Question

Show that $\mathbf{[}\left(Q\sqrt{3},i\right)\mathbf{:}\mathbf{Q}\mathbf{]}\mathbf{=}\mathbf{4}$ .

Short answer

$[(\mathrm{Q}\sqrt{3},\mathrm{i}):\mathrm{Q}]=4$

Step-by-step solution

Definition of irreducible polynomial.

Irreducible polymial is a non constant polynomial that cannot be factored into the product of two non constant polynomials. The property of irreducibility depends on the field or the ring to which the coefficients are considered to belong.

Step 2:Showing that  [(Q3,i):Q]=4

Let the polynomial $x^2+1$over $Q\sqrt{3}$

$f\left(x\right)=x^2+1$

First find its zeros as follows

$$\begin{gathered} f\left(x\right)=0 \\ {x}^2+1=0 \\ {x}^2=-1 \\ x=\sqrt{i} \end{gathered}$$

As $\sqrt{i}\notin Q\sqrt{3}$

Therefore $f\left(x\right)$ has no root in $Q\sqrt{3}$. Hence $f\left(x\right)$ is irreducible over$Q\sqrt{3}$ .

Thus $x^2+1$ is the minimal polynomial of i over$Q\sqrt{3}$ . Hence

$\left[\left(Q\sqrt{3},i\right):Q\sqrt{3}\right]=2$

And

$$\begin{gathered} \left[\left(Q\sqrt{3},i\right):Q\right]=\left[\left(Q\sqrt{3},i\right):Q\sqrt{3}\right]\left[\left(Q\sqrt{3},i\right):Q\right] \\ \left[\left(Q\sqrt{3},i\right):Q\right]=2.2 \\ \left[\left(Q\sqrt{3},i\right):Q\right]=4 \end{gathered}$$

Hence the required value $\left[\left(Q\sqrt{3},i\right):Q\right]=4$.