Question

Let I be a nonzero ideal in $\mathbf{\mathbb{Z}}\mathbf{\left[}\mathbf{i}\mathbf{\right]}$. Show that the quotient ring$\mathit{\mathbb{Z}}\mathbf{\left[}\mathbf{i}\mathbf{\right]}\mathbf{/}\mathit{I}$ is finite.

Short answer

It is shown that the quotient ring$\mathbb{Z}\left[i\right]/I$is finite.

Step-by-step solution

As given in the question

Considering I be a non-zero ideal in $\mathbb{Z}\left[i\right]$.

Taking any arbitrary $a+bi\in \mathbb{Z}\left[i\right]$, and $I=\left(r+si\right)$.

Proving that the quotient ring ℤ[i]/I is finite

Since$I=\left(r+si\right)$ and Iis a non-zero ideal in localid="1659541034135" $\mathbb{Z}\left[i\right]$, for any localid="1659541046699" $a+bi\in \mathbb{Z}\left[i\right]$, by division algorithm, we can write $a+bi=\left(r+si\right)\left(p+qi\right)+\left(m+in\right)$.

Since $\left(r+si\right)=0$ in I and $\delta \left(m+ni\right)<\delta \left(r+si\right),a+bi\cong m+ni.$

In localid="1659541068576" $\mathbb{Z}\left[i\right]/I,\delta \left(a+bi\right)=\delta \left(m+ni\right)<\delta \left(r+si\right)$.

Therefore, for every element, localid="1659541084182" $x+iy\in \mathrm{\mathbb{Z}}\left[i\right]/I,\delta \left(x+iy\right)<\delta \left(r+si\right)$.

That is,$x^2+y^2<r^2+s^2$ implying that there are only finitely many values of x and y, which satisfy the requirement.

Hence, it is proved that the quotient ringlocalid="1659541100869" $\mathrm{\mathbb{Z}}\left[i\right]/I$is finite.