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Let $F$ be a field with nine elements. Suppose $i \in F$ such that $i^2+1=0$. What is $(i+1)^{-1}?$

As $i^2 = -1$ I get that $i=-i^{-1}$. So $i+1=-i^{-1}+1$, but it's not very helpful for as I just end up with $(-i^{-1}+1)^{-1}$. What can be done here?

Stef
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Walker
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  • Well, with a field with just nine elements you could brute force the solution. – lisyarus Aug 19 '22 at 05:15
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    Alternatively, this inverse must be in the subfield generated by $i$, so we can solve $(i+1)(a+bi)=1$ which leads to a linear system $a-b=1$ and $a+b=0$. The solution involves dividing by 2, which is possible because the field has $\operatorname{char}=3$. – lisyarus Aug 19 '22 at 05:22
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    Since $(i+1)^8=1$, that means $(i+1)^{-1}=(i+1)^7$, so just calculate $(i+1)^7$ by the binomial theorem or otherwise. I get $i-1$. Check: $(i+1)(i-1)=i^2-1^2=-1-1=-2=1$. – bof Aug 19 '22 at 07:06
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    In the field $F$ we have $3=0$. Therefore $(1+i)^2=1^2+2i+i^2=2i=-i$. Surely you know the inverse of $-i$ from elsewhere, and can take advantage. Comes to much the same thing as @bof's hint, but I like to look for tricks like these. – Jyrki Lahtonen Aug 19 '22 at 08:18
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    It works exactly the same way as in the reals: note that $(a+ib)(a-ib)=a^2+b^2$, so the inverse of $a+ib$ is $\frac{a-ib}{a^2+b^2}$. In your case, $a=b=1$, and the denominator can be removed by using $2\cdot2=1$ in fields of characteristic $3$. – Vercassivelaunos Aug 19 '22 at 08:33
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    As in the linked dupes: we can invert algebraic numbers by rationalizing the denominator

    $$\dfrac{1}{1+i} = \dfrac{1}{1+i},\dfrac{1-i}{1-i} = \dfrac{1-i}{\color{#c00}2} = i-1\ \ {\rm by}\ \ \color{#c00}2 = -1$$

     – Bill Dubuque Aug 19 '22 at 08:35

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