Question

Assume that F is infinite that $\mathbf{v}\mathbf{,}\mathbf{w}\mathbf{\in }\mathbf{K}$are algebraic over F and that w is the root of a separable polynomial in $\mathbf{F}\left[x\right]$. Prove that $\mathbf{F}\left(v,w\right)$ is a simple extension over F .

Short answer

F(v,x) is a simple extension over F.

Step-by-step solution

Definition of minimal polynomial

The monic polynomial p(x) over a field F of least degree such that $\mathrm{p}\left(\mathrm{\alpha }\right)=0$for an algebraic element $\alpha$ over a field F.

Showing that F(v,w) is a simple extension over F  

Let $K=F\left(v,w\right)$and F is a infinite and $p\left(x\right)\in F\left[x\right]$ be the minimal polynomial of v and $q\left(x\right)\in F\left[x\right]$be the minimal polynomial.of w.

Now let L be the splititing field of $p\left(x\right),q\left(x\right)$over F and $w=w_1,w_2....w_n$be the root of $q\left(x\right)$in $L_i$also $w_i$are distinct.

Similarly $v=v_1,v_2,...v_n$be the root of the $p\left(x\right)$in $L_i$also $v_i$are distinct. Since F is infinite there exist $c\in F$such that

$Fc\ne \frac{v_i-v}{w-w_i}$for all$1\le i\le m,1\le j\le n\left(1\right)$

Let $u=v+cw$and $w\in F\left(u\right)$and $h\left(x\right)=p\left(u-cx\right)\in F\left(u\right)\left[x\right]$. Now w is a root of h(x). So,

$\begin{array}{rcl}\mathrm{h}\left(\mathrm{w}\right)& =& \mathrm{p}\left(\mathrm{u}-\mathrm{cw}\right)\\ & =& {\mathrm{p}}_{\mathrm{v}}\\ & =& 0_{\mathrm{F}}\end{array}$

Suppose some $w_j$is also a root of h(x). Then $p\left(u-c{w}_j\right)=0_F$so that $u-c{w}_j$is one of the root of p(x).

Let,

$\begin{array}{rcl}\mathrm{u}-{\mathrm{cw}}_{\mathrm{j}}& =& {\mathrm{v}}_{\mathrm{j}}\\ \mathrm{u}& =& {\mathrm{v}}_{\mathrm{j}}+{\mathrm{cw}}_{\mathrm{j}}\end{array}$

Therefore,

$$\begin{gathered} \mathrm{v}+\mathrm{cw}-{\mathrm{cw}}_{\mathrm{j}}={\mathrm{v}}_{\mathrm{j}} \\ \mathrm{c}=\frac{{\mathrm{v}}_{\mathrm{j}}-\mathrm{v}}{\mathrm{w}-{\mathrm{w}}_{\mathrm{j}}} \end{gathered}$$

Now this is a contradicts the equation 1. Therefore w is the only common root of q(x) and h(x). Let r(x) be the minimal polynomial of w over F(u) . Then r(x) divides q(x), so that every root of r(x) is a root of q(x). But as r(x) also divides q(x) so all its root are root of h(x) and r(x) has a single root w in L .

Therefore $r\left(x\right)\in F\left[u\right]\left(x\right)$must have degree 1 hence its root w is in F(u). Since $v=u-cw$with $u,w\in F\left(u\right)$and also $v\in F\left(u\right)$. Hence $K=F\left(v,w\right)\subseteq F\left(u\right)$ but

$$\begin{gathered} u=v+cw\in K \\ F\left(u\right)\subseteq K \\ F\left(u\right)=K \end{gathered}$$

Now let F is finite. Then the multiplicative group of nonzero element of K is cyclic. If u is generator of this group then the subfield of F(u) contain 0_F and all power of u and hence contain every element of K. So, $F\left(u\right)=K$.