Question

Show that $\mathbf{\mathbb{Q}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{/}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{)}$is a field.

Short answer

It is proved that$\mathbb{Q}\left[x\right]/(x^2-2)$ is a field.

Step-by-step solution

General concept 

Observe that, $\mathbf{\mathbb{Q}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{/}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{)}$ is the set of all congruence classes modulo $\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{)}$. By the division algorithm, each equivalence class in the set$\mathbb{Q}\left[x\right]/(x^2-2)$ is expressed as a rational linear polynomial function. That is, all classes in the set$\mathbb{Q}\left[x\right]/(x^2-2)$ are of the form role="math" localid="1654238412089" $[ax+b]$.

Show that ℚ[x]/(x2−2) is a field

Consider a field $F$and a non-constant polynomial $p\left(x\right)$in $F\left[x\right]$.

If $g\left(x\right)\in F\left[x\right]$and $g\left(x\right)$is relatively prime to $p\left(x\right)$, then $\left[g\left(x\right)\right]$is a unit in $F\left[x\right]/p\left(x\right)$.

Find the roots of the polynomial $x^2-2$.

$$\begin{gathered} x^2-2=0 \\ x=\pm \sqrt{2} \end{gathered}$$

The root $\pm \sqrt{2}$ is irrational. That is $\pm \sqrt{2}\notin Q$. This means that the polynomial is irreducible over the field $Q$.

This implies that all non-zero rational linear polynomials of the form role="math" localid="1654238948336" $ax+b$are relatively prime to $x^2-2$. Therefore, all classes of the form role="math" localid="1654238930563" $[ax+b]$are units in $\mathbb{Q}\left[x\right]/(x^2-2)$by the above mentioned result.

Therefore,$\mathbb{Q}\left[x\right]/(x^2-2)$ is a field.