Prove that a homomorphism is surjective

Question

Let $R$ be the Gaussian integer and $J = (2 + 3i)Z[i]$ a ideal of R. Let $\phi : \mathbb{Z} \rightarrow R/J$ be a homomorphism defined by $\phi(n) = n + J$ for all $n \in \mathbb{Z}$.

I am having difficult to prove that $\phi$ is surjective.

Answer 1

Take $a+bi+J\in\mathcal R/J$. Then $\phi (a-5b)=a-5b+J=a+bi+J$.

This is because $a-5b-(a+bi)=(b-bi)(2+3i)$.

Thus $\phi$ is surjective.

Answer 2

Let $p:\mathbb{Z}[i]\rightarrow\mathbb{Z}[i]/(2+3i)$ be the quotient map; $(2+3i)(1-i)=5+i$ implies that $\phi(5-(2+3i)(1-i))=-\phi(5)=p(i)$, since $\phi(1)=p(1)$, we deduce that $\phi$ is surjective.