Question

Show that $x^2+1$is irreducible in$\mathrm{\mathbb{Q}}\left[x\right]$ . [Hint: If not, it must factor as$\left(ax+b\right)\left(cx+d\right)$ with$\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{,}\mathit{d}\mathbf{\in }\mathbf{\mathbb{Q}}$ ; show that is impossible.]

Short answer

It is proved that $x^2+1$ is irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$.

Step-by-step solution

Irreducible element 

It is known that the degree of a polynomial is the sum of the product of two polynomials.

If $x^2+1$ is irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$,then all roots of $x^2+1$do not exist in$\mathrm{\mathbb{Q}}\left[x\right]$.

Steps for irreducible element

The roots of the polynomial $x^2+1$is:

$$\begin{gathered} x^2+1=0 \\ {x}^2=-1 \\ x=\sqrt{-1} \\ x=\pm i \end{gathered}$$

Where the roots are $\pm i\notin \mathrm{\mathbb{Q}}$.

Hence, $x^2+1$ is irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$.