$ that means $\displaystyle t,$ as an

element of the multiplicative group $\displaystyle F^{\times}=F-\{0 \},$ has order $\displaystyle 7.$ thus $\displaystyle F^{\times}=<t>=\{t^k: \ 0 \leq t \leq 6 \}.$ clearly the inverse of $\displaystyle t^k$ is $\displaystyle t^{7-k}.$

Note: (some extra explanations!) i probably don't have to mention that since $\displaystyle t^3=t^2+1,$ we'll have $\displaystyle t^4=t^2+t, \ t^5=t^2+t+1, \ t^6=t^2+1$ and $\displaystyle t^7=1.$ so, for example, the inverse of $\displaystyle t^2+t=t^4$
is $\displaystyle t^{7-4}=t^3=t+1,$ etc. this method can be used in any finite field because we know that the multiplicative group of any finite finite field is cyclic.