# Thread: Inverses over a field F

1. ## Inverses over a field F

I'm given the field $Z_2[t]/(t^3 + t + 1)$ 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 $t^2 + 1$. 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 $Z_p[x]$.

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

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

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

Is this correct?

2. Originally Posted by jamix
I'm given the field $Z_2[t]/(t^3 + t + 1)$ 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 $t^2 + 1$. 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 $Z_p[x]$.

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

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

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

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

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

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