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### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624

# 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 .

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

See the step by step solution

## Step 1: 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.

## Step 2: 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,

$\mathrm{v}+\mathrm{cw}-{\mathrm{cw}}_{\mathrm{j}}={\mathrm{v}}_{\mathrm{j}}\phantom{\rule{0ex}{0ex}}\mathrm{c}=\frac{{\mathrm{v}}_{\mathrm{j}}-\mathrm{v}}{\mathrm{w}-{\mathrm{w}}_{\mathrm{j}}}\phantom{\rule{0ex}{0ex}}$

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

$u=v+cw\in K\phantom{\rule{0ex}{0ex}}F\left(u\right)\subseteq K\phantom{\rule{0ex}{0ex}}F\left(u\right)=K$

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 0F and all power of u and hence contain every element of K. So, $F\left(u\right)=K$ .