Question

Find a splitting field of${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$ over ${\mathit{Z}}_{\mathbf{3}}$.

Short answer

$\frac{Z_3\left(x\right)}{(x^2+1)}$is the splitting field of $x^2+1$ with $\frac{Z_3\left(x\right)}{(x^2+1)}=\{0,1,x,x,x+1,x+2,2x,2x+1,2x+2\}$

Step-by-step solution

Definition of splitting field

A splitting field of a polynomial with coefficient in a smallest field over which the polynomial split or split into linear facots.

Find the splitting field

Let$Z_3=[0,1,2]$. Gives polynomial$x^2+1$has no zero in$Z_3$as

$0^2+1\ne 0$

And

$1^2+1\ne 0$

So by lemma gives. Let$F$be a field and let$f\left(x\right)\in F\left[x\right]$be a polynomial of degree 2 or 3. Then$f\left(x\right)$is irreducible in$F\left[x\right]$if and only if$f\left(x\right)$has no root in$F$.

Therefore given polynomial$x^2+1$is irreducible over$Z_3$. Thus the splitting field of$x^2+1$irreducible over$Z_3$is

$\frac{Z_3\left(x\right)}{(x^2+1)}=\{0,1,x,x,x+1,x+2,2x,2x+1,2x+2\}$