Prove that (3) is a maximal ideal in $\mathbb{Z}[i]$. [duplicate]

Question is to verify that $Z[i]/(3)$ is a field. If it is identify it's elements. I verify it follows. Since Z[i] is PID, and 3 is irreducible , so 3 becomes prime hence (3) is prime ideal ,so it is maximal ideal hence given quotient ring is field. I know number of elements in it by formula which will give 9. But how can identify elements?

Solution 1:

You can use the (second and third) isomorphism theorems to conclude $$\mathbb{Z}[i]/(3) := (\mathbb{Z}[x]/(x^2+1))/(3) \cong \mathbb{Z}[x]/(3,x^2+1) \cong (\mathbb{Z}/3\mathbb{Z} [x])/(x^2+1) =: (\mathbb{Z}/3\mathbb{Z})[i].$$ Its elements can thus be thought of as the degree-$1$ polynomials in $i$ over $\mathbb{Z}/3\mathbb{Z}$, i.e., the elements of the form $a+bi$ with $a,b\in \mathbb{Z}/3\mathbb{Z}$, with addition and multiplication defined in the obvious way.