Factoring isomorphism

Question

I have $\mathbb Z[i\sqrt2] = ${$a+bi\sqrt2; a,b \in \mathbb Z, i^2=-1$} and $I =\{a+bi: a,b \in \mathbb Z, i^2=-1, 11\mid a+3b\}$. My task was to prove that $I$ is an ideal in $\mathbb Z[i\sqrt2]$ by constructing a homomorphism $\phi:\mathbb Z[i\sqrt2] \to K$ to some ring $K$ such that $\ker\phi = I$. In my homomorphism, $K = \mathbb Z_{11}$, and $\phi(a+bi\sqrt2) = [a+3b]_{11}$. $\phi$ is also surjective, so $\mathbb Z_{11} $ is isomorphic to the factorring $\mathbb Z[i\sqrt2] /I$, and now I am curious about this factorring. How do I build this isomorphism? From which elements do $\mathbb Z[i\sqrt2] /I$ consist?

Answer 1

$$\phi:a+bi\sqrt{2}\in\mathbb{Z}\left[i\sqrt{2}\right]\longmapsto\overline{a+3b}\in\mathbb{Z}/11\mathbb{Z} $$ is a morphism : for any $a+bi\sqrt{2},u+vi\sqrt{2}\in\mathbb{Z}\left[i\sqrt{2}\right]$ with $a,b,u,v\in\mathbb{Z}$ , one has$$\phi\left(\left(a+bi\sqrt{2}\right)\left(u+vi\sqrt{2}\right)\right)=\phi\left(\left(au-2bv\right)+\left(av+bu\right)i\sqrt{2}\right)=\overline{au-2bv+3av+3bu}$$ and$$\phi\left(a+bi\sqrt{2}\right)\phi\left(u+vi\sqrt{2}\right)=\left(\overline{a+3b}\right)\left(\overline{u+3v}\right)=\overline{au+3av+3bu+9bv}$$ and both are equal since $-2\equiv9\left[11\right]$ . The additivity is clear and $\phi\left(1\right)=\bar{1}$ is ok. Now, for any $0\leq k\leq10$ , one has $$\phi\left(k+0i\sqrt{2}\right)=\bar{k}$$ whence the surjectivity. Finally, $$a+bi\sqrt{2}\in\ker\phi\Longleftrightarrow\overline{a+3b}=\bar{0}\Longleftrightarrow a+3b\equiv0\left[11\right]\Longleftrightarrow11\mid a+3b\Longleftrightarrow a+bi\sqrt{2}\in I.$$ By the factorization theorem, one has$$\mathbb{Z}\left[i\sqrt{2}\right]/I\simeq\mathbb{Z}/11\mathbb{Z}$$ and since $\mathbb{Z}/11\mathbb{Z}$ is a field (because $11$ is a prime number), $\mathbb{Z}\left[i\sqrt{2}\right]/I$ is also a field, so $I$ is a maximal ideal of $\mathbb{Z}\left[i\sqrt{2}\right]$ . One has$$I=\left\{ \left(u+vi\sqrt{2}\right)+\left(a+bi\sqrt{2}\right);11\mid a+3b\right\} .$$