Here is my work, I am self learning basic algebra. Firstly, by definition: $Z[i]/(5)= \{a+bi + (5): a,b \in \mathbb{Z}\}$ where (5) is the principal ideal generated by 5. Let I=(5). Now we claim the coset $2+i+I$ has no inverse. Here's my justification. If $2+i+ I$ had an inverse, then there would be $a+bi$ such that $(2+i)I (a+bi) I= 1 + I$, in other words we would have $(2+i)(a+bi) -1 \in I$ or equivalently $(2+i)(a+bi)-1= 5(c+di)$ for some $c,d \in \mathbb{Z}$. Then $2a-b + (2b+a) i -1= 5c+5di$ and so matching real and imaginary parts, we have $2a-b-1= 5c$ and $2b+a=5d$ and so $b= 2a-5c-1$ which we subtitute into $2b+a=5d$ to get $2(2a-5c-1)+a= 5d$ which implies $5(a-d-2c)=2$. But 5 does not divide 2, contradiction.
Is this correct or have I overlooked something? Thanks.
$\mathbb Z[i]/(5) \cong \mathbb Z[x]/(x^2+1,5) \cong \mathbb Z_5[x]/(x^2+1) \cong \mathbb Z_5[x]/(x+2)(x+3) \cong \mathbb Z_5[x]/(x+2) \times \mathbb Z_5[x]/(x+3) \cong \mathbb Z_5 \times \mathbb Z_5 $