<s>[INST] Compute the characteristic of $R = Z[j]/(2-5j),$ where $j^3 = 1$ and $j^2 \ne 1.$ Compute the characteristic of $R = Z[j]/(2-5j),$ where $j^3 = 1$ and $j^2 \ne 1.$
Here i provide proof found in book. But i don't understand a lot from this proof.

$\overline x$ denotes an element in the quotient group involved.
Here we have $(2 − 5j)(2 − 5j^2) = 4 − 10(j + j^2) + 25j^3 = 4 + 10 + 25 = 39.$ Hence $39\cdot \overline 1 = \overline {39} = \overline{(2 − 5j) · (2 − 5j^2)} = \overline 0.$ Then the characteristic of $R$ is finite and divides
  $39.$ Therefore the characteristic of $R$ is $1, 3, 13$ or $39.$ Now
  let $c = \mbox{char}(R) > 0.$ Since $c \cdot 1_R$ lies in the ideal
  $(2 − 5j),$ then $c = (2 − 5j)(a + bj)$ for some $a, b, \in Z.$ Hence
  $|c|^2 = |2 − 5j|^2|a + bj|^2,$ so $$c^2 = 39(a^2 + b^2 − ab)$$ and
  therefore $39|c^2.$ The only value (among $1, 3, 13$ and $39$) for
  which it is possible is $c = 39.$ Thus $\mbox{char}(R) = 39.$

Why $j+j^2 = -1?$
Why $|2 − 5j|^2|a + bj|^2= 39(a^2 + b^2 − ab)?$
 [/INST] $j^3 = 1$ impies $0=j^3-1=(j-1)(j^2+j+1)$.
$j^2 \ne 1$ implies $j \ne 1$ and so $j^2+j+1=0$.
$|\cdot|^2$ here is just the complex number absolute value, $|z|^2 = z \bar z$, which implies $|zw| = |z| \, |w|$.
Using that $\bar j = j^2 = -1-j$, it is easy to compute $|a + bj|^2 = a^2 + b^2 − ab$.
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