Question

Does the ring Z7[x]/<x^2+3> have nilpotent elements? justify.

Answer 1

Using the necessary condition for nilpotent elements - they cannot be invertible, we fix the element of the ring ax+b and try to solve this equation:

$(ax+b)(cx+d)=1\\ acx^2+(ad+bc)x+bd=1\\ (ad+bc)x+(bd-3ac)=1\\ \begin{cases} ad+bc=0\\ bd-3ac=1 \end{cases}\\ Because\ \mathbb{Z}_7 \ is \ a \ field\ from \ first\ equation:\\ c=-\frac{ad}{b}\\ Put\ in\ second:\\ bd+\frac{3a^2d}{b}=1$

It is easy to see that this equation is solvable for b not equal to 0

If b=0:

$ax(cx+d)=1\\ adx-3ac=1$

And this equation is easily solvable for a not equal to 0, which means that there are no nilpotent elements in the ring