Elements of $S=R/2R$ where $R= \mathbb Z[\sqrt{2}]$?

Question

My approach so far is that for elements ${a + b\sqrt{2}}$ in $S$, $a$ and $b$ just can be $0$ or $1$ (because $a$ and $b$ must be in $\mathbb{2Z}$). So we will have 4 elements: $0,1,\sqrt{2}$ and $1+\sqrt{2}$. Is this right? Is there any proof that shows this better?

Answer 1

Yes, $\,\Bbb Z[\sqrt 2]\cong \Bbb Z[x]/f,\ f =x^2\!-\!2,\,$ so employing quotient reciprocity yields

$$\begin{align} \Bbb Z[\sqrt 2]/2 &\,\cong\, (\Bbb Z[x]/f)\,/\,(2,f)/f\\[.2em] &\,\cong\, \ \Bbb Z[x]/(2,f)\\[.2em] &\,\cong\, (\Bbb Z[x]/2)\,/\,(f,2)/2\\[.2em] &\,\cong\ \ \Bbb F_2[x]/f \\[.2em] &\,\cong \ \ \Bbb F_2[x]/x^2 \end{align}\qquad$$

This is the algebra of dual numbers over $\Bbb F_2,\,$ used as an algebraic model of tangent and jet spaces.