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?
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2You could write down about 20 elements of $\Bbb Z[i]$, like $1 + 3i, 2-i, 0+2i, \ldots$, and then see which fall into the same equivalence classes mod $(3)$. Pretty soon you'll notice a pattern (I hope). – John Hughes Mar 26 '19 at 23:32
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You have to mention that $\mathbf Z[i]$ is a P.I.D. (actually it is Euclidean domain for the norm). Beware it is not true that rings of quadratic integers are all P.I.D.s, so that in general, irreducible does not necessarily imply prime. – Bernard Mar 26 '19 at 23:42
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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.
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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.
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