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I don't understand why $(1-i)$ is in the set of element of $R_2$ in the solution.

My understanding is that $\mathbb{Z}/2\mathbb{Z} = \{0,1\}$, endowed with addition and multiplication mod 2. So wouldn't the elements of $R_2$ be $\{0,1,i,1+i\}$? And shouldn't the multiplication table of $R_2$ be something like this?

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José Carlos Santos
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ensbana
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    Use your table, but also take into account that $i^2=-1$ and $2i=0$. – GEdgar May 12 '18 at 10:27
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    $\Bbb Z/2\Bbb Z$ is the quotient of $\Bbb Z$ under the equivalence $a\sim b$ iff $a\equiv b$ (mod $2$). Its elements are equivalence classes of this relation: $[0]$ the set of even numbers and $[1]$ the set of odd numbers. Then $[-1]=[1]$ as $-1\sim 1$. Writing lots of brackets is tiresome, so we abuse notation by writing $1$ for $[1]$ and $-1$ for $[-1]$ etc., so that $-1=1$ insider $R_2$. – Angina Seng May 12 '18 at 10:31
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    The $p=2$ case is answered in MSE question 2777107 "Addition and multiplication in F_4". – Somos May 12 '18 at 11:21

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Yes, you are right, the elements of $R_2$ are $0$, $1$, $i$, and $1+i$. And since, in $\mathbb{Z}/2\mathbb{Z}$, $1=-1$, $1+i=1-i$ there.

José Carlos Santos
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  • @ensbana $7+2i=7-2i\iff2i=-2i\iff4i=0$, which is false. – José Carlos Santos May 17 '18 at 16:07
  • My last question was actually not very well though-out. Sorry! What I meant to ask is that whether there is a general way to do this sort of manipulation? For example, I'm considering $\mathbb{Z}/53\mathbb{Z}$, and want to know whether $7-2i$ is a member of the field, and if it is, how I can rewrite it in the standard form as defined in the definition of $\mathbb{Z}/53\mathbb{Z}$. – ensbana May 17 '18 at 16:11
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    @ensbana In this case, it is not a field, since $(23+i)(23-i)=23^2+1=530=0$. – José Carlos Santos May 17 '18 at 16:24
  • Thanks. But the hint for this exercise is $53 = 7^2 + 2^2$, so I thought I might be able to use an idea similar to the previous exercise, i.e. seeing whether $(7-ki)(7+ki) = 0$ has a positive integer solution k. But this require knowing if $7 - ki$ is a member of the field, given that $7 + ki$ is one. – ensbana May 17 '18 at 16:30