Math4302 Modern Algebra (Lecture 37)

Rings

Field Extension

Theorem

If $F\subset E$ is a field extension, the set of elements of $e$ algebraic over $f$ is a subfield of $E$ .

$\alpha,\beta$ is algebraic over $F$ , then $\alpha\pm\beta,\alpha\beta,\frac{\alpha}{\beta}$ are also algebraic over $F$ .

Proposition for irreducible polynomial and field extension

If $\alpha\in E$ and $\deg(\alpha,F)=n$ , $\operatorname{irr}(\alpha,F)=p(x)$

$p(x)\in F[x]$
$p(x)$ is monic and irreducible
$(p(x))=\{h(x)|h(x)\in F[x],h(\alpha)=0\}$
$p(\alpha)=0$
$\deg(p(x),F)=n$

Then $F(\alpha)=\{c_0+c_1\alpha+\cdots+c_{n-1}\alpha^{n-1}|c_0,c_1,\cdots,c_{n-1}\in F\}$ is a vector space of dimension $n$ over $F$ .

Proof

Let $A=\{c_0+c_1\alpha+\cdots+c_{n-1}\alpha^{n-1}|c_0,c_1,\cdots,c_{n-1}\in F\}$ .

Clearly $A\subseteq F(\alpha)$ ,

if $\frac{f(\alpha)}{g(\alpha)}\in F(\alpha)$ , then $g(\alpha)\neq 0$ , so $g(x)\notin (p(x))$ , Since $p(x)$ is irreducible, so $I=(p(x))$ is maximal so $F[x]/I$ is a field.

Since $g\notin I$ , $g+I$ has a multiplicative inverse $h+I$ .

Therefore

\begin{aligned} (g+I)(h+I)=1+I&\implies gh-1\in I\\ &\implies (gh-1)(\alpha)=0\\ &\implies g(\alpha)h(\alpha)-1=0\\ &\implies \frac{1}{g(\alpha)}=h(\alpha)\\ \end{aligned}

So $\frac{f(\alpha)}{g(\alpha)}=f(\alpha)h(\alpha)=r(\alpha)$ by division algorithm $\exists r(\alpha)\in F(\alpha)$ such that $fh=pq+r$ and $\deg r\leq n-1$ .

So $F(\alpha)\subseteq A$ .

Example

Let $x^2-2\in \mathbb Q[x]$ , $\operatorname{irr}(\sqrt{2},\mathbb Q)=x^2-2$

So $\mathbb Q\subset \mathbb Q(\sqrt{2})\subset \mathbb C$

$\mathbb Q(\sqrt{2})=\{a+b\sqrt{2}|a,b\in \mathbb Q\}$

need to prove inverse is also in $\mathbb Q(\sqrt{2})$

Proposition for tower of field extensions

If $F\subset E\subset G$ , then $[K:F]=[E:F]\times [K:E]$

Proof

Suppose $\alpha,\beta\in E$ are algebraic over $F$ , set $n=\deg(\alpha,F)$ , $m=\deg(\beta,F)$ .

Then $F\subset F(\alpha)\subset F(\alpha)(\beta)\subset E$ .

F(\alpha)(\beta)=\{\alpha,\beta,\alpha\beta,\frac{\alpha}{\beta}\}

$[F(\alpha):F]=n$ , $[F(\alpha)(\beta):F(\alpha)]=\deg (\beta,F(\alpha))\leq \deg(\beta,F)=m$ .

By Proposition of irreducible polynomial and field extension, $[F(\alpha)(\beta):F]\leq n\times m$ .

Tip

If $F\subseteq K$ is a finite extension, $[K:F]=r$ then every element $\gamma\in K$ is algebraic over $F$ , $1,\gamma,\gamma^2,\cdots,\gamma^{r-1},\gamma^r$ are linearly dependent, therefore a nontrivial linear combination of them is zero, i.e. $\exists c_0,c_1,\cdots,c_{r-1}\in F$ such that $\gamma=c_0+c_1\gamma+\cdots+c_{r-1}\gamma^{r-1}=0$ .

So $[F(\alpha)(\beta):F]\leq n\times m$ , and every element $F(\alpha)(\beta)$ is algebraic over $F$ .

Galois

Setups

Definition of splitting field

Let $p(x)\in F[x]$ , $p(x)$ is irreducible over $F$ .

Then $F(\alpha)$ is a splitting field of $p(x)$ over $F$ if $p(\alpha)=0$ $F\subset K$ where K $is a splitting field of$ p(x) $over$ F $, then$ K=F(\alpha)$.

Automorphism group of splitting field

Consider $\operatorname{Aut}(K/F)$ be the set of group of isomorphism fixing $F$ , $K\to K$ .

Example

Consider $\mathbb Q(\sqrt{2})\overset{\phi}{\to}\mathbb \mathbb Q(\sqrt{2})$

$\phi(a)=a$ where $a\in \mathbb Q$ .

$\phi(\sqrt{2})=\pm \sqrt{2}$ . so $|\operatorname{Aut}(\mathbb Q(\sqrt{2}))|=2$

Galois theory

$f(x)$ is solvable in radicals $\iff$ the Galois group of $f(x)$ is a solvable group.

$S^5$ is not solvable.