I am self studying algebra, and have come across this exercise:

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> Show that $\mathbb{Z}[i]/(2+3i)$ has 13 elements and is a field.
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My proof: Since Z[i] is a euclidian domain with euclidian norm function $f(c+id)= c^2+d^2$, for any element $a+bi \in Z[i]$, we can "divide" it by $2+3i$ to get $a+bi= q (2+3i) + r$ where either $r=0$ or $r=c+di$ for some integers $c,d$ not both $0$ with $c^2 + d^2=f(c+di) < f(2+3i)= 13$.
But then $a+bi$ belongs to the coset $ r + (2+3i) $. So every element in the quotient ring belongs to some $r + (2+3i)$ for $r=0$ or $r=c+di$ with $c^2 + d^2<13$. Now we can count all the possible pairs of integers satisfying this condition, ie $(0,0),(0,1),(0,2),(0,3), ..., (3,1)$ to get 13 and so there are 13 elements in the quotient ring. 

Is this a correct argument? Also is there an elegant way of showing it is a field rather than showing its an integral domain by showing each element has an inverse by brute force? Thanks in advance!