Question

Let $\mathit{F}\mathbf{=}\mathit{Q}\left({\pi }^4\right)$and $\mathit{K}\mathbf{=}\mathit{Q}\left(\pi \right)$. Show that $\mathit{\pi }$is algebraic over F and find a basis of K over F.

Short answer

$\left\{1_{F,}\pi \pi ,{\pi }^2,......\right\}$ is a basis of K over F.

Step-by-step solution

Definition of basis.

A space consisting of vectors, together with the associative and commutative operations of addition of vectors and the associative operation of multiplication of vectors by scalers.

Showing that {1F,ππ,π2,......} is a basis.

Let $F=Q\left({\pi }^4\right)$and $K=Q\left(\pi \right)$. Define a polynomial $p\left(X\right)\in F\left[X\right]=Q\left({\pi }^4\right)$as,

$p\left(x\right)=x^4-{\pi }^4$

Replace x by $\pi$and if the polynomial $p\left(x\right)$ over F become zero that means, $\pi$is algebraic overF.

Therefore evaluate p(x) at $x=\pi$as follows

$$\begin{gathered} p\left(\pi \right)={\pi }^4={\pi }^4 \\ =0 \end{gathered}$$

Since the polynomial p(x) at $x=\pi$is zero.

This gives is $\pi$a root of the polynomial. This means there exist a polynomial $p\left(x\right)$ over F such that$p\left(\pi \right)=0$

Hence $\pi$ is algebraic over $F$. Thus the polynomial is minimal polynomial also. Now by the theorem given below,

Let $K$ be an extension field of $F$ and $u\in K$ is algebraic element over F with minimal polynomial of degree n . Then$\left\{1_{F,}\pi \pi ,{\pi }^2,......\right\}$ is a basic of the space F(u) over F. Thus apply the theorem mentioned above.

Therefore $\left\{1_{F,}\pi \pi ,{\pi }^2,......\right\}$ is basis of K over F.