Can someone explain why $\displaystyle \mathbb{Z}[x]/(x^2+3,5) \cong \mathbb{F}_{25}$
$\displaystyle Z[x] \to Z_{5}[x] $ by f(x) $\displaystyle \in Z[x] = a_nx^n + ... + a_0 \to (a_n) mod 5 x^n +... + (a_0) mod 5 $. is a homomorphism.
Then $\displaystyle Z[x]/<x^2+3,5> \to Z_{5}[x]/<x^2+3> $ since x^2 + 3 is irreducible in $\displaystyle Z_{5}[x]$ (no zeros are there). $\displaystyle Z_{5}[x]/<x^2+3>$ is a field containing $\displaystyle 5^2$ elements.