Question

1. Prove that $\mathbf{R}\mathbf{=}\left\{\left.\mathbf{a}+\mathbf{bi}\right|\mathbf{a}\mathbf{,}\mathbf{b}\in \mathbb{Z}\right\}$is a subring of $\mathit{ℂ}$and that $\mathbf{M}\mathbf{=}\left\{\left.\mathbf{a}+\mathbf{bi}\right|\left.3\right|\mathbf{a}\mathbf{and}\left.3\right|\mathbf{b}\right\}$ is a maximal ideal in R. [Hint: If $\mathbf{r}+\mathbf{si}\notin \mathbf{M}$, then $\mathbf{3}\cancel{\mathbf{|}}\mathbf{r}$or . Show that 3 does not divide ${\mathbf{r}}^2+{\mathbf{s}}^2=\left(r+\mathrm{si}\right)\left(\mathbf{r}-\mathbf{si}\right)$. Then show that any ideal containing $\mathbf{r}+\mathbf{si}$ and M also contains 1.]
2. Show that $\mathbf{R}/\mathbf{M}$is a field with nine elements.

Short answer

It is proved that$\mathbf{R}/\mathbf{M}$ is a field with nine elements.

Step-by-step solution

Statement of theorem 6.15

Consider that F is a field and $f\left(x\right),g\left(x\right)\in F\left[x\right]$with $g\left(x\right)\ne 0$ . Then, there is a unique polynomial q(x) and r(x) so that $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ and either-or$\mathrm{deg}r\left(x\right)<\mathrm{deg}g\left(x\right)$ .

Show that  R/M is a field with nine elements

Here, R/M would be a field according to a preceding item and theorem 6.15. With an additive abelian group, R/M would be isomorphic to $\left(\mathbb{Z}\times \mathbb{Z}\right)/\left(\left(3\right)\times \left(3\right)\right)\cong {\mathbb{Z}}_3\times {\mathbb{Z}}_3$that has 9 elements.

Hence, it is proved that R/M is a field with nine elements.