Question

Show that $\mathbf{ℝ}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{/}\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}$is a field by verifying that every nonzero congruence

class [ax + b] is a unit. [Hint: Show that the inverse of [ax + b] is [cx + d], where c = a/(a² + b²) and d = b/(a² + b²).]

Short answer

$\left[a_0+a_{1}x\right]\in \mathrm{ℝ}\left[x\right]/(x^2+1)$verifying that$\left[\frac{a_0}{{a_0}^2+{a_1}^2}-\frac{a_1}{{a_0}^2+{a_1}^2}x\right]$ is an inverse applying this formula.

Step-by-step solution

Step 1:

Consider $\left[a_0+a_{1}x\right]\in \mathrm{ℝ}\left[x\right]/(x^2+1)$, verifying that $\left[\frac{a_0}{{a_0}^2+{a_1}^2}-\frac{a_1}{{a_0}^2+{a_1}^2}x\right]$is an inverse applying this formula.

$$\begin{gathered} \left[a_0+a_{1}x\right]·\left[\frac{a_0}{{a_0}^2+{a_1}^2}+\frac{-a_1}{{a_0}^2+{a_1}^2}x\right]=\left[\frac{{a_0}^2+{a_1}^2}{{a_0}^2+{a_1}^2}+\frac{-a_0{a}_1+a_0{a}_1}{{a_0}^2+{a_1}^2}x\right] \\ =\left[1\right] \end{gathered}$$.