To identify field $Z[i]/(3)$

Question

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?

Answer 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.

Answer 2

Note that $2=-1$ is not a square in $\mathbb{F}_3$, the three-element field. Let $j$ be a root of $x^2+1$ and consider $\mathbb{F}_3[j]$. Since $j^2=-1$, you can define a surjective ring homomorphism $\mathbb{Z}[i]\to\mathbb{F}_3[j]$ by $\varphi(a+bi)=a+bj$.

Note that $3\in\ker\phi$; since $3$ is irreducible in $\mathbb{Z}[i]$, which is a Euclidean domain, we have $(3)=\ker\varphi$ and therefore $$ \mathbb{Z}[i]/(3)\cong\mathbb{F}_3[j] $$ Since the elements of $\mathbb{F}_3[j]$ are $$ \{0,1,2,j,1+j,2+j,2j,1+2j,2+2j\} $$ you get a complete description of $\mathbb{Z}[i]/(3)$ by taking one counter image for each element.