Show that $\Bbb{Z}[X] /\langle2,X^2+1\rangle $ is a ring with 4 elements but it is not isomorphic with $\Bbb{Z}_2 \times \Bbb{Z}_2$

Question

I have tried to rewrite the generators $\langle 2 ,x^2 +1\rangle = \langle 2,x^2+1,x^2-1\rangle = \langle2,x^2+1,(x+1)(x-1)\rangle = \langle2 , x^2+x,x+1,x-1\rangle $ etc.

But this was to no avail. Can someone help me determine the elements of the ring?

Answer 1

Doing the computation with quotient rings (rather than with the ideal) is often easier: ${\Bbb Z}[X]/(2,X^2+1) \cong {\Bbb F}_2[X]/(X^2+1) = {\Bbb F}_2[X]/((X+1)^2) \cong {\Bbb F_2}[X]/(X^2)$. Writing $\epsilon$ for the residue class of $X$, this ring has elements $0, 1, \epsilon, \epsilon + 1$ and it has $\epsilon^2 = 0$.